NCERT Solutions for Class 12 Mathematics Chapter 7: Integrals – Free PDF Download

Chapter 7 of Class 12 Maths is Integrals, the reverse process of differentiation. It covers indefinite and definite integrals, integration techniques, and the Fundamental Theorem of Calculus.

Last Updated: September 23, 2026

Integration as the Inverse of Differentiation

If d/dx[F(x)]=f(x), then ∫f(x)dx=F(x)+C, where C is the constant of integration. This is the indefinite integral.

Methods of Integration

Substitution method: change variables to simplify the integrand. Integration by partial fractions: used for rational functions, splitting into simpler fractions. Integration by parts: ∫u dv=uv−∫v du, useful for products of functions (using the ILATE rule to choose u).

Standard Integrals

Key results: ∫xⁿdx=xⁿ⁺¹/(n+1)+C (n≠−1); ∫1/x dx=log|x|+C; ∫e^xdx=e^x+C; ∫sin x dx=−cos x+C; ∫cos x dx=sin x+C; ∫1/√(a²−x²) dx=sin⁻¹(x/a)+C.

Definite Integrals

The definite integral ∫_a^bf(x)dx=F(b)−F(a), representing the (signed) area under the curve between x=a and x=b, evaluated via the Fundamental Theorem of Calculus.

Properties of Definite Integrals

Key properties used to simplify evaluation: ∫_a^bf(x)dx=−∫_b^af(x)dx; ∫_a^bf(x)dx=∫_a^bf(a+b−x)dx; ∫₀^2af(x)dx=2∫₀^af(x)dx if f(2a−x)=f(x), else 0.

Exercise 7.1 Solutions (All 22 Questions)

1. Find an antiderivative (or integral) of sin 2x by the method of inspection.
Ans: Since d/dx(−cos 2x/2)=sin 2x, an antiderivative is −(cos 2x)/2.

2. Find an antiderivative (or integral) of cos 3x by the method of inspection.
Ans: Since d/dx(sin 3x/3)=cos 3x, an antiderivative is (sin 3x)/3.

3. Find an antiderivative (or integral) of e²ˣ by the method of inspection.
Ans: Since d/dx(e²ˣ/2)=e²ˣ, an antiderivative is e²ˣ/2.

4. Find an antiderivative (or integral) of (ax+b)² by the method of inspection.
Ans: Since d/dx[(ax+b)³/(3a)]=(ax+b)², an antiderivative is (ax+b)³/(3a).

5. Find an antiderivative (or integral) of sin 2x−4e³ˣ by the method of inspection.
Ans: Combining the antiderivatives of each term, an antiderivative is −(cos 2x)/2−(4/3)e³ˣ.

6. Find the integral: ∫(4e³ˣ+1)dx.
Ans: ∫(4e³ˣ+1)dx=(4/3)e³ˣ+x+C.

7. Find the integral: ∫x²(1−1/x²)dx.
Ans: Simplifying the integrand: x²(1−1/x²)=x²−1. So ∫(x²−1)dx=x³/3−x+C.

8. Find the integral: ∫(ax²+bx+c)dx.
Ans: ∫(ax²+bx+c)dx=ax³/3+bx²/2+cx+C.

9. Find the integral: ∫(2x²+eˣ)dx.
Ans: ∫(2x²+eˣ)dx=2x³/3+eˣ+C.

10. Find the integral: ∫(√x−1/√x)²dx.
Ans: Expanding: (√x−1/√x)²=x−2+1/x. So ∫(x−2+1/x)dx=x²/2−2x+log|x|+C.

11. Find the integral: ∫(x³+5x²−4)/x² dx.
Ans: Dividing term by term: (x³+5x²−4)/x²=x+5−4/x². So ∫(x+5−4/x²)dx=x²/2+5x+4/x+C.

12. Find the integral: ∫(x³+3x+4)/√x dx.
Ans: Dividing term by term: (x³+3x+4)/√x=x^5/2+3x^1/2+4x^−1/2. So the integral is (2/7)x^7/2+2x^3/2+8x^1/2+C.

13. Find the integral: ∫(x³−x²+x−1)/(x−1) dx.
Ans: Factoring the numerator by grouping: x³−x²+x−1=x²(x−1)+(x−1)=(x−1)(x²+1). So the integrand simplifies to x²+1, and ∫(x²+1)dx=x³/3+x+C.

14. Find the integral: ∫(1−x)√x dx.
Ans: Expanding: (1−x)√x=x^1/2−x^3/2. So the integral is (2/3)x^3/2−(2/5)x^5/2+C.

15. Find the integral: ∫√x(3x²+2x+3)dx.
Ans: Expanding: √x(3x²+2x+3)=3x^5/2+2x^3/2+3x^1/2. So the integral is (6/7)x^7/2+(4/5)x^5/2+2x^3/2+C.

16. Find the integral: ∫(2x−3cos x+eˣ)dx.
Ans: ∫(2x−3cos x+eˣ)dx=x²−3sin x+eˣ+C.

17. Find the integral: ∫(2x²−3sin x+5√x)dx.
Ans: ∫(2x²−3sin x+5√x)dx=2x³/3+3cos x+(10/3)x^3/2+C.

18. Find the integral: ∫sec x(sec x+tan x)dx.
Ans: Expanding: sec x(sec x+tan x)=sec²x+sec x tan x. Since d/dx(tan x)=sec²x and d/dx(sec x)=sec x tan x, the integral is tan x+sec x+C.

19. Find the integral: ∫sec²x/cosec²x dx.
Ans: sec²x/cosec²x=sec²x·sin²x=(sin²x/cos²x)=tan²x=sec²x−1. So ∫(sec²x−1)dx=tan x−x+C.

20. Find the integral: ∫(2−3sin x)/cos²x dx.
Ans: Splitting the integrand: (2−3sin x)/cos²x=2sec²x−3sec x tan x. So the integral is 2tan x−3sec x+C.

21. The antiderivative of (√x+1/√x) equals (A) (1/3)√x+2√x+C (B) (2/3)x^3/2+2x^1/2+C (C) (2/3)x^3/2+(1/2)x^1/2+C (D) (3/2)x^3/2+(1/2)x^1/2+C.
Ans: Writing √x+1/√x=x^1/2+x^−1/2 and integrating term by term gives (2/3)x^3/2+2x^1/2+C. The answer is (B).

22. If d/dx f(x)=4x³−3/x⁴, such that f(2)=0, then f(x) is (A) x⁴+1/x³−129/8 (B) x³+1/x⁴+129/8 (C) x⁴+1/x³+129/8 (D) x³+1/x⁴−129/8.
Ans: Integrating: f(x)=x⁴+1/x³+C (since ∫(−3/x⁴)dx=1/x³). Using f(2)=0: 16+1/8+C=0, so C=−129/8. Hence f(x)=x⁴+1/x³−129/8. The answer is (A).

Exercise 7.2 Solutions (All 39 Questions)

1. Integrate the function: 2x/(1+x²).
Ans: Let t=1+x², so dt=2x dx. The integral becomes ∫dt/t=log|t|+C=log(1+x²)+C.

2. Integrate the function: (log x)²/x.
Ans: Let t=log x, so dt=dx/x. The integral becomes ∫t²dt=t³/3+C=(log x)³/3+C.

3. Integrate the function: 1/(x+x log x).
Ans: Writing the denominator as x(1+log x), let t=1+log x, so dt=dx/x. The integral becomes ∫dt/t=log|t|+C=log|1+log x|+C.

4. Integrate the function: sin x·sin(cos x).
Ans: Let t=cos x, so dt=−sin x dx. The integral becomes ∫sin(t)(−dt)=−(−cos t)+C=cos t+C=cos(cos x)+C.

5. Integrate the function: sin(ax+b)cos(ax+b).
Ans: Let t=sin(ax+b), so dt=a cos(ax+b)dx. The integral becomes ∫t dt/a=t²/(2a)+C=sin²(ax+b)/(2a)+C.

6. Integrate the function: √(ax+b).
Ans: Let t=ax+b, so dt=a dx. The integral becomes ∫√t dt/a=(1/a)(2/3)t^3/2+C=(2/(3a))(ax+b)^3/2+C.

