Here we provide RD Sharma Class 12 Ex 20.2 Solutions Chapter 20 Definite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 20.2 Solutions Chapter 20 Definite Integrals book pdf download. Now you will get step-by-step solutions to each question.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 20 |

Exercise | 20.2 |

Category | RD Sharma Solutions |

**RD Sharma Class 12 Ex 20.2 Solutions Chapter 20 Definite Integrals**

**Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed **

**Evaluate the following definite integrals:**

**Question 1. **

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 2. **

**Solution:**

We have,

I =

Let 1 + log x = t, so we have,

=> (1/x) dx = 2t dt

Now, the lower limit is, x = 1

=> t = 1 + log x

=> t = 1 + log 1

=> t = 1 + 0

=> t = 1

Also, the upper limit is, x = 2

=> t = 1 + log x

=> t = 1 + log 2

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 3. **

**Solution:**

We have,

I =

Let 9x

^{2}– 1 = t, so we have,=> 18x dx = dt

=> 3x dx = dt/6

Now, the lower limit is, x = 1

=> t = 9x

^{2}– 1=> t = 9 (1)

^{2}– 1=> t = 9 – 1

=> t = 8

Also, the upper limit is, x = 2

=> t = 9x

^{2}– 1=> t = 9 (2)

^{2}– 1=> t = 36 – 1

=> t = 35

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value ofis.

**Question 4. **

**Solution:**

We have,

I =

On putting sin x = and cos x = , we get

I =

I =

Let tan x/2 = t. So, we have

=> = dt

Now, the lower limit is, x = 0

=> t = tan x/2

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = tan x/2

=> t = tan π/4

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 5. **

**Solution:**

We have,

I =

Let a

^{2}+ x^{2}= t^{2}. So, we have=> 2x dx = 2t dt

=> x dx = t dt

Now, the lower limit is, x = 0

=> t

^{2}= a^{2 }+ x^{2}=> t

^{2}= a^{2 }+ 0^{2}=> t

^{2}= a^{2}=> t = a

Also, the upper limit is, x = a

=> t

^{2}= a^{2}+ x^{2}=> t

^{2 }= a^{2}+ a^{2}=> t

^{2}= 2a^{2}=> t = √2 a

So, the equation becomes,

I =

I =

I =

I = √2a – a

I = a (√2 – 1)

Therefore, the value ofis a (√2 – 1).

**Question 6. **

**Solution:**

We have,

I =

Let e

^{x}= t. So, we have=> e

^{x}dx = dtNow, the lower limit is, x = 0

=> t = e

^{x}=> t = e

^{0}=> t = 1

Also, the upper limit is, x = a

=> t = e

^{x}=> t = e

^{1}=> t = e

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value ofis.

**Question 7. **

**Solution:**

We have,

I =

Let x

^{2}= t. So, we have=> 2x dx = dt

Now, the lower limit is, x = 0

=> t = x

^{2}=> t = 0

^{2}=> t = 0

Also, the upper limit is, x = 1

=> t = x

^{2}=> t = 1

^{2}=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 8. **

**Solution:**

We have,

I =

Let log x = t. So, we have

=> (1/x) dx = dt

Now, the lower limit is, x = 1

=> t = log x

=> t = log 1

=> t = 0

Also, the upper limit is, x = 3

=> t = log x

=> t = log 3

So, the equation becomes,

I =

I =

I = sin (log 3) – sin 0

I = sin (log 3) – 0

I = sin (log 3)

Therefore, the value ofis sin (log 3).

