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Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 20 |

Exercise | 20.1 |

Category | RD Sharma Solutions |

Table of Contents

**RD Sharma Class 12 Ex 20.1 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 = 2[√9 – √4 ]

I = 2 (3 − 2)

I = 2 (1)

I = 2

Therefore, the value ofis 2.

### Question 2.

**Solution:**

We have,

I =

I =

I = log (3 + 7) − log (−2 + 7)

I = log 10 − log 5

I =

I = log 2

Therefore, the value ofis log 2.

### Question 3.

**Solution:**

We have,

I =

Let x = sin t, so we have,

=> dx = cos t dt

Now, the lower limit is,

=> x = 0

=> sin t = 0

=> t = 0

Also, the upper limit is,

=> x = 1/2

=> sin t = 1/2

=> t = π/6

So, the equation becomes,

I =

I =

I =

I =

I =

I =

π/6 – 0I =

π/6

Therefore, the value ofis π/6.

### Question 4.

**Solution:**

We have,

I =

I =

I =

I =

I = π/4

Therefore, the value ofis π/4.

### Question 5.

**Solution:**

We have,

I =

Let x

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

=> x dx = dt/2

Now, the lower limit is, x = 2

=> t = x

^{2}+ 1=> t = (2)

^{2}+ 1=> t = 4 + 1

=> t = 5

Also, the upper limit is, x = 3

=> t = x

^{2}+ 1=> t = (3)

^{2 }+ 1=> t = 9 + 1

=> t = 10

So, the equation becomes,

I =

I =

I =

I = 1/2[log10 – log5]

I = 1/2[log10/5]

I = 1/2[log2]

I = log√2

Therefore, the value ofis log√2.

### Question 6.

**Solution:**

We have,

I =

I =

I =

I =

I =

I = 1/ab[tan

^{-1}∞ – tan^{-1}0]I = 1/ab[π/2 – 0]

I = 1/ab[π/2]

I = π/2ab

Therefore, the value ofis π/2ab.

### Question 7.

**Solution:**

We have,

I =

I =

I = [tan

^{-1}1 – tan^{-1}(-1)]I = [π/4 – (-π/4)]

I = [π/4 + π/4]

I = 2π/4

I = π/2

Therefore, the value ofis π/2.

### Question 8.

**Solution:**

We have,

I =

I =

I = -e

^{–}^{∞}– (-e^{0})I = − 0 + 1

I = 1

Therefore, the value ofis 1.

### Question 9.

**Solution:**

We have,

I =

I =

I =

I =

I =

I = [1 − 0] − [log(1 + 1) − log(0 + 1)]

I = 1 − [log2 − log1]

I = 1 – log2/1

I = 1 − log 2

I = log e − log 2

I = loge/2

Therefore, the value ofis loge/2.

### Question 10.

**Solution:**

We have,

I =

I =

I =

I = [-cosπ/2 + cos0] + [sinπ/2 – sin0]

I = [−0 + 1] + 1

I = 1 + 1

I = 2

Therefore, the value ofis 2.

### Question 11.

**Solution:**

We have,

I =

I =

I = log(sinπ/2) – log(sinπ/4)

I = log1 – log1/√2

I =

I = log√2

Therefore, the value ofis log√2.

### Question 12.

**Solution:**

We have,

I =

I =

I = log(secπ/4 + tanπ/4 – log(sec0 + tan0)

I = log(√2 + 1) – log(1 + 0)

I =

I = log(√2 + 1)

Therefore, the value ofis log(√2 + 1).

### Question 13.

**Solution:**

We have,

I =

I =

I = [log|cosecπ/4 – cotπ/4|] – [log|cosecπ/6 – cotπ/6|]

I = [log|√2 – 1|] – [log|2 – √3|]

I =

Therefore, the value ofis.