7. Integrate the function: x√(x+2).
Ans: Let t=x+2, so x=t−2 and dt=dx. The integral becomes ∫(t−2)√t dt=∫(t^3/2−2t^1/2)dt=(2/5)t^5/2−(4/3)t^3/2+C=(2/5)(x+2)^5/2−(4/3)(x+2)^3/2+C.

8. Integrate the function: x√(1+2x²).
Ans: Let t=1+2x², so dt=4x dx. The integral becomes ∫√t dt/4=(1/4)(2/3)t^3/2+C=(1/6)(1+2x²)^3/2+C.

9. Integrate the function: (4x+2)√(x²+x+1).
Ans: Let t=x²+x+1, so dt=(2x+1)dx, i.e. (4x+2)dx=2dt. The integral becomes ∫√t·2dt=2(2/3)t^3/2+C=(4/3)(x²+x+1)^3/2+C.

10. Integrate the function: 1/(x−√x).
Ans: Let √x=t, so x=t² and dx=2t dt. The integral becomes ∫2t dt/(t²−t)=∫2dt/(t−1)=2log|t−1|+C=2log|√x−1|+C.

11. Integrate the function: x/√(x+4), x>0.
Ans: Let t=x+4, so x=t−4 and dt=dx. The integral becomes ∫(t−4)/√t dt=∫(t^1/2−4t^−1/2)dt=(2/3)t^3/2−8t^1/2+C=(2/3)(x+4)^3/2−8(x+4)^1/2+C.

12. Integrate the function: (x³−1)^1/3x⁵.
Ans: Let t=x³−1, so dt=3x²dx and x³=t+1. Then x⁵dx=x³·x²dx=(t+1)dt/3. The integral becomes ∫t^1/3(t+1)dt/3=(1/3)[(3/7)t^7/3+(3/4)t^4/3]+C=(1/7)(x³−1)^7/3+(1/4)(x³−1)^4/3+C.

13. Integrate the function: x²/(2+3x³)³.
Ans: Let t=2+3x³, so dt=9x²dx. The integral becomes ∫dt/(9t³)=(1/9)·t⁻²/(−2)+C=−1/(18t²)+C=−1/(18(2+3x³)²)+C.

14. Integrate the function: 1/[x(log x)^m], x>0.
Ans: Let t=log x, so dt=dx/x. The integral becomes ∫t^−mdt=t^1−m/(1−m)+C=(log x)^1−m/(1−m)+C.

15. Integrate the function: x/(9−4x²).
Ans: Let t=9−4x², so dt=−8x dx. The integral becomes ∫−dt/(8t)=−(1/8)log|t|+C=−(1/8)log|9−4x²|+C.

16. Integrate the function: e^2x+3.
Ans: Let t=2x+3, so dt=2dx. The integral becomes ∫eᵗdt/2=(1/2)eᵗ+C=(1/2)e^2x+3+C.

17. Integrate the function: x/e^x².
Ans: Let t=−x², so dt=−2x dx. The integral becomes ∫eᵗ(−dt/2)=−(1/2)eᵗ+C=−(1/2)e^−x²+C.

18. Integrate the function: e^tan⁻¹x/(1+x²).
Ans: Let t=tan⁻¹x, so dt=dx/(1+x²). The integral becomes ∫eᵗdt=eᵗ+C=e^tan⁻¹x+C.

19. Integrate the function: (e²ˣ−1)/(e²ˣ+1).
Ans: Dividing numerator and denominator by eˣ, the integrand becomes (eˣ−e⁻ˣ)/(eˣ+e⁻ˣ). Let t=eˣ+e⁻ˣ, so dt=(eˣ−e⁻ˣ)dx. The integral becomes ∫dt/t=log|t|+C=log(eˣ+e⁻ˣ)+C, which is the same as log(e²ˣ+1)−x+C.

20. Integrate the function: (e²ˣ−e⁻²ˣ)/(e²ˣ+e⁻²ˣ).
Ans: Let t=e²ˣ+e⁻²ˣ, so dt=2(e²ˣ−e⁻²ˣ)dx. The integral becomes ∫dt/(2t)=(1/2)log|t|+C=(1/2)log(e²ˣ+e⁻²ˣ)+C.

21. Integrate the function: tan²(2x−3).
Ans: Using tan²θ=sec²θ−1 with θ=2x−3: the integral becomes ∫[sec²(2x−3)−1]dx=(1/2)tan(2x−3)−x+C.

22. Integrate the function: sec²(7−4x).
Ans: Let t=7−4x, so dt=−4dx. The integral becomes ∫sec²t(−dt/4)=−(1/4)tan t+C=−(1/4)tan(7−4x)+C.

23. Integrate the function: sin⁻¹x/√(1−x²).
Ans: Let t=sin⁻¹x, so dt=dx/√(1−x²). The integral becomes ∫t dt=t²/2+C=(sin⁻¹x)²/2+C.

24. Integrate the function: (2cos x−3sin x)/(6cos x+4sin x).
Ans: Let t=6cos x+4sin x, so dt=(−6sin x+4cos x)dx=2(2cos x−3sin x)dx. The integral becomes ∫(dt/2)/t=(1/2)log|t|+C=(1/2)log|6cos x+4sin x|+C.

25. Integrate the function: 1/[cos²x(1−tan x)²].
Ans: Let t=1−tan x, so dt=−sec²x dx=−dx/cos²x. The integral becomes ∫−dt/t²=1/t+C=1/(1−tan x)+C.

26. Integrate the function: cos√x/√x.
Ans: Let t=√x, so dt=dx/(2√x), i.e. dx/√x=2dt. The integral becomes ∫cos t·2dt=2sin t+C=2sin√x+C.

27. Integrate the function: √(sin 2x)cos 2x.
Ans: Let t=sin 2x, so dt=2cos 2x dx. The integral becomes ∫√t dt/2=(1/2)(2/3)t^3/2+C=(1/3)(sin 2x)^3/2+C.

28. Integrate the function: cos x/√(1+sin x).
Ans: Let t=1+sin x, so dt=cos x dx. The integral becomes ∫dt/√t=2√t+C=2√(1+sin x)+C.

29. Integrate the function: cot x·log(sin x).
Ans: Let t=log(sin x), so dt=cot x dx. The integral becomes ∫t dt=t²/2+C=(log sin x)²/2+C.

30. Integrate the function: sin x/(1+cos x).
Ans: Let t=1+cos x, so dt=−sin x dx. The integral becomes ∫−dt/t=−log|t|+C=−log|1+cos x|+C.

31. Integrate the function: sin x/(1+cos x)².
Ans: Let t=1+cos x, so dt=−sin x dx. The integral becomes ∫−dt/t²=1/t+C=1/(1+cos x)+C.

32. Integrate the function: 1/(1+cot x).
Ans: Writing 1/(1+cot x)=sin x/(sin x+cos x), and splitting sin x=(1/2)(sin x+cos x)+(1/2)(sin x−cos x), the integral splits as (1/2)∫dx+(1/2)∫(sin x−cos x)/(sin x+cos x)dx. Let t=sin x+cos x, so dt=(cos x−sin x)dx=−(sin x−cos x)dx. The second part becomes (1/2)∫−dt/t=−(1/2)log|t|+C. So the integral is x/2−(1/2)log|sin x+cos x|+C.

33. Integrate the function: 1/(1−tan x).
Ans: Writing 1/(1−tan x)=cos x/(cos x−sin x), and splitting cos x=(1/2)(cos x−sin x)+(1/2)(sin x+cos x), the integral splits as (1/2)∫dx+(1/2)∫(sin x+cos x)/(cos x−sin x)dx. Let t=cos x−sin x, so dt=(−sin x−cos x)dx=−(sin x+cos x)dx. The second part becomes (1/2)∫−dt/t=−(1/2)log|t|+C. So the integral is x/2−(1/2)log|cos x−sin x|+C.

34. Integrate the function: √(tan x)/(sin x cos x).
Ans: Dividing numerator and denominator by cos²x, the integrand becomes √(tan x)·sec²x/tan x=sec²x/√(tan x). Let t=tan x, so dt=sec²x dx. The integral becomes ∫dt/√t=2√t+C=2√(tan x)+C.

35. Integrate the function: (1+log x)²/x.
Ans: Let t=1+log x, so dt=dx/x. The integral becomes ∫t²dt=t³/3+C=(1+log x)³/3+C.

