# RD Sharma Class 12 Ex 20.1 Solutions Chapter 20 Definite Integrals

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## RD Sharma Class 12 Ex 20.1 Solutions Chapter 20 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 of is 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 of is 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 – 0

I = π/6​

Therefore, the value of is π/6.

### Question 4.

Solution:

We have,

I =

I =

I =

I =

I = π/4

Therefore, the value of is π/4.

### Question 5.

Solution:

We have,

I =

Let x2 + 1 = t, so we have,

=> 2x dx = dt

=> x dx = dt/2

Now, the lower limit is, x = 2

=> t = x2 + 1

=> t = (2)2 + 1

=> t = 4 + 1

=> t = 5

Also, the upper limit is, x = 3

=> t = x2 + 1

=> t = (3)+ 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 of is log√2.

### Question 6.

Solution:

We have,

I =

I =

I =

I =

I =

I = 1/ab[tan-1∞ – tan-10]

I = 1/ab[π/2 – 0]

I = 1/ab[π/2]

I = π/2ab

Therefore, the value of is π/2ab.

### Question 7.

Solution:

We have,

I =

I =

I = [tan-11 – tan-1(-1)]

I = [π/4 – (-π/4)]

I = [π/4 + π/4]

I = 2π/4

I = π/2

Therefore, the value of is π/2.

### Question 8.

Solution:

We have,

I =

I =

I = -e – (-e0

I = − 0 + 1

I = 1

Therefore, the value of is 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 of  is 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 of is 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 of is 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 of is 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 of is .

### 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 of is 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 of is 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 of is 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 of  is π/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 of  is 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 of  is 5/12.

### 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 of  is 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 of  is -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 of is 3π/16.

### Question 23.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I = [(a2 + b2)/2][π/2]

I = π(a2 + b2)/4

Therefore, the value of is π(a2 + b2)/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 of  is 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 of is 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 of is 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 of is π/2 – 1.

### Question 28.

Solution:

We have,

I =

By using integration by parts, we get,

I = x2sinx – ∫(2x∫(cosx)dx)dx

I = x2sinx – ∫(2xsinx)dx

I = x2sinx – 2[-xcosx – ∫(1∫sinxdx)dx]

I = x2sinx – 2[-xcosx + ∫sinxdx]

I = x2sinx – 2[-xcosx + sinx]

I = x2sinx + 2xcosx – 2sinx

So we get,

I =

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

I = [π2/4 + 0 – 2 – 0 – 0 + 0]

I = π2/4 – 2

Therefore, the value of  is π2/4 – 2.

### Question 29.

Solution:

We have,

I =

By using integration by parts, we get,

I = -x2cosx – ∫(2x∫sinxdx)dx

I = -x2cosx + ∫(2xcosx)dx

I = -x2cosx + 2[xsinx – ∫(∫cosxdx)dx]

I = -x2cosx + 2[xsinx – ∫sinxdx]

I = -x2cosx + 2[xsinx + cosx]

I = -x2cosx + 2xsinx + 2cosx

So we get,

I =

I = -(π/4)2cosπ/4 + 2π/4sinπ/4 + 2cosπ/4 + 0 – 0 – 2

I = –π2/16(1/ √2) + π/2(1/√2) + 2(1/√2) + 0 – 0 – 2

I = –π2/16√2 + π/2√2 + √2 –  2

Therefore, the value of is -π2/16√2 + π/2√2 + √2 –  2.

### Question 30.

Solution:

We have,

I =

By using integration by parts, we get,

I = 1/2x2sin2x – ∫(2x∫cos2xdx)dx

I = 1/2x2sin2x – ∫(xsin2x)dx

I = 1/2x2sin2x – [-1/2xcos2x – ∫(∫sin2xdx)dx]

I = 1/2x2sin2x – [-1/2xcos2x + ∫1/2 cos2xdx]

I = 1/2x2sin2x – [-1/2xcos2x + 1/4sin2xdx]

I = 1/2x2sin2x + 1/2xcos2x – 1/4sin2xdx

So we get,

I =

I = [1/2(π2/4)sinπ + 1/2(π/2)cosπ – 0 – 0 – 0 + 0]

I = -π/4

Therefore, the value of is -π/4.

### Question 31.

Solution:

We have,

I =

I =

I =

I =

By using integration by parts, we get,

I = 1/2[x3/3] + x2sin2x/2 – [x ∫sin2x – ∫(∫sin2xdx)dx]

I = 1/2[x3/3] + x2sin2x/2 + xcosx/2 – sin2x/4

So we get,

I =

I = [1/6[π3/8] + 0 + 0 – π/8]

I = π3/48 – π/8

Therefore, the value of  is π3/48 – 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 of is 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 of is 3/4log3 – log2.

### Question 34.

Solution:

We have,

I =

I =

I =

By using integration by parts, we get,

I =

I = exlogx

So we get,

I =

I = eeloge – e1log1

I = ee (1) – 0

I = ee

Therefore, the value of is ee.

### 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 of is 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 = e2/2 – e

Therefore, the value of is e2/2 – 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 of is 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 of  is 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 of  is .

### 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 of is 2/√7[tan-1(5/√7) – tan-1(1/√7)].

### Question 41.

Solution:

We have,

I =

Let x = sin2 t, so we have,

=> dx = 2 sin t cos t dt

Now, the lower limit is, x = 0

=> sin2 t = 0

=> sin t = 0

=> t = 0