### Question 14.

**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 = 0

=> 2t = π/2

=> t = π/4

Also, the upper limit is,

=> x = 1

=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

Let cos t = z, so we have,

=> – sin t dt = dz

=> sin t dt = – dz

Now, the lower limit is,

=> t = 0

=> z = cos t

=> z = cos 0

=> z = 1

Also, the upper limit is,

=> t = π/4

=> z = cos t

=> z = cos π/4

=> z = 1/√2

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I = -4[(log1/√2 – 1/2(2)) – (log1 – 1/2)]

I = -4[(log1/√2 – 1/4) – (0 – 1/2)]

I = -4[log1/√2 – 1/4 – 0 + 1/2]

I = -4[-log√2 + 1/4]

I = 4log√2 – 1

I = 4 × 1/2log2 – 1

I = 2log2 – 1

Therefore, the value ofis 2log2 – 1.

### Question 15.

**Solution: **

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I = [tan π – tan0] – [sec π – sec 0]

I = [0 – 0] – [–1 – 1]

I = 0 – (–2)

I = 2

Therefore, the value ofis 2.

### Question 16.

**Solution: **

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I = [tan π/4 – tan(–π/4)] – [sec π/4 – sec (–π/4)]

I = [1 – (–1)] – [sec π/4 – sec (π/4)]

I = 2 – 0

I = 2

Therefore, the value ofis 2.

### Question 17.

**Solution: **

We have,

I =

I =

I =

I =

I =

I =

I = 1/2[π/2 – 0] + 1/4[sinπ – sin0]

I = 1/2[π/2] + 1/4[0 – 0]

I = π/4

Therefore, the value ofisπ/4.

### Question 18.

**Solution: **

We have,

I =

I =

I =

I =

I =

I =

I = 1/12 [-1 – 0] + 3/4[1 – 0]

I = 3/4 – 1/12

I = (9 – 1)/12

I = 8/12

I = 2/3

Therefore, the value ofis 2/3.

### Question 19.

**Solution: **

We have,

I =

I =

I =

I =

I =

I =

I = 1/6[sinπ/2 – sin0] + 1/2[sinπ/6 – sin0]

I = 1/6[1 – 0] + 1/2[1/2 – 0]

I = 1/6 + 1/4

I = (4 + 6)/24

I = 10/24

I = 5/12

Therefore, the value ofis5/12.

### Question 20.

**Solution: **

We have,

I =

I =

I =

I =

I =

I =

I = 1/2[sinπ/2 – sin0] – 1/6[sin3π/2 – sin0]

I = 1/2[1 – 0] – 1/6[-1 – 0]

I = 1/2 – 1/6(-1)

I = 1/2 + 1/6

I = (6 + 2)/12

I = 8/12

I = 2/3

Therefore, the value ofis 2/3.

### Question 21.

**Solution: **

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I = 2[-cotπ/2 + cot2π/3]

I = 2[-1/√3 – 0]

I = -2/√3

Therefore, the value ofis -2/√3.

### Question 22.

**Solution: **

We have,

I =

I =

I =

I =

I =

I =

I =

I = 1/4[π/2 + π/4 + 0 + 0 – 0 – 0 – 0 – 0]

I = 1/4[3π/4]

I = 3π/16

Therefore, the value ofis3π/16.

### Evaluate the following definite integrals:

### Question 23.

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I = [(a

^{2}+ b^{2})/2][π/2]I = π(a

^{2}+ b^{2})/4

Therefore, the value ofis π(a^{2}+ b^{2})/4.

### Question 24.

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I = 2[sinπ/4 – cosπ/4 – 0 + 1]

I = 2[1/√2 – 1/√2 – 0 + 1]

I = 2 (1)

I = 2

Therefore, the value ofis 2.

### Question 25.

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I = 2√2[sinπ/4 – sin0]

I = 2√2[1/√2- sin0]

I = 2√2[1/√2]

I = 2

Therefore, the value ofis 2.

### Question 26.

**Solution:**

We have,

I =

By using integration by parts, we get,

I = x ∫sinxdx – ∫(∫sin x (1)dx)dx

I = -xcosx – ∫(∫sin xdx)dx

I = -xcosx + ∫cosxdx

I = -xcosx + sinx

So we get,

I =

I = [-π/2cosπ/2 + sinπ/2 + 0 – 0]

I = 0 + 1 + 0 – 0

I = 1

Therefore, the value ofis 1.