36. Integrate the function: (x+1)(x+log x)²/x.
Ans: Writing (x+1)/x=1+1/x, let t=x+log x, so dt=(1+1/x)dx. The integral becomes ∫t²dt=t³/3+C=(x+log x)³/3+C.

37. Integrate the function: x³sin(tan⁻¹x⁴)/(1+x⁸).
Ans: Let t=tan⁻¹(x⁴), so dt=4x³/(1+x⁸)dx, i.e. x³dx/(1+x⁸)=dt/4. The integral becomes ∫sin t·dt/4=−(1/4)cos t+C=−(1/4)cos(tan⁻¹x⁴)+C.

38. The antiderivative of (10x⁹+10ˣlog_e10)/(x¹⁰+10ˣ) equals (A) 10ˣ−x¹⁰+C (B) 10ˣ+x¹⁰+C (C) (10ˣ−x¹⁰)⁻¹+C (D) log(10ˣ+x¹⁰)+C.
Ans: Let t=x¹⁰+10ˣ, so dt=(10x⁹+10ˣlog_e10)dx (using d/dx(10ˣ)=10ˣlog_e10). The integral becomes ∫dt/t=log|t|+C=log(x¹⁰+10ˣ)+C. The answer is (D).

39. If ∫dx/(sin²x cos²x)=tan x+λcot x+C, then λ equals (A) 1 (B) −1 (C) 2 (D) −2.
Ans: Since 1/(sin²x cos²x)=sec²x+cosec²x, the integral is ∫(sec²x+cosec²x)dx=tan x−cot x+C. Comparing with tan x+λcot x+C gives λ=−1. The answer is (B).

Exercise 7.3 Solutions (All 24 Questions)

1. Find the integral of sin²(2x+5).
Ans: Using sin²θ=(1−cos2θ)/2 with θ=2x+5: sin²(2x+5)=(1−cos(4x+10))/2. The integral is x/2−sin(4x+10)/8+C.

2. Find the integral of sin 3x cos 4x.
Ans: Using sinA cosB=(1/2)[sin(A+B)+sin(A−B)]: sin3x cos4x=(1/2)[sin7x−sinx]. The integral is (1/2)[−cos7x/7+cosx]+C=−cos7x/14+cosx/2+C.

3. Find the integral of cos 2x cos 4x cos 6x.
Ans: Using the product-to-sum identity twice: cos2x cos4x=(1/2)(cos6x+cos2x), and multiplying by cos6x and simplifying using cos²6x=(1+cos12x)/2 and cos2x cos6x=(1/2)(cos8x+cos4x), the integrand becomes (1/4)[1+cos12x+cos8x+cos4x]. The integral is x/4+sin12x/48+sin8x/32+sin4x/16+C.

4. Find the integral of sin³(2x+1).
Ans: Using sin³θ=(3sinθ−sin3θ)/4 with θ=2x+1: the integral becomes (1/4)∫[3sin(2x+1)−sin(6x+3)]dx=(1/4)[−3cos(2x+1)/2+cos(6x+3)/6]+C=−3cos(2x+1)/8+cos(6x+3)/24+C.

5. Find the integral of sin³x cos³x.
Ans: Writing sinx cosx=sin2x/2, sin³x cos³x=(sin2x/2)³=sin³2x/8. Using sin³θ=(3sinθ−sin3θ)/4 with θ=2x: sin³2x=(3sin2x−sin6x)/4, so the integrand is (3sin2x−sin6x)/32. The integral is (1/32)[−3cos2x/2+cos6x/6]+C=−3cos2x/64+cos6x/192+C.

6. Find the integral of sin x sin 2x sin 3x.
Ans: Using sinx sin3x=(1/2)(cos2x−cos4x), then multiplying by sin2x and applying the product-to-sum identity again, the integrand simplifies to (1/4)[sin4x+sin2x−sin6x]. The integral is −cos4x/16−cos2x/8+cos6x/24+C.

7. Find the integral of sin 4x sin 8x.
Ans: Using sinA sinB=(1/2)[cos(A−B)−cos(A+B)]: sin4x sin8x=(1/2)[cos4x−cos12x]. The integral is (1/2)[sin4x/4−sin12x/12]+C=sin4x/8−sin12x/24+C.

8. Find the integral of (1−cos x)/(1+cos x).
Ans: Using the half-angle identity (1−cos x)/(1+cos x)=tan²(x/2), the integral becomes ∫[sec²(x/2)−1]dx=2tan(x/2)−x+C.

9. Find the integral of cos x/(1+cos x).
Ans: Writing cos x/(1+cos x)=1−1/(1+cos x), and using 1/(1+cos x)=(1/2)sec²(x/2), the integral becomes ∫[1−(1/2)sec²(x/2)]dx=x−tan(x/2)+C.

10. Find the integral of sin⁴x.
Ans: Using sin⁴x=(3−4cos2x+cos4x)/8, the integral is 3x/8−sin2x/4+sin4x/32+C.

11. Find the integral of cos⁴2x.
Ans: Using cos⁴θ=(3+4cos2θ+cos4θ)/8 with θ=2x: cos⁴2x=(3+4cos4x+cos8x)/8. The integral is 3x/8+sin4x/8+sin8x/64+C.

12. Find the integral of sin²x/(1+cos x).
Ans: Since sin²x=1−cos²x=(1−cos x)(1+cos x), the integrand simplifies to 1−cos x. The integral is x−sin x+C.

13. Find the integral of (cos 2x−cos 2α)/(cos x−cos α).
Ans: Using cos2θ=2cos²θ−1: cos2x−cos2α=2(cos²x−cos²α)=2(cos x−cos α)(cos x+cos α), so the integrand simplifies to 2(cos x+cos α). The integral is 2sin x+2x cos α+C.

14. Find the integral of (cos x−sin x)/(1+sin 2x).
Ans: Since 1+sin2x=(sin x+cos x)², let t=sin x+cos x, so dt=(cos x−sin x)dx. The integral becomes ∫dt/t²=−1/t+C=−1/(sin x+cos x)+C.

15. Find the integral of tan³2x sec 2x.
Ans: Let t=sec 2x, so dt=2sec 2x tan 2x dx. Writing tan³2x sec 2x=(sec²2x−1)(sec 2x tan 2x), the integral becomes ∫(t²−1)dt/2=(1/2)(t³/3−t)+C=sec³2x/6−sec 2x/2+C.

16. Find the integral of tan⁴x.
Ans: Writing tan⁴x=tan²x(sec²x−1)=tan²x sec²x−(sec²x−1), the integral becomes ∫tan²x sec²x dx−∫sec²x dx+∫dx=tan³x/3−tan x+x+C.

17. Find the integral of (sin³x+cos³x)/(sin²x cos²x).
Ans: Splitting the fraction: sin³x/(sin²x cos²x)=sin x/cos²x and cos³x/(sin²x cos²x)=cos x/sin²x. Since d/dx(sec x)=sin x/cos²x and d/dx(−cosec x)=cos x/sin²x, the integral is sec x−cosec x+C.

18. Find the integral of (cos 2x+2sin²x)/cos²x.
Ans: Since cos2x=1−2sin²x, the numerator cos2x+2sin²x=1, so the integrand simplifies to 1/cos²x=sec²x. The integral is tan x+C.

19. Find the integral of 1/(sin x cos³x).
Ans: Writing 1=sin²x+cos²x in the numerator: 1/(sin x cos³x)=sin x/cos³x+1/(sin x cos x)=tan x sec²x+cosec x sec x. Since ∫tan x sec²x dx=tan²x/2 (letting u=tan x) and ∫cosec x sec x dx=log|tan x| (as d/dx log|tan x|=1/(sin x cos x)), the integral is tan²x/2+log|tan x|+C.

20. Find the integral of cos 2x/(cos x+sin x)².
Ans: Since cos2x=cos²x−sin²x=(cos x−sin x)(cos x+sin x), the integrand simplifies to (cos x−sin x)/(cos x+sin x). Let t=cos x+sin x, so dt=(cos x−sin x)dx. The integral becomes ∫dt/t=log|t|+C=log|sin x+cos x|+C.

