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Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 20 |

Exercise | 20.2 |

Category | RD Sharma Solutions |

Table of Contents

**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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