### Question 27.

**Solution:**

We have,

I =

By using integration by parts, we get,

I = x∫cosxdx – ∫(∫cos x (1)dx)dx

I = xsinx – ∫(∫cosxdx)dx

I = xsinx – ∫sinxdx

I = x sin x + cos x

So we get,

I =

I = [π/2sinπ/2 + cosπ/2 – 0 – cos0]

I = π/2 + 0 – 0 – 1

I = π/2 – 1

Therefore, the value ofis π/2 – 1.

### Question 28.

**Solution:**

We have,

I =

By using integration by parts, we get,

I = x

^{2}sinx – ∫(2x∫(cosx)dx)dxI = x

^{2}sinx – ∫(2xsinx)dxI = x

^{2}sinx – 2[-xcosx – ∫(1∫sinxdx)dx]I = x

^{2}sinx – 2[-xcosx + ∫sinxdx]I = x

^{2}sinx – 2[-xcosx + sinx]I = x

^{2}sinx + 2xcosx – 2sinxSo we get,

I =

I = [(π/2)

^{2}sinπ/2 + 2(π/2)cosπ/2 – 2sinπ/2 – 0 – 0 + sin0]I = [π

^{2}/4 + 0 – 2 – 0 – 0 + 0]I = π

^{2}/4 – 2

Therefore, the value ofis π^{2}/4 – 2.

### Question 29.

**Solution:**

We have,

I =

By using integration by parts, we get,

I = -x

^{2}cosx – ∫(2x∫sinxdx)dxI = -x

^{2}cosx + ∫(2xcosx)dxI = -x

^{2}cosx + 2[xsinx – ∫(∫cosxdx)dx]I = -x

^{2}cosx + 2[xsinx – ∫sinxdx]I = -x

^{2}cosx + 2[xsinx + cosx]I = -x

^{2}cosx + 2xsinx + 2cosxSo we get,

I =

I = -(π/4)

^{2}cosπ/4 + 2π/4sinπ/4 + 2cosπ/4 + 0 – 0 – 2I = –π

^{2}/16(1/ √2) + π/2(1/√2) + 2(1/√2) + 0 – 0 – 2I = –π

^{2}/16√2 + π/2√2 + √2 – 2

Therefore, the value ofis -π^{2}/16√2 + π/2√2 + √2 – 2.

### Question 30.

**Solution:**

We have,

I =

By using integration by parts, we get,

I = 1/2x

^{2}sin2x – ∫(2x∫cos2xdx)dxI = 1/2x

^{2}sin2x – ∫(xsin2x)dxI = 1/2x

^{2}sin2x – [-1/2xcos2x – ∫(∫sin2xdx)dx]I = 1/2x

^{2}sin2x – [-1/2xcos2x + ∫1/2 cos2xdx]I = 1/2x

^{2}sin2x – [-1/2xcos2x + 1/4sin2xdx]I = 1/2x

^{2}sin2x + 1/2xcos2x – 1/4sin2xdxSo we get,

I =

I = [1/2(π

^{2}/4)sinπ + 1/2(π/2)cosπ – 0 – 0 – 0 + 0]I = -π/4

Therefore, the value ofis -π/4.

### Question 31.

**Solution:**

We have,

I =

I =

I =

I =

By using integration by parts, we get,

I = 1/2[x

^{3}/3] + x^{2}sin2x/2 – [x ∫sin2x – ∫(∫sin2xdx)dx]I = 1/2[x

^{3}/3] + x^{2}sin2x/2 + xcosx/2 – sin2x/4So we get,

I =

I = [1/6[π

^{3}/8] + 0 + 0 – π/8]I = π

^{3}/48 – π/8

Therefore, the value ofis π^{3}/48 – π/8.

### Question 32.

**Solution:**

We have,

I =

By using integration by parts, we get,

I =

I = xlogx – ∫1dx

I = xlogx – x

So we get,

I =

I = 2log2 – 2 – log1 + 1

I = 2 log 2 – 1

Therefore, the value ofis 2 log 2 – 1.