21. Find the integral of sin⁻¹(cos x).
Ans: Since cos x=sin(π/2−x), sin⁻¹(cos x)=π/2−x on the relevant domain. The integral is ∫(π/2−x)dx=πx/2−x²/2+C.

22. Find the integral of 1/[cos(x−a)cos(x−b)].
Ans: Using sin[(x−a)−(x−b)]=sin(b−a)=sin(x−a)cos(x−b)−cos(x−a)sin(x−b), dividing both sides by cos(x−a)cos(x−b) gives 1/[cos(x−a)cos(x−b)]=[tan(x−a)−tan(x−b)]/sin(b−a). The integral is (1/sin(b−a))[−log|cos(x−a)|+log|cos(x−b)|]+C=(1/sin(b−a))log|cos(x−b)/cos(x−a)|+C.

23. (sin²x−cos²x)/(sin²x cos²x) is equal to (A) tan x+cot x+C (B) tan x+cosec x+C (C) −tan x+cot x+C (D) tan x+sec x+C.
Ans: Writing (sin²x−cos²x)/(sin²x cos²x)=1/cos²x−1/sin²x=sec²x−cosec²x, whose antiderivative is tan x−(−cot x)=tan x+cot x. The answer is (A).

24. eˣ(1+x)/cos²(eˣx) is equal to (A) −cot(xeˣ)+C (B) tan(xeˣ)+C (C) tan(eˣ)+C (D) cot(eˣ)+C.
Ans: Let t=xeˣ, so dt=(eˣ+xeˣ)dx=eˣ(1+x)dx. The integral becomes ∫sec²t dt=tan t+C=tan(xeˣ)+C. The answer is (B).

Exercise 7.4 Solutions (All 25 Questions)

1. Integrate the function: 3x²/(x⁶+1).
Ans: Let t=x³, so dt=3x²dx. The integral becomes ∫dt/(t²+1)=tan⁻¹t+C=tan⁻¹(x³)+C.

2. Integrate the function: 1/√(1+4x²).
Ans: Using the standard form ∫dx/√(x²+a²)=log|x+√(x²+a²)|+C, the integral is (1/2)log|2x+√(1+4x²)|+C.

3. Integrate the function: 1/√[(2−x)²+1].
Ans: Let u=2−x, so du=−dx. The integral becomes −∫du/√(u²+1)=−log|u+√(u²+1)|+C=−log|(2−x)+√[(2−x)²+1]|+C.

4. Integrate the function: 1/√(9−25x²).
Ans: Writing 9−25x²=25(9/25−x²), and using ∫dx/√(a²−x²)=sin⁻¹(x/a)+C with a=3/5, the integral is (1/5)sin⁻¹(5x/3)+C.

5. Integrate the function: 3x/(1+2x⁴).
Ans: Let t=x², so dt=2x dx, i.e. 3x dx=(3/2)dt. The integral becomes (3/2)∫dt/(1+2t²)=(3/(2√2))tan⁻¹(√2 t)+C=(3√2/4)tan⁻¹(√2 x²)+C.

6. Integrate the function: x²/(1−x⁶).
Ans: Let t=x³, so dt=3x²dx. The integral becomes (1/3)∫dt/(1−t²)=(1/6)log|(1+t)/(1−t)|+C=(1/6)log|(1+x³)/(1−x³)|+C.

7. Integrate the function: (x−1)/√(x²−1).
Ans: Splitting the numerator: (x−1)/√(x²−1)=x/√(x²−1)−1/√(x²−1). Since ∫x/√(x²−1)dx=√(x²−1) and ∫1/√(x²−1)dx=log|x+√(x²−1)|, the integral is √(x²−1)−log|x+√(x²−1)|+C.

8. Integrate the function: x²/√(x⁶+a⁶).
Ans: Let t=x³, so dt=3x²dx. The integral becomes (1/3)∫dt/√(t²+a⁶)=(1/3)log|t+√(t²+a⁶)|+C=(1/3)log|x³+√(x⁶+a⁶)|+C.

9. Integrate the function: sec²x/√(tan²x+4).
Ans: Let t=tan x, so dt=sec²x dx. The integral becomes ∫dt/√(t²+4)=log|t+√(t²+4)|+C=log|tan x+√(tan²x+4)|+C.

10. Integrate the function: 1/√(x²+2x+2).
Ans: Completing the square, x²+2x+2=(x+1)²+1, so the integral is log|(x+1)+√(x²+2x+2)|+C.

11. Integrate the function: 1/(9x²+6x+5).
Ans: Completing the square, 9x²+6x+5=9(x+1/3)²+4. The integral is (1/6)tan⁻¹((3x+1)/2)+C.

12. Integrate the function: 1/√(7−6x−x²).
Ans: Completing the square, 7−6x−x²=16−(x+3)². The integral is sin⁻¹((x+3)/4)+C.

13. Integrate the function: 1/√[(x−1)(x−2)].
Ans: Writing (x−1)(x−2)=x²−3x+2=(x−3/2)²−1/4, the integral is log|(x−3/2)+√(x²−3x+2)|+C.

14. Integrate the function: 1/√(8+3x−x²).
Ans: Completing the square, 8+3x−x²=41/4−(x−3/2)². The integral is sin⁻¹((2x−3)/√41)+C.

15. Integrate the function: 1/√[(x−a)(x−b)].
Ans: Writing (x−a)(x−b)=[x−(a+b)/2]²−[(a−b)/2]², the integral is log|x−(a+b)/2+√[(x−a)(x−b)]|+C.

16. Integrate the function: (4x+1)/√(2x²+x−3).
Ans: Since d/dx(2x²+x−3)=4x+1 exactly matches the numerator, the integral is 2√(2x²+x−3)+C.

17. Integrate the function: (x+2)/√(x²−1).
Ans: Splitting the numerator: (x+2)/√(x²−1)=x/√(x²−1)+2/√(x²−1). Since ∫x/√(x²−1)dx=√(x²−1) and ∫1/√(x²−1)dx=log|x+√(x²−1)|, the integral is √(x²−1)+2log|x+√(x²−1)|+C.

18. Integrate the function: (5x−2)/(1+2x+3x²).
Ans: Writing 5x−2=(5/6)(6x+2)−11/3, where 6x+2 is the derivative of 3x²+2x+1, the integral splits as (5/6)∫(6x+2)/(3x²+2x+1)dx−(11/3)∫dx/(3x²+2x+1). The first part is (5/6)log|3x²+2x+1|. Completing the square in the second, 3x²+2x+1=3(x+1/3)²+2/3, giving (1/√2)tan⁻¹((3x+1)/√2). The integral is (5/6)log|3x²+2x+1|−(11√2/6)tan⁻¹((3x+1)/√2)+C.

19. Integrate the function: (6x+7)/√[(x−5)(x−4)].
Ans: Writing 6x+7=3(2x−9)+34, where 2x−9 is the derivative of x²−9x+20, the integral splits as 3∫(2x−9)/√(x²−9x+20)dx+34∫dx/√(x²−9x+20). The first part is 6√(x²−9x+20). Completing the square in the second, x²−9x+20=(x−9/2)²−1/4, giving log|(x−9/2)+√(x²−9x+20)|. The integral is 6√(x²−9x+20)+34log|x−9/2+√(x²−9x+20)|+C.

20. Integrate the function: (x+2)/√(4x−x²).
Ans: Let u=x−2, so 4x−x²=4−u² and the numerator x+2=u+4. The integral becomes ∫u/√(4−u²)du+4∫du/√(4−u²)=−√(4−u²)+4sin⁻¹(u/2)+C=−√(4x−x²)+4sin⁻¹((x−2)/2)+C.

21. Integrate the function: (x+2)/√(x²+2x+3).
Ans: Writing x+2=(1/2)(2x+2)+1, where 2x+2 is the derivative of x²+2x+3, the integral splits as (1/2)∫(2x+2)/√(x²+2x+3)dx+∫dx/√(x²+2x+3)=√(x²+2x+3)+log|(x+1)+√(x²+2x+3)|+C.