**Question 9. **

**Solution:**

We have,

I =

Let x

^{2}= t. So, we have=> 2x dx = dt

Now, the lower limit is, x = 0

=> t = x

^{2}=> t = 0

^{2}=> t = 0

Also, the upper limit is, x = 1

=> t = x

^{2}=> t = 1

^{2}=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 10. **

**Solution:**

We have,

I =

Let x = a sin t. So, we have

=> dx = a cos t dt

Now, the lower limit is, x = 0

=> a sin t = x

=> a sin t = 0

=> sin t = 0

=> t = 0

Also, the upper limit is, x = a

=> a sin t = a

=> a sin t = a

=> sin t = 1

=> t = π/2

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 11. **

**Solution:**

We have,

I =

I =

Let sin Ø = t. So, we have

=> cos Ø dØ = dt

Now, the lower limit is, Ø = 0

=> t = sin Ø

=> t = sin 0

=> t = 0

Also, the upper limit is, Ø = π/2

=> t = sin Ø

=> t = sin π/2

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 12. **

**Solution:**

We have,

I =

Let sin x = t. So, we have

=> cos x dx = dt

Now, the lower limit is, x = 0

=> t = sin x

=> t = sin 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = sin x

=> t = sin π/2

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 13. **

**Solution:**

We have,

I =

Let 1 + cos θ = t

^{2}. So, we have=> – sin θ dθ = 2t dt

=> sin θ dθ = –2t dt

Now, the lower limit is, θ = 0

=> t

^{2}= 1 + cos θ=> t

^{2}= 1 + cos 0=> t

^{2}= 1 + 1=> t

^{2 }= 2=> t = √2

Also, the upper limit is, θ = π/2

=> t

^{2}= 1 + cos θ=> t

^{2 }= 1 + cos π/2=> t

^{2}= 1 + 0=> t

^{2}= 1=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 14. **

**Solution:**

We have,

I =

Let 3 + 4 sin x = t. So, we have

=> 0 + 4 cos x dx = dt

=> 4 cos x dx = dt

=> cos x dx = dt/4

Now, the lower limit is, x = 0

=> t = 3 + 4 sin x

=> t = 3 + 4 sin 0

=> t = 3 + 0

=> t = 3

Also, the upper limit is, x = π/3

=> t = 3 + 4 sin x

=> t = 3 + 4 sin π/3

=> t = 3 + 4 (√3/2)

=> t = 3 + 2√3

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value ofis.

**Question 15. **

**Solution:**

We have,

I =

Let tan

^{–1}x = t. So, we have=> (1/1+x

^{2}) dx = dtNow, the lower limit is, x = 0

=> t = tan

^{–1}x=> t = tan

^{–1}0=> t = 0

Also, the upper limit is, x = 1

=> t = tan

^{–1}t=> t = tan

^{–1 }1=> t = π/4

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 16. **

**Solution:**

We have,

I =

Let x + 2 = t

^{2}. So, we have=> dx = 2t dt

Now, the lower limit is, x = 0

=> t

^{2}= x + 2=> t

^{2}= 0 + 2=> t

^{2}= 2=> t = √2

Also, the upper limit is, x = 2

=> t

^{2}= x + 2=> t

^{2}= 2 + 2=> t

^{2 }= 4=> t = 2

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 17. **

**Solution:**

We have,

I =

Let x = tan t. So, we have

=> dx = sec

^{2}t dtNow, the lower limit is, x = 0

=> tan t = x

=> tan x = 0

=> x = 0

Also, the upper limit is, x = 1

=> tan t = x

=> tan x = 1

=> x = π/4

So, the equation becomes,

I =

I =

I =

I =

On applying integration by parts method, we get

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 18. **

**Solution:**

We have,

I =

Let sin

^{2}x = t. So, we have=> 2 sin x cos x = dt

=> sin x cos x = dt/2

Now, the lower limit is, x = 0

=> t = sin

^{2}x=> t = sin

^{2}0=> t = 0

Also, the upper limit is, x = π/2

=> t = sin

^{2}x=> t = sin

^{2}π/2=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 19. **

**Solution:**

We have,

I =

On putting cos x = and sin x = , we get,

I =

Let tan x/2 = t. So, we have

=> 1/2 sec

^{2}x/2 dx = dt=> sec

^{2}x/2 dx = 2 dtNow, the lower limit is, x = 0

=> t = tan x/2

=> t = tan 0/2

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = tan x/2

=> t = tan π/4

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 20. **

**Solution:**

We have,

I =

On putting sin x = , we get

I =

I =

I =

I =

I =

I =

Let tan x/2 = t. So, we have

=> 1/2 sec

^{2}x/2 dx = dt=> sec

^{2 }x/2 dx = 2 dtNow, the lower limit is, x = 0

=> t = tan x/2

=> t = tan 0/2

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = tan x/2

=> t = tan π/4

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Evaluate the following definite integrals:**