### Question 33.

**Solution:**

We have,

I =

By using integration by parts, we get,

I =

I =

I =

I =

So we get,

I =

I = -log3/4 + log3 – log4 + log1/2 – log1 + log2

I = log3(1 – 1/4) – 2log2 + 0 – 0 + log2

I = 3/4log3 – log2

Therefore, the value ofis 3/4log3 – log2.

### Question 34.

**Solution:**

We have,

I =

I =

I =

By using integration by parts, we get,

I =

I = e

^{x}logxSo we get,

I =

I = e

^{e}loge – e^{1}log1I = e

^{e}(1) – 0I = e

^{e}

Therefore, the value ofis e^{e}.

### Question 35.

**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 = e

=> t = log x

=> t = log e

=> t = 1

So, the equation becomes,

I =

I =

I =

I = 1/2 – 0/2

I = 1/2

Therefore, the value ofis 1/2.

### Question 36.

**Solution:**

We have,

I =

I =

By using integration by parts, we get,

I =

I =

I =

I = x/logx

So we get,

I =

I =

I =

I = e

^{2}/2 – e

Therefore, the value ofis e^{2}/2 – e.

### Question 37.

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I =

I = 1/2[3log2 – log4 + log3]

I = 1/2[3log2 – 2log2 + log3]

I = 1/2[log 2 – log 3]

I = 1/2[log6]

I = log6/2

Therefore, the value ofis log6/2.

### Question 38.

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I = [1/5log6 + 3/√5tan

^{-1}(√5) – 1/5log1 – 3/√5tan^{-1}(0)]I = [1/5 log6 + 3√5 tan

^{-1}(√5) – 0 – 0]I = 1/5 log6 + 3√5 tan

^{-1}(√5)

Therefore, the value ofis 1/5 log6 + 3√5 tan^{-1}(√5).

### Question 39.

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

Let x – 1/2 = t, so we have,

=> dx = dt

Now, the lower limit is, x = 0

=> t = x – 1/2

=> t = 0 – 1/2

=> t = 1/2

Also, the upper limit is, x = 2

=> t = x – 1/2

=> t = 2 – 1/2

=> t = 3/2

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

### Question 40.

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I = 4/2√7[tan

^{-1}(5/√7) – tan^{-1}(1/√7)]I = 2/√7[tan

^{-1}(5/√7) – tan^{-1}(1/√7)]

Therefore, the value ofis 2/√7[tan^{-1}(5/√7) – tan^{-1}(1/√7)].

### Question 41.

**Solution:**

We have,

I =

Let x = sin

^{2}t, so we have,=> dx = 2 sin t cos t dt

Now, the lower limit is, x = 0

=> sin

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

=> t = 0

Also, the upper limit is, x = 1

=> sin

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

=> t = π/2

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I = 1/4[π/2 – 0] – 1/16[sin2π – 0]

I = 1/4[π/2] – 1/16[0 – 0 ]

I = π/8

Therefore, the value ofis π/8.

### Question 42.

**Solution:**

We have,

I =

I =

I =

I =

I =

I = [sin

^{-1}(1/2) – sin^{-1}(-1/2)]I = π/6 -(-π/6)

I = π/6 + π/6

I = π/3

Therefore, the value ofisπ/3.

**Question 43. **

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I = [sin

^{-1}(2/2) – sin^{-1}(-2/2)]I = sin

^{-1}1 – sin^{-1}(-1)I = π/2 – (-π/2)

I = π/2 + π/2

I = π

Therefore, the value ofis π.

### Question 44.

**Solution:**

We have,

I =

I =

I =

Let x + 1 = t, so we have,

=> dx = dt

Now, the lower limit is, x = –1

=> t = x + 1

=> t = – 1 + 1

=> t = 0

Also, the upper limit is, x = 1

=> t = x + 1

=> t = 1 + 1

=> t = 2

So, the equation becomes,

I =

I =

I = 1/2tan

^{-1}2/2 – 1/2tan^{-1}0/2I = 1/2tan

^{-1}1 – 1/2tan^{-1}0I = 1/2(π/4) – 0

I = π/8

Therefore, the value ofis π/8.