22. Integrate the function: (x+3)/(x²−2x−5).
Ans: Writing x+3=(1/2)(2x−2)+4, where 2x−2 is the derivative of x²−2x−5, the integral splits as (1/2)∫(2x−2)/(x²−2x−5)dx+4∫dx/(x²−2x−5). The first part is (1/2)log|x²−2x−5|. Completing the square in the second, x²−2x−5=(x−1)²−6, giving (1/(2√6))log|(x−1−√6)/(x−1+√6)|. The integral is (1/2)log|x²−2x−5|+(√6/3)log|(x−1−√6)/(x−1+√6)|+C.

23. Integrate the function: (5x+3)/√(x²+4x+10).
Ans: Writing 5x+3=(5/2)(2x+4)−7, where 2x+4 is the derivative of x²+4x+10, the integral splits as (5/2)∫(2x+4)/√(x²+4x+10)dx−7∫dx/√(x²+4x+10)=5√(x²+4x+10)−7log|(x+2)+√(x²+4x+10)|+C.

24. 1/(x²+2x+2) equals (A) x tan⁻¹(x+1)+C (B) tan⁻¹(x+1)+C (C) (x+1)tan⁻¹x+C (D) tan⁻¹x+C.
Ans: Since x²+2x+2=(x+1)²+1, the integral is tan⁻¹(x+1)+C. The answer is (B).

25. 1/√(9x−4x²) equals (A) (1/9)sin⁻¹((9x−8)/8)+C (B) (1/2)sin⁻¹((8x−9)/9)+C (C) (1/3)sin⁻¹((9x−8)/8)+C (D) (1/2)sin⁻¹((9x−8)/9)+C.
Ans: Writing 9x−4x²=81/16−4(x−9/8)², the integral becomes (1/2)∫du/√((9/8)²−u²) with u=x−9/8, giving (1/2)sin⁻¹(u/(9/8))+C=(1/2)sin⁻¹((8x−9)/9)+C. The answer is (B).

Exercise 7.5 Solutions (All 23 Questions)

1. Integrate the function: x/[(x+1)(x+2)].
Ans: By partial fractions, x/[(x+1)(x+2)]=−1/(x+1)+2/(x+2). The integral is −log|x+1|+2log|x+2|+C.

2. Integrate the function: 1/(x²−9).
Ans: By partial fractions, 1/(x²−9)=(1/6)[1/(x−3)−1/(x+3)]. The integral is (1/6)log|(x−3)/(x+3)|+C.

3. Integrate the function: (3x−1)/[(x−1)(x−2)(x−3)].
Ans: By partial fractions, the integrand equals 1/(x−1)−5/(x−2)+4/(x−3). The integral is log|x−1|−5log|x−2|+4log|x−3|+C.

4. Integrate the function: x/[(x−1)(x−2)(x−3)].
Ans: By partial fractions, the integrand equals (1/2)/(x−1)−2/(x−2)+(3/2)/(x−3). The integral is (1/2)log|x−1|−2log|x−2|+(3/2)log|x−3|+C.

5. Integrate the function: 2x/(x²+3x+2).
Ans: Since x²+3x+2=(x+1)(x+2), partial fractions give the integrand as −2/(x+1)+4/(x+2). The integral is −2log|x+1|+4log|x+2|+C.

6. Integrate the function: (1−x²)/[x(1−2x)].
Ans: Since the degrees of numerator and denominator are equal, dividing first and then applying partial fractions gives the integrand as 1/2+1/x−(3/2)/(2x−1). The integral is x/2+log|x|−(3/4)log|2x−1|+C.

7. Integrate the function: x/[(x²+1)(x−1)].
Ans: By partial fractions, the integrand equals (1/2)(1−x)/(x²+1)+(1/2)/(x−1). Using ∫1/(x²+1)dx=tan⁻¹x and ∫x/(x²+1)dx=(1/2)log(x²+1), the integral is (1/2)tan⁻¹x−(1/4)log(x²+1)+(1/2)log|x−1|+C.

8. Integrate the function: x/[(x−1)²(x+2)].
Ans: By partial fractions, the integrand equals (2/9)/(x−1)+(1/3)/(x−1)²−(2/9)/(x+2). The integral is (2/9)log|x−1|−1/[3(x−1)]−(2/9)log|x+2|+C.

9. Integrate the function: (3x+5)/(x³−x²−x+1).
Ans: Since x³−x²−x+1=(x−1)²(x+1), partial fractions give the integrand as −(1/2)/(x−1)+4/(x−1)²+(1/2)/(x+1). The integral is −(1/2)log|x−1|−4/(x−1)+(1/2)log|x+1|+C.

10. Integrate the function: (2x−3)/[(x²−1)(2x+3)].
Ans: By partial fractions, the integrand equals −(1/10)/(x−1)+(5/2)/(x+1)−(24/5)/(2x+3). The integral is −(1/10)log|x−1|+(5/2)log|x+1|−(12/5)log|2x+3|+C.

11. Integrate the function: 5x/[(x−1)(x²−4)].
Ans: By partial fractions, the integrand equals −(5/3)/(x−1)+(5/2)/(x−2)−(5/6)/(x+2). The integral is −(5/3)log|x−1|+(5/2)log|x−2|−(5/6)log|x+2|+C.

12. Integrate the function: (x³+x+1)/(x²−1).
Ans: Dividing first, the integrand equals x+(2x+1)/(x²−1), and partial fractions give (2x+1)/(x²−1)=(3/2)/(x−1)+(1/2)/(x+1). The integral is x²/2+(3/2)log|x−1|+(1/2)log|x+1|+C.

13. Integrate the function: 2/[(1−x)(1+x²)].
Ans: By partial fractions, the integrand equals 1/(1−x)+(x+1)/(1+x²). The integral is −log|1−x|+(1/2)log(1+x²)+tan⁻¹x+C.

14. Integrate the function: (3x−1)/(x+2)².
Ans: Let u=x+2. Since 3x−1=3u−7, the integrand equals 3/u−7/u². The integral is 3log|x+2|+7/(x+2)+C.

15. Integrate the function: 1/(x⁴−1).
Ans: Since x⁴−1=(x−1)(x+1)(x²+1), partial fractions give the integrand as (1/4)/(x−1)−(1/4)/(x+1)−(1/2)/(x²+1). The integral is (1/4)log|(x−1)/(x+1)|−(1/2)tan⁻¹x+C.

16. Integrate the function: 1/[x(xⁿ+1)].
Ans: Multiplying numerator and denominator by xⁿ⁻¹, let t=xⁿ, so dt=nxⁿ⁻¹dx. The integral becomes (1/n)∫dt/[t(t+1)]=(1/n)log|t/(t+1)|+C=(1/n)log|xⁿ/(xⁿ+1)|+C.

17. Integrate the function: cos x/[(1−sin x)(2−sin x)].
Ans: Let t=sin x, so dt=cos x dx. By partial fractions, 1/[(1−t)(2−t)]=1/(1−t)−1/(2−t), so the integral becomes ∫[1/(1−t)−1/(2−t)]dt=−log|1−t|+log|2−t|+C=log|(2−sin x)/(1−sin x)|+C.

18. Integrate the function: (x²+1)(x²+2)/[(x²+3)(x²+4)].
Ans: Writing y=x² and dividing (since numerator and denominator have equal degree in y), the integrand simplifies to 1+2/(x²+3)−6/(x²+4). The integral is x+(2√3/3)tan⁻¹(x/√3)−3tan⁻¹(x/2)+C.

19. Integrate the function: 2x/[(x²+1)(x²+3)].
Ans: Let t=x², so dt=2x dx. By partial fractions, 1/[(t+1)(t+3)]=(1/2)/(t+1)−(1/2)/(t+3), so the integral is (1/2)log[(x²+1)/(x²+3)]+C.

20. Integrate the function: 1/[x(x⁴−1)].
Ans: By partial fractions (using x⁴−1=(x−1)(x+1)(x²+1)), the integrand equals −1/x+(1/4)/(x−1)+(1/4)/(x+1)+(1/2)x/(x²+1). The integral is −log|x|+(1/4)log|x⁴−1|+C.

21. Integrate the function: 1/(eˣ−1).
Ans: Let t=eˣ, so dt=eˣdx, i.e. dx=dt/t. The integral becomes ∫dt/[t(t−1)]=log|t−1|−log|t|+C=log|eˣ−1|−x+C.