**Question 21. **

**Solution:**

We have,

I =

Let sin x = A (sin x + cos x) + B

=> sin x = A (sin x + cos x) + B (cos x – sin x)

=> sin x = sin x (A – B) + cos x (A + B)

On comparing both sides, we get

A – B = 1 and A + B = 0

On solving, we get A = 1/2 and B = –1/2.

Therefore, the expression becomes,

I =

I =

I =

I =

Therefore, the value ofis.

**Question 22. **

**Solution:**

We have,

I =

On putting cos x = and sin x = , we get,

I =

I =

I =

I =

Let tan x/2 = t. So, we have

=> 1/2 sec

^{2}x/2 dx = dt=> sec

^{2}x/2 dx = 2 dtNow, the lower limit is, x = 0

=> t = tan x/2

=> t = tan 0/2

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π

=> t = tan x/2

=> t = tan π/2

=> t = ∞

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 23. **

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 24. **

**Solution:**

We have,

I =

Let sin

^{–1}x = t. So, we have=> = dt

Now, the lower limit is, x = 0

=> t = sin

^{–1}x=> t = sin

^{–1}0=> t = 0

Also, the upper limit is, x = 1/2

=> t = sin

^{–1}x=> t = sin

^{–1}1/2=> t = π/6

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 25. **

**Solution:**

We have,

I =

I =

I =

I =

I =

Let sinx – cosx = t. So, we have

=> (cos x + sin x) dx = dt

Now, the lower limit is, x = 0

=> t = sinx – cosx

=> t = sin 0 – cos 0

=> t = 0 – 1

=> t = –1

Also, the upper limit is, x = π/4

=> t = sinx – cosx

=> t = sin π/4 – cos π/4

=> t = sin π/4 – sin π/4

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 26. **

**Solution:**

We have,

I =

I =

I =

I =

Let tan x = t. So, we have

=> sec

^{2}x dx = dtNow, the lower limit is, x = 0

=> t = tan x

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/4

=> t = tan x

=> t = tan π/4

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value ofis.

**Question 27. **

**Solution:**

We have,

I =

On putting cos x = , we get

I =

I =

I =

Let tan x/2 = t. So, we have

=> 1/2 sec

^{2}x/2 dx = dt=> sec

^{2}x/2 dx = 2 dtNow, the lower limit is, x = 0

=> t = tan x/2

=> t = tan 0/2

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π

=> t = tan x/2

=> t = tan π/2

=> t = ∞

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 28. **

**Solution:**

We have,

I =

I =

I =

I =

Let tan x = t. So, we have

=> sec

^{2 }x dx = dtNow, the lower limit is, x = 0

=> t = tan x

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = tan x

=> t = tan π/2

=> t = ∞

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 29. **

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 30. **

**Solution:**

We have,

I =

Let tan

^{–1}x = t. So, we have=> = dt

Now, the lower limit is, x = 0

=> t = tan

^{–1}x=> t = tan

^{–1}0=> t = 0

Also, the upper limit is, x = 1

=> t = tan

^{–1}x=> t = tan

^{–1}1=> t = π/4

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value ofis.

**Question 31. **

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 32. **

**Solution:**

We have,

I =

On using integration by parts, we get

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 33. **

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 34. **

**Solution:**

We have,

I =

Let 1 + x

^{2}= t. So, we have=> 2x dx = dt

Now, the lower limit is, x = 0

=> t = 1 + x

^{2}=> t = 1 + 0

^{2}=> t = 1 + 0

=> t = 1

Also, the upper limit is, x = π

=> t = 1 + x

^{2}=> t = 1 + 1

^{2}=> t = 1 + 1

=> t = 2

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I = 1

Therefore, the value ofis 1.