### Evaluate the following definite integrals:

### Question 45.

**Solution:**

We have,

I =

Let 2x + 1 = t

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

=> dx = t dt

Now, the lower limit is, x = 1

=> t

^{2}= 2x + 1=> t

^{2}= 2(1) + 1=> t

^{2}= 3=> t = √3

Also, the upper limit is, x = 4

=> t

^{2 }= 2x + 1=> t

^{2}= 2(4) + 1=> t

^{2}= 9=> t = 3

So, the equation becomes,

I =

I =

I =

I =

I =

I = 1/4 [3

^{5}/5 – 3 – (√3)^{5}/5 + √3]I = 1/4[243/5 – 3 – 9√3/5 + √3]

I = 1/4((243 – 15 – 9√3 + 5√3)/5)

I = 1/4[(228 – 4√3)/5]

I = 1/4[4(57 – √3)/5]

I = (57 – √3)/5

Therefore, the value ofis (57 – √3)/5.

### Question 46.

**Solution:**

We have,

I =

By using binomial theorem in the expansion of (1 – x)

^{5}, we get,I =

I =

I =

I =

I = 1/2 – 5/3 + 10/4 – 10/5 + 5/6 – 1/7

I = 1/2 – 5/3 + 5/3 – 2 + 5/6 – 1/7

I = 1/2 – 2 + 5/6 – 1/7

I = 1/42

Therefore, the value ofis 1/42.

### Question 47.

**Solution:**

We have,

I =

I =

I =

I =

By using integration by parts, we get,

I =

I =

I = e

^{x}/xSo we get,

I =

I = e

^{2}/2 – e^{1}/1I = e

^{2}/2 – e

Therefore, the value ofise^{2}/2 – e.

### Question 48.

**Solution:**

We have,

I =

By using integration by parts in first integral, we get,

I =

I = xe

^{2x}/2 – (1/2)(e^{2x}/2) + 2/π[1 – 0]I = xe

^{2x}/2 – e^{2x}/4 + 2/πSo we get,

I =

I = [e

^{2}/2 + e^{2}/4 – 0 + 1/4] + 2/πI = e

^{2}/4 + 1/4 + 2/π

Therefore, the value ofis e^{2}/4 + 1/4 + 2/π.

### Question 49.

**Solution:**

We have,

I =

By using integration by parts in first integral, we get,

I =

I =

I =

I =

So we get,

I =

I =

I = [e

^{1}(1 – 1) – e^{0}(0 – 1)] + 2√2/πI = [0 – (-1)] + 2√2/π

I = 1 + 2√2/π

Therefore, the value ofis 1 + 2√2/π.

### Question 50.

**Solution:**

We have,

I =

I =

I =

I =

I = -e

^{π}cotπ/2 + e^{π/2 }cotπ/4I = 0 + e

^{π/2}(1)I = e

^{π/2}

Therefore, the value ofis e^{π/2}.

### Question 51.

**Solution:**

We have,

I =

I =

I =

I =

By using integration by parts in first integral, we get,

I =

I =

I =

I = 1/√2[sinπ(2e

^{π}) – 0]I = 1/√2[0 – 0]

I = 0

Therefore, the value ofis 0.

### Question 52.

**Solution:**

We have,

I =

By using integration by parts, we get,

I = e^{x}cos(x/2 + π/4) + 1/2∫e^{x}sin(x/2 + π/4)

I = e^{x }cos(x/2 + π/4) + 1/2[ e^{x}sin(x/2 + π/4) – 1/2 ∫e^{x}cos(x/2 + π/4)dx]

I = e^{x}cos(x/2 + π/4) + 1/2e^{x}sin(x/2 + π/4) – 1/4I

5I/4 = -3/ 2√2(e^{2π }+ 1)

I = -3√2/5(e^{2π} + 1)

**Therefore, the value of ****is -3√2/5(e ^{2π} + 1).**

### Question 53.

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I = 2/3[2

^{3/2}– 1] + 2/3[1 – 0]I =

I = 2

^{5/2}/3

Therefore, the value ofis 2^{5/2}/3.