22. ∫x dx/[(x−1)(x−2)] equals (A) log|(x−1)²/(x−2)|+C (B) log|(x−2)²/(x−1)|+C (C) log|(x−1)²(x−2)|+C (D) log|(x−1)(x−2)|+C.
Ans: By partial fractions, x/[(x−1)(x−2)]=−1/(x−1)+2/(x−2). The integral is −log|x−1|+2log|x−2|+C=log|(x−2)²/(x−1)|+C. The answer is (B).

23. ∫dx/[x(x²+1)] equals (A) log|x|−(1/2)log(x²+1)+C (B) log|x|+(1/2)log(x²+1)+C (C) −log|x|+(1/2)log(x²+1)+C (D) (1/2)log|x|+log(x²+1)+C.
Ans: By partial fractions, 1/[x(x²+1)]=1/x−x/(x²+1). The integral is log|x|−(1/2)log(x²+1)+C. The answer is (A).

Exercise 7.6 Solutions (All 24 Questions)

Integrate the functions in Exercises 1 to 22.

1. x sin x
Ans: sin x – x cos x + C

2. x sin 3x
Ans: (sin 3x)/9 – (x cos 3x)/3 + C

3. x²eˣ
Ans: eˣ(x² – 2x + 2) + C

4. x log x
Ans: (x²/2) log x – x²/4 + C

5. x log 2x
Ans: (x²/2) log 2x – x²/4 + C

6. x² log x
Ans: (x³/3) log x – x³/9 + C

7. x sin⁻¹x
Ans: ((2x² – 1)/4) sin⁻¹x + (x√(1-x²))/4 + C

8. x tan⁻¹x
Ans: ((x² + 1)/2) tan⁻¹x – x/2 + C

9. x cos⁻¹x
Ans: ((2x² – 1)/4) cos⁻¹x – (x√(1-x²))/4 + C

10. (sin⁻¹x)²
Ans: x(sin⁻¹x)² – 2x + 2√(1-x²) sin⁻¹x + C

11. (x cos⁻¹x)/√(1-x²)
Ans: -√(1-x²) cos⁻¹x – x + C

12. x sec²x
Ans: x tan x + log|cos x| + C

13. tan⁻¹x
Ans: x tan⁻¹x – (1/2) log(1 + x²) + C

14. x(log x)²
Ans: (x²/2)(log x)² – (x²/2) log x + x²/4 + C

15. (x² + 1) log x
Ans: (x³/3) log x – x³/9 + x log x – x + C

16. eˣ(sin x + cos x)
Ans: eˣ sin x + C

17. (xeˣ)/(1 + x)²
Ans: eˣ/(1 + x) + C

18. eˣ((1 + sin x)/(1 + cos x))
Ans: eˣ tan(x/2) + C

19. eˣ(1/x – 1/x²)
Ans: eˣ/x + C

20. ((x – 3)eˣ)/(x – 1)³
Ans: eˣ/(x – 1)² + C

21. e^2x sin x
Ans: (e^2x(2 sin x – cos x))/5 + C

22. sin⁻¹(2x/(1 + x²))
Ans: 2x tan⁻¹x – log(1 + x²) + C

23. Choose the correct answer: ∫x²e^x³dx equals
(A) (1/3)e^x³ + C
(B) (1/3)e^x² + C
(C) (1/2)e^x³ + C
(D) (1/2)e^x² + C
Ans: (A) (1/3)e^x³ + C

24. Choose the correct answer: ∫eˣsec x(1 + tan x)dx equals
(A) eˣcos x + C
(B) eˣsec x + C
(C) eˣsin x + C
(D) eˣtan x + C
Ans: (B) eˣsec x + C

Exercise 7.7 Solutions (All 11 Questions)

Integrate the functions in Exercises 1 to 9.

1. √(4 – x²)
Ans: (x/2)√(4 – x²) + 2 sin⁻¹(x/2) + C

2. √(1 – 4x²)
Ans: (x/2)√(1 – 4x²) + (1/4) sin⁻¹(2x) + C

3. √(x² + 4x + 6)
Ans: ((x + 2)/2)√(x² + 4x + 6) + log|(x + 2) + √(x² + 4x + 6)| + C

4. √(x² + 4x + 1)
Ans: ((x + 2)/2)√(x² + 4x + 1) – (3/2) log|(x + 2) + √(x² + 4x + 1)| + C

5. √(1 – 4x – x²)
Ans: ((x + 2)/2)√(1 – 4x – x²) + (5/2) sin⁻¹((x + 2)/√5) + C

6. √(x² + 4x – 5)
Ans: ((x + 2)/2)√(x² + 4x – 5) – (9/2) log|(x + 2) + √(x² + 4x – 5)| + C

7. √(1 + 3x – x²)
Ans: ((2x – 3)/4)√(1 + 3x – x²) + (13/8) sin⁻¹((2x – 3)/√13) + C

8. √(x² + 3x)
Ans: ((2x + 3)/4)√(x² + 3x) – (9/8) log|(x + 3/2) + √(x² + 3x)| + C

9. √(1 + x²/9)
Ans: (x/6)√(9 + x²) + (3/2) log|x + √(9 + x²)| + C

10. Choose the correct answer: ∫√(1 + x²) dx is equal to
(A) (x/2)√(1 + x²) + (1/2) log|x + √(1 + x²)| + C
(B) (2/3)(1 + x²)^3/2 + C
(C) (2/3)x(1 + x²)^3/2 + C
(D) (x³/2)√(1 + x²) + (x²/2) log|x + √(1 + x²)| + C
Ans: (A) (x/2)√(1 + x²) + (1/2) log|x + √(1 + x²)| + C

11. Choose the correct answer: ∫√(x² – 8x + 7) dx is equal to
(A) (1/2)(x – 4)√(x² – 8x + 7) + 9 log|x – 4 + √(x² – 8x + 7)| + C
(B) (1/2)(x + 4)√(x² – 8x + 7) + 9 log|x + 4 + √(x² – 8x + 7)| + C
(C) (1/2)(x – 4)√(x² – 8x + 7) – 3√2 log|x – 4 + √(x² – 8x + 7)| + C
(D) (1/2)(x – 4)√(x² – 8x + 7) – (9/2) log|x – 4 + √(x² – 8x + 7)| + C
Ans: (D) (1/2)(x – 4)√(x² – 8x + 7) – (9/2) log|x – 4 + √(x² – 8x + 7)| + C

Exercise 7.8 Solutions (All 22 Questions)

Evaluate the following definite integrals in Exercises 1 to 20.

1. ∫_-1¹ (x + 1) dx
Ans: [x²/2 + x]_-1¹ = (1/2 + 1) – (1/2 – 1) = 2

2. ∫₂³ (1/x) dx
Ans: [log x]₂³ = log 3 – log 2 = log(3/2)

3. ∫₁² (4x³ – 5x² + 6x + 9) dx
Ans: [x⁴ – (5/3)x³ + 3x² + 9x]₁² = 64/3

4. ∫₀^π/4 sin 2x dx
Ans: [-(cos 2x)/2]₀^π/4 = 1/2

5. ∫₀^π/2 cos 2x dx
Ans: [(sin 2x)/2]₀^π/2 = 0

6. ∫₄⁵ eˣ dx
Ans: [eˣ]₄⁵ = e⁵ – e⁴

7. ∫₀^π/4 tan x dx
Ans: [-log|cos x|]₀^π/4 = (1/2) log 2

8. ∫_π/6^π/4 cosec x dx
Ans: [log|tan(x/2)|]_π/6^π/4 = log(tan(π/8)) – log(tan(π/12))

9. ∫₀¹ dx/√(1 – x²)
Ans: [sin⁻¹x]₀¹ = π/2

10. ∫₀¹ dx/(1 + x²)
Ans: [tan⁻¹x]₀¹ = π/4

11. ∫₂³ dx/(x² – 1)
Ans: [(1/2) log|(x – 1)/(x + 1)|]₂³ = (1/2) log(3/2)

12. ∫₀^π/2 cos²x dx
Ans: π/4

13. ∫₂³ x dx/(x² + 1)
Ans: [(1/2) log(x² + 1)]₂³ = (1/2) log 2

14. ∫₀¹ (2x + 3)/(5x² + 1) dx
Ans: [(1/5) log(5x² + 1) + (3/√5) tan⁻¹(√5 x)]₀¹ = (1/5) log 6 + (3/√5) tan⁻¹√5