**Question 35. **

**Solution:**

We have,

I =

Let x – 4 = t

^{3}. So, we have=> dx = 3t

^{2}dtNow, the lower limit is, x = 4

=> t

^{3}= x – 4=> t

^{3}= 4 – 4=> t

^{3}= 0=> t = 0

Also, the upper limit is, x = 12

=> t

^{3}= x – 4=> t

^{3}= 12 – 4=> t

^{3}= 8=> t = 2

So, the equation becomes,

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 36. **

**Solution:**

We have,

I =

On using integration by parts, we get

I =

I =

I =

I =

I =

I = π + 0 – 0 – 0 – 2

I = π – 2

Therefore, the value ofis π – 2.

**Question 37. **

**Solution:**

We have,

I =

Let x = cos 2t. So, we have

=> dx = – 2 sin 2t dt

Now, the lower limit is, x = 0

=> cos 2t = x

=> cos 2t = 0

=> 2t = π/2

=> t = π/4

Also, the upper limit is, x = 1

=> cos 2t = x

=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 38. **

**Solution:**

We have,

I =

I =

I =

Let x + 1/x = t. So, we have

=> (1 – 1/x

^{2})dx = dtNow, the lower limit is, x = 0

=> t = x + 1/x

=> t = ∞

Also, the upper limit is, x = 1

=> t = x + 1/x

=> t = 1 + 1

=> t = 2

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 39. **

**Solution:**

We have,

I =

Let x

^{5}+ 1 = t. So, we have=> 5x

^{4}dx = dtNow, the lower limit is, x = –1

=> t = x

^{5}+ 1=> t = (–1)

^{5}+ 1=> t = –1 + 1

=> t = 0

Also, the upper limit is, x = 1

=> t = x

^{5}+ 1=> t = (1)

^{5}+ 1=> t = 1 + 1

=> t = 2

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 40. **

**Solution:**

We have,

I =

I =

Let tan x = t. So, we have

=> sec

^{2 }x dx = dtNow, the lower limit is, x = 0

=> t = tan x

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = tan x

=> t = tan π/2

=> t = ∞

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 41. **

**Solution:**

We have,

I =

Let sin 2t = u. So, we have

=> 2 cos 2t dt = du

=> cos 2t dt = du/2

Now, the lower limit is, x = 0

=> u = sin 2t

=> u = sin 0

=> u = 0

Also, the upper limit is, x = π/4

=> u = sin 2t

=> u = sin π/2

=> u = 1

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value ofis.

**Evaluate the following definite integrals:**

**Question 42. **

**Solution:**

We have,

I =

Let 5 – 4 cos θ = t. So, we have

=> 4 sin θ dθ = dt

=> sin θ dθ = dt/4

Now, the lower limit is, θ = 0

=> t = 5 – 4 cos θ

=> t = 5 – 4 cos 0

=> t = 5 – 4

=> t = 1

Also, the upper limit is, θ = π

=> t = 5 – 4 cos θ

=> t = 5 – 4 cos π

=> t = 5 + 4

=> t = 9

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I = 9√3 – 1

Therefore, the value ofis 9√3 – 1.

**Question 43. **

**Solution:**

We have,

I =

I =

I =

I =

I =

Let tan 2θ = t. So, we have

=> 2 sec

^{2}2θ dθ = dt=> sec

^{2}2θ dθ = dt/2Now, the lower limit is, θ = 0

=> t = tan 2θ

=> t = tan 0

=> t = 0

Also, the upper limit is, θ = π/6

=> t = tan 2θ

=> t = tan π/3

=> t = √3

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value ofis.