### Question 54.

**Solution:**

We have,

I =

I =

I =

I =

I = -log3 + log2 + 2[log4 – log3]

I = -log3 + log2 + 2[2log2 – log3]

I = -log3 + log2 + 4log2 – 2log3

I = 5log2 – 3log3

I = log2

^{5 }– log3^{3}I = log32 – log27

I = log32/27

Therefore, the value ofis log32/27.

### Question 55.

**Solution:**

We have,

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 = [0 – 1/3] – [0 – 1]

I = [-1/3] – [-1]

I = -1/3 + 1

I = 2/3

Therefore, the value ofis 2/3.

### Question 56.

**Solution:**

We have,

I =

I =

I =

I =

I = -sinπ + sin0

I = 0

Therefore, the value ofis 0.

### Question 57.

**Solution:**

We have,

I =

Let 2x = t, so we have,

=> 2x dx = dt

Now, the lower limit is, x = 1

=> t = 2x

=> t = 2(1)

=> t = 2

Also, the upper limit is, x = 2

=> t = 2x

=> t = 2(2)

=> t = 4

So, the equation becomes,

I =

I =

I =

By using integration by parts in first integral, we get,

I =

I =

I =

I = e

^{4}/4 – e^{2}/2

Therefore, the value ofis e^{4}/4 – e^{2}/2.

### Question 58.

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I = [sin

^{-1}(1) – sin^{-1}(-1)]I = π/2 – (-π/2)

I = π/2 + π/2

I = π

Therefore, the value ofis π.

### Question 59. If , find the value of k.

**Solution:**

We have,

=>

=>

=>

=>

=> tan

^{-1}2k/4 – tan^{-1}0 = π/16=> tan

^{-1}2k/4 – 0 = π/16=> tan

^{-1}2k/4 = π/16=> tan

^{-1}2k = π/4=> 2k = tanπ

/4=> 2k = 1

=> k = 1/2

Therefore, the value of k is 1/2.

### Question 60. If , find the value of k.

**Solution:**

We have,

=>

=>

=>

=>

=> a

^{3}– 0 = 8=> a

^{3}= 8=> a = 2

Therefore, the value of a is 2.

### Question 61.

**Solution:**

We have,

I =

I =

I =

I =

I = -[√2cos3

π/2 – √2cosπ]I = -(-√2 – 0)

I = √2

Therefore, the value ofis √2.

### Question 62.

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I =

I = [-4cosπ/2 + 4cos0] + [4sinπ/2 – 4sin0]

I = 0 + 4 + 4 – 0

I = 8

Therefore, the value ofis 8.

### Question 63.

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I = (

π/8 – 1/4) – (3/4(π/8 – 1/4) – 1/16)I =

π/8 – 1/4 – (3π/32 – 3/16 – 1/16)I = π/8 – 1/4 – (3π/32 – 1/4)

I = π/8 – 1/4 – 3π/32 + 1/4

I = π/8 – 3π/32

I = (4π – 3π)/32

I = π/32

Therefore, the value ofisπ/32.

### Question 64.

**Solution:**

We have,

I =

By using integration by parts we get,

I =

I =

I =

I =

I =

So we get,

I =

I = log3/2 – 1/8log3

I = 3/8log3

Therefore, the value ofis 3/8log3.

### Question 65.

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I =

I = [tanπ/3 – tanπ/6] + [-cotπ/3 + cotπ/6]

I = [√3 – 1/√3] + [- 1/√3 – √3]

I = 2[√3 – 1/√3]

I = 4/√3

Therefore, the value ofis 4/√3.

### Question 66.

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

### Question 67.

**Solution:**

We have,

I =

I =

I =

I =

I = -log2/4 + log2/2 – 1/4 + 1/2

I = log2/4 + 1/4

Therefore, the value ofis log2/4 + 1/4.

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