15. ∫₀¹ xe^x² dx
Ans: [(1/2)e^x²]₀¹ = (e – 1)/2

16. ∫₁² 5x²/(x² + 4x + 3) dx
Ans: [5x + (5/2) log|x + 1| – (45/2) log|x + 3|]₁² = 5 + (5/2) log(3/2) – (45/2) log(5/4)

17. ∫₀^π/4 (2sec²x + x³ + 2) dx
Ans: [2tan x + x⁴/4 + 2x]₀^π/4 = 2 + π⁴/1024 + π/2

18. ∫₀^π (sin²(x/2) – cos²(x/2)) dx
Ans: Since sin²(x/2) – cos²(x/2) = -cos x, the integral is [-sin x]₀^π = 0

19. ∫₀² (6x + 3)/(x² + 4) dx
Ans: [3 log(x² + 4) + (3/2) tan⁻¹(x/2)]₀² = 3 log 2 + 3π/8

20. ∫₀¹ (xeˣ + sin(πx/4)) dx
Ans: [eˣ(x – 1) – (4/π) cos(πx/4)]₀¹ = 1 + (4 – 2√2)/π

21. Choose the correct answer: ∫₁^√3 dx/(1 + x²) is equal to
(A) π/3
(B) 2π/3
(C) π/6
(D) π/12
Ans: (D) π/12

22. Choose the correct answer: ∫₀^2/3 dx/(4 + 9x²) is equal to
(A) π/6
(B) π/12
(C) π/24
(D) π/4
Ans: (C) π/24

Exercise 7.9 Solutions (All 10 Questions)

Evaluate the definite integrals in Exercises 1 to 8.

1. ∫₀¹ x/(x² + 1) dx
Ans: Let t = x² + 1. [(1/2) log(x² + 1)]₀¹ = (1/2) log 2

2. ∫₀^π/2 √(sin φ) cos⁵φ dφ
Ans: Put t = sin φ, so cos⁴φ = (1 – t²)². The integral becomes ∫₀¹ √t (1 – t²)² dt = 64/231

3. ∫₀¹ sin⁻¹(2x/(1 + x²)) dx
Ans: Since sin⁻¹(2x/(1 + x²)) = 2tan⁻¹x for 0 ≤ x ≤ 1, the integral is [2x tan⁻¹x – log(1 + x²)]₀¹ = π/2 – log 2

4. ∫₀² x√(x + 2) dx
Ans: Put t = x + 2, so x = t – 2. ∫₂⁴ (t – 2)√t dt = (32 + 16√2)/15

5. ∫₀^π/2 sin x/(1 + cos²x) dx
Ans: Put t = cos x, so dt = -sin x dx. ∫₀¹ dt/(1 + t²) = π/4

6. ∫₀² dx/(x + 4 – x²)
Ans: Since x + 4 – x² = 17/4 – (x – 1/2)², the integral evaluates to (1/√17) log[(5 + √17)/(5 – √17)]

7. ∫_-1¹ dx/(x² + 2x + 5)
Ans: Since x² + 2x + 5 = (x + 1)² + 4, [(1/2) tan⁻¹((x + 1)/2)]_-1¹ = π/8

8. ∫₁² (1/x – 1/(2x²))e^2x dx
Ans: Using the f + f′ pattern with f(x) = 1/(2x): [e^2x/(2x)]₁² = e⁴/4 – e²/2

9. Choose the correct answer: The value of ∫_1/3¹ (x – x³)^1/3/x⁴ dx is
(A) 6
(B) 0
(C) 3
(D) 4
Ans: (A) 6

10. Choose the correct answer: If f(x) = ∫₀^x t sin t dt, then f′(x) is
(A) cos x + x sin x
(B) x sin x
(C) x cos x
(D) sin x + x cos x
Ans: (B) x sin x

Miscellaneous Exercise on Chapter 7 (All 44 Questions)

Integrate the functions in Exercises 1 to 24.

1. ∫ dx/(x – x³)
Ans: Since 1/(x – x³) = 1/x + x/(1 – x²) by partial fractions, the integral is log|x| – (1/2) log|1 – x²| + C = (1/2) log|x²/(1 – x²)| + C

2. ∫ dx/(√(x+a) + √(x+b))
Ans: Rationalizing by multiplying by (√(x+a) – √(x+b)), the denominator becomes a – b. The integral is [2/(3(a-b))][(x+a)^3/2 – (x+b)^3/2] + C

3. ∫ dx/(x√(ax – x²)) [Hint: Put x = a/t]
Ans: The integral is -(2/a)√((a-x)/x) + C

4. ∫ dx/[x²(x⁴+1)^3/4]
Ans: Let t = 1 + 1/x⁴, so dt = -4/x⁵ dx. The integral becomes -(1+1/x⁴)^1/4 + C

5. ∫ dx/(√x + x^1/3) [Hint: Put x = t⁶]
Ans: Put x = t⁶, so dx = 6t⁵dt, √x = t³, x^1/3 = t². The integral becomes 6∫t³dt/(t+1) = 6∫(t² – t + 1 – 1/(t+1))dt = 2t³ – 3t² + 6t – 6log|t+1| + C = 2√x – 3x^1/3 + 6x^1/6 – 6log(1+x^1/6) + C

6. ∫ 5x dx/[(x+1)(x²+9)]
Ans: By partial fractions, 5x/[(x+1)(x²+9)] = -1/[2(x+1)] + (x+9)/[2(x²+9)]. The integral is -(1/2)log|x+1| + (1/4)log(x²+9) + (3/2)tan⁻¹(x/3) + C

7. ∫ sin x/sin(x-a) dx
Ans: Writing sin x = sin[(x-a)+a] = sin(x-a)cos a + cos(x-a)sin a, the integrand becomes cos a + sin a · cos(x-a)/sin(x-a). The integral is x cos a + sin a · log|sin(x-a)| + C

8. ∫ [e^{5 log x} – e^{4 log x}]/[e^{3 log x} – e^{2 log x}] dx
Ans: Since e^{k log x} = x^k, the integrand simplifies to (x⁵ – x⁴)/(x³ – x²) = x². The integral is x³/3 + C

9. ∫ cos x/√(4 – sin²x) dx
Ans: This is of the form f′(x)/√(a²-f(x)²) with f(x) = sin x, a = 2. The integral is sin⁻¹(sin x/2) + C

10. ∫ (sin⁸x – cos⁸x)/(1 – 2sin²x cos²x) dx
Ans: Since sin⁸x – cos⁸x = (sin²x – cos²x)(sin⁴x + cos⁴x) = -cos 2x · (1 – 2sin²x cos²x), the integrand simplifies to -cos 2x. The integral is -(sin 2x)/2 + C

11. ∫ dx/[cos(x+a)cos(x+b)]
Ans: Using sin[(x+a)-(x+b)] = sin(a-b) = sin(x+a)cos(x+b) – cos(x+a)sin(x+b), dividing by cos(x+a)cos(x+b) gives 1/[cos(x+a)cos(x+b)] = [tan(x+a) – tan(x+b)]/sin(a-b). The integral is [1/sin(a-b)] log|cos(x+b)/cos(x+a)| + C

12. ∫ x³dx/√(1 – x⁸)
Ans: Let t = x⁴, so dt = 4x³dx. The integral becomes (1/4)∫dt/√(1-t²) = (1/4)sin⁻¹(x⁴) + C

13. ∫ eˣdx/[(1+eˣ)(2+eˣ)]
Ans: Let t = eˣ, so dt = eˣdx. By partial fractions, 1/[(1+t)(2+t)] = 1/(1+t) – 1/(2+t). The integral is log|(eˣ+1)/(eˣ+2)| + C

14. ∫ dx/[(x²+1)(x²+4)]
Ans: By partial fractions, 1/[(x²+1)(x²+4)] = (1/3)/(x²+1) – (1/3)/(x²+4). The integral is (1/3)tan⁻¹x – (1/6)tan⁻¹(x/2) + C

15. ∫ cos³x · e^{log sin x} dx
Ans: Since e^{log sin x} = sin x, let t = cos x, so dt = -sin x dx. The integral becomes -∫t³dt = -cos⁴x/4 + C

16. ∫ e^{3 log x}(x⁴+1)⁻¹ dx
Ans: Since e^{3 log x} = x³, let t = x⁴+1, so dt = 4x³dx. The integral is (1/4)log(x⁴+1) + C

17. ∫ f′(ax+b)[f(ax+b)]ⁿ dx
Ans: Let t = f(ax+b), so dt = a · f′(ax+b)dx. The integral becomes (1/a)∫tⁿdt = [f(ax+b)]ⁿ⁺¹/[a(n+1)] + C

18. ∫ dx/√(sin³x · sin(x+α))
Ans: The integral is -(2/sinα)√(cosα + sinα · cot x) + C

19. ∫ [sin⁻¹√x – cos⁻¹√x]/[sin⁻¹√x + cos⁻¹√x] dx, x∈[0,1]
Ans: Since sin⁻¹√x + cos⁻¹√x = π/2 (constant), the integrand simplifies to (4/π)sin⁻¹√x – 1. The integral is [2(2x-1)/π]sin⁻¹√x + (2/π)√(x-x²) – x + C

20. ∫ √[(1-√x)/(1+√x)] dx
Ans: Putting x = cos²θ and simplifying, the integral is -2√(1-x) + cos⁻¹√x + √(x-x²) + C

21. ∫ [(2+sin 2x)/(1+cos 2x)]eˣ dx
Ans: Since (2+sin2x)/(1+cos2x) = sec²x + tan x, which has the form f(x)+f′(x) with f(x) = tan x, the integral is eˣtan x + C

22. ∫ (x²+x+1)/[(x+1)²(x+2)] dx
Ans: By partial fractions, the integrand equals -2/(x+1) + 1/(x+1)² + 3/(x+2). The integral is -2log|x+1| – 1/(x+1) + 3log|x+2| + C

23. ∫ tan⁻¹√[(1-x)/(1+x)] dx
Ans: Putting x = cosθ, tan⁻¹√[(1-cosθ)/(1+cosθ)] = θ/2 = (1/2)cos⁻¹x. Integrating by parts, the integral is (1/2)[x cos⁻¹x – √(1-x²)] + C

24. ∫ √(x²+1)[log(x²+1) – 2log x]/x⁴ dx
Ans: Let t = 1 + 1/x². Simplifying via this substitution, the integral is -(1/3)(1+1/x²)^3/2[log(1+1/x²) – 2/3] + C

Evaluate the definite integrals in Exercises 25 to 33.

25. ∫_π/2^π eˣ · (1-sin x)/(1-cos x) dx
Ans: Using 1-sin x = (sin(x/2)-cos(x/2))² and 1-cos x = 2sin²(x/2), the integral evaluates to e^π/2

26. ∫₀^π/4 sin x cos x/(cos⁴x+sin⁴x) dx
Ans: Dividing numerator and denominator by cos⁴x and substituting t = tan²x, the integral evaluates to π/8

27. ∫₀^π/2 cos²x/(cos²x+4sin²x) dx
Ans: Dividing by cos²x and substituting t = tan x, the integral evaluates to π/6

28. ∫_π/6^π/3 (sin x+cos x)/√(sin 2x) dx
Ans: Let t = sin x – cos x, so dt = (cos x+sin x)dx and sin2x = 1-t². The integral becomes ∫dt/√(1-t²) = sin⁻¹t, which evaluates to 2sin⁻¹((√3-1)/2)

29. ∫₀¹ dx/(√(1+x)-√x)
Ans: Rationalizing, 1/(√(1+x)-√x) = √(1+x)+√x. The integral is [(2/3)(1+x)^3/2+(2/3)x^3/2]₀¹ = 4√2/3

30. ∫₀^π/4 (sin x+cos x)/(9+16sin 2x) dx
Ans: Let t = sin x – cos x, so dt = (cos x+sin x)dx and sin2x = 1-t². The integral becomes ∫_-1⁰ dt/(25-16t²) = (1/40)log 9

31. ∫₀^π/2 sin 2x · tan⁻¹(sin x) dx
Ans: Integrating by parts with u = tan⁻¹(sin x), the integral evaluates to π/2 – 1

32. ∫₀^π x tan x/(sec x+tan x) dx
Ans: Using ∫₀^af(x)dx = ∫₀^af(a-x)dx with a = π, the integral evaluates to π²/2 – π

33. ∫₁⁴ [|x-1|+|x-2|+|x-3|] dx
Ans: Splitting the interval at x = 1, 2, 3 and integrating each piece, the integral evaluates to 19/2

Prove the following (Exercises 34 to 39).

34. ∫₁³ dx/[x²(x+1)] = 2/3 + log(2/3)
Ans: By partial fractions, 1/[x²(x+1)] = 1/x² – 1/x + 1/(x+1). Integrating and evaluating from 1 to 3 gives 2/3 + log(2/3)

35. ∫₀¹ x eˣ dx = 1
Ans: By integration by parts, [xeˣ-eˣ]₀¹ = 0-(-1) = 1

36. ∫_-1¹ x¹⁷cos⁴x dx = 0
Ans: Since x¹⁷ is an odd function and cos⁴x is an even function, their product is odd; the integral of an odd function over the symmetric interval [-1,1] is 0

37. ∫₀^π/2 sin³x dx = 2/3
Ans: Writing sin³x = sin x(1-cos²x) and substituting t = cos x, the integral evaluates to 2/3

38. ∫₀^π/4 2tan³x dx = 1 – log 2
Ans: Writing tan³x = tan x(sec²x-1), the integral evaluates to 1 – log 2

39. ∫₀¹ sin⁻¹x dx = π/2 – 1
Ans: By integration by parts, [x sin⁻¹x + √(1-x²)]₀¹ = π/2 – 1

Evaluate the following integral as the limit of a sum (Exercise 40).

40. ∫₀¹ e^2-3x dx
Ans: Using the limit-of-a-sum definition, the integral evaluates to [-e^2-3x/3]₀¹ = (1/3)(e² – 1/e)

Choose the correct answer in Exercises 41 to 44.

41. ∫ dx/(eˣ+e⁻ˣ) is equal to
(A) tan⁻¹(eˣ)+C
(B) tan⁻¹(e⁻ˣ)+C
(C) log(eˣ-e⁻ˣ)+C
(D) log(eˣ+e⁻ˣ)+C
Ans: Let t = eˣ, so dt = eˣdx. The integral becomes ∫dt/(t²+1) = tan⁻¹t = tan⁻¹(eˣ). The answer is (A)

42. ∫ cos 2x/(sin x+cos x)² dx is equal to
(A) -1/(sin x+cos x)+C
(B) log|sin x+cos x|+C
(C) log|sin x-cos x|+C
(D) 1/(sin x+cos x)²+C
Ans: Since cos2x = cos²x-sin²x = (cos x-sin x)(cos x+sin x), the integrand simplifies to (cos x-sin x)/(sin x+cos x). The integral is log|sin x+cos x|+C. The answer is (B)

43. If f(a+b-x) = f(x), then ∫_a^b x · f(x)dx equals
(A) (a+b)/2 ∫f(b-x)dx
(B) (a+b)/2 ∫f(b+x)dx
(C) (b-a)/2 ∫_a^bf(x)dx
(D) (a+b)/2 ∫_a^bf(x)dx
Ans: Using ∫_a^bx f(x)dx = ∫_a^b(a+b-x)f(a+b-x)dx = ∫_a^b(a+b-x)f(x)dx and adding both forms gives 2∫_a^bx f(x)dx = (a+b)∫_a^bf(x)dx. The answer is (D)

44. The value of ∫₀¹ tan⁻¹[(2x-1)/(1+x-x²)] dx is
(A) 1
(B) 0
(C) -1
(D) π/4
Ans: Substituting x = 1/2+u makes the integrand an odd function of u over the symmetric interval [-1/2,1/2]. The answer is (B) 0

Frequently Asked Questions

Why do we add a constant C in indefinite integration?
Because differentiation of a constant is zero, so infinitely many antiderivatives differ only by a constant.

What is the ILATE rule?
A priority order (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential) for choosing u in integration by parts.

What does a definite integral represent geometrically?
The signed area between the curve and the x-axis over the given interval.

Quick visual: a worked diagram from the full Extra Questions page, for reference.

Written by Satish

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