**Question 44. **

**Solution:**

We have,

I =

Let = t. So, we have

=> = dt

Now, the lower limit is, x = 0

=> t =

=> t =

=> t = 0

Also, the upper limit is, x =

=> t =

=> t =

=> t = π

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 45. **

**Solution:**

We have,

I =

Let 1 + log x = t. So, we have

=> 1/x dx = dt

Now, the lower limit is, x = 0

=> t = 1 + log x

=> t = 1 + log 0

=> t = 1

Also, the upper limit is, x = 2

=> t = 1 + log x

=> t = 1 + log 2

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value ofis.

**Question 46. **

**Solution:**

We have,

I =

I =

Let sin x = t. So, we have

=> cos x dx = dt

Now, the lower limit is, x = 0

=> t = sin x

=> t = sin 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = sin x

=> t = sin π/2

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 47. **

**Solution:**

We have,

I =

Let 30 – x

^{3/2}= t. So, we have=> = dt

=> = – dt

Now, the lower limit is, x = 4

=> t = 30 – x

^{3/2}=> t = 30 – 4

^{3/2}=> t = 30 – 8

=> t = 22

Also, the upper limit is, x = 9

=> t = 30 – x

^{3/2}=> t = 30 – 9

^{3/2}=> t = 30 – 27

=> t = 3

So, the equation becomes,

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 48. **

**Solution:**

We have,

I =

Let cos x = t. So, we have

=> – sin x dx = dt

=> sin x dx = –dt

Now, the lower limit is, x = 0

=> t = cos x

=> t = cos 0

=> t = 1

Also, the upper limit is, x = π

=> t = cos x

=> t = cos π

=> t = –1

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 49. **

**Solution:**

We have,

I =

Let sin x = t. So, we have

=> cos x dx = dt

Now, the lower limit is, x = 0

=> t = sin x

=> t = sin 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = sin x

=> t = sin π/2

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value ofis.

**Question 50. **

**Solution:**

We have,

I =

I =

Let sin x = t. So, we have

=> cos x dx = dt

Now, the lower limit is, x = 0

=> t = sin x

=> t = sin 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = sin x

=> t = sin π/2

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value ofis.

**Question 51. **

**Solution:**

We have,

I =

On using integration by parts, we get,

I =

I =

Let cos

^{-1}x = t. So, we have=> = dt

Now, the lower limit is, x = 0

=> t = cos

^{-1}x=> t = cos

^{-1}0=> t = π/2

Also, the upper limit is, x = 1

=> t = cos

^{-1}x=> t = cos

^{-1}1=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I = π – 2

Therefore, the value ofis π – 2.

**Question 52. **

**Solution:**

We have,

I =

Let x = a tan

^{2}t. So, we have=> dx = 2a tan t sec

^{2 }t dtNow, the lower limit is, x = 0

=> a tan

^{2}t = x=> a tan

^{2}t = 0=> tan t = 0

=> t = 0

Also, the upper limit is, x = a

=> a tan

^{2}t = x=> a tan

^{2}t = a=> tan

^{2}t = 1=> tan t = 1

=> t = π/4

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 53. **

**Solution:**

We have,

I =

I =

I =

I =

I =

Let cot x/2 = t. So, we have

=> = dt

Now, the lower limit is, x = π/3

=> t = cot x/2

=> t = cot π/6

=> t = √3

Also, the upper limit is, x = π/2

=> t = cot x/2

=> t = cot π/4

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I = 1

Therefore, the value ofis 1.

**Question 54. **

**Solution:**

We have,

I =

Let x

^{2}= a^{2}cos 2t. So, we have=> 2x dx = – 2a

^{2}sin 2t dtNow, the lower limit is, x = 0

=> a

^{2}cos 2t = x^{2}=> a

^{2}cos 2t = 0=> cos 2t = 0

=> 2t = π/2

=> t = π/4

Also, the upper limit is, x = a

=> a

^{2}cos 2t = x^{2}=> a

^{2}cos 2t = a^{2}=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 55. **

**Solution:**

We have,

I =

Let x = a cos 2t. So, we have

=> dx = –2a sin 2t

Now, the lower limit is, x = –a

=> a cos 2t = x

=> a cos 2t = –a

=> cos 2t = –1

=> 2t = π

=> t = π/2

Also, the upper limit is, x = a

=> a cos 2t = x

=> a cos 2t = a

=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I = πa

Therefore, the value ofis πa.

**Question 56. **

**Solution:**

We have,

I =

Let cos x = t. So, we have

=> – sin x dx = dt

=> sin x dx = –dt

Now, the lower limit is, x = 0

=> t = cos x

=> t = cos 0

=> t = 1

Also, the upper limit is, x = π/2

=> t = cos x

=> t = cos π/2

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I = – log 2 + 2 log 3 + 0 – 2 log 2

I = 2 log 3 – 3 log 2

I = log 9 – log 8

I = log 9/8

Therefore, the value ofis log 9/8.

**Question 57. **

**Solution:**

We have,

I =

I =

I =

Let sin

^{2}x = t. So, we have=> 2 sin x cos x dx = dt

Now, the lower limit is, x = 0

=> t = sin

^{2}x=> t = sin

^{2}0=> t = 0

Also, the upper limit is, x = π/2

=> t = sin

^{2}x=> t = sin

^{2}π/2=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 58. **

**Solution:**

We have,

I =

Let x = sin t. So, we have

=> dx = cos t dt

Now, the lower limit is, x = 0

=> sin t = x

=> sin t = 0

=> t = 0

Also, the upper limit is, x = 1/2

=> sin t = x

=> sin t = 1/2

=> t = π/6

So, the equation becomes,

I =

I =

I =

I =

I =

Let tan t = u. So, we have

=> sec

^{2}t dt = duNow, the lower limit is, t = 0

=> u = tan t

=> u = tan 0

=> t = 0

Also, the upper limit is, t = π/6

=> u = tan t

=> u = tan π/6

=> t = 1/√3

So, the equation becomes,

I =

I =

I =

Therefore, the value ofis.

**Question 59. **

**Solution:**

We have,

I =

Let 1/x

^{2}– 1 = t. So, we have=> –2/x

^{3}dx = dtNow, the lower limit is, x = 1/3

=> t = 1/x

^{2}– 1=> t = 9 – 1

=> t = 8

Also, the upper limit is, x = 1

=> t = 1/x

^{2}– 1=> t = 1 – 1

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

I = 6

Therefore, the value ofis 6.

**Question 60. **

**Solution:**

We have,

I =

I =

Let tan x = t. So, we have

=> sec

^{2}x dx = dtNow, the lower limit is, x = 0

=> t = tan t

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/4

=> t = tan t

=> t = tan π/4

=> t = 1

So, the equation becomes,

I =

Let t

^{3 }= u. So, we have=> 3t

^{2}dt = du=> t

^{2}dt = du/3Now, the lower limit is, t = 0

=> u = t

^{3}=> u = 0

^{3}=> u = 0

Also, the upper limit is, t = 1

=> u = t

^{3}=> u = 1

^{3}=> u = 1

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 61. **

**Solution:**

We have,

I =

I =

I =

I =

I =

Let cos x = t. So, we have

=> – sin x dx = dt

Now, the lower limit is, x = 0

=> t = cos x

=> t = cos 0

=> t = 1

Also, the upper limit is, x = π/2

=> t = cos x

=> t = cos π/2

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 62. **

**Solution:**

We have,

I =

I =

I =

Let cos x/2 + sin x/2 = t. So, we have

=> (cos x/2 – sin x/2) dx = 2 dt

Now, the lower limit is, x = 0

=> t = cos x/2 + sin x/2

=> t = cos 0 + sin 0

=> t = 1 + 0

=> t = 1

Also, the upper limit is, x = π/2

=> t = cos x/2 + sin x/2

=> t = cos π/2 + sin π/2

=> t = 1/√2 + 1/√2

=> t = √2

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value ofis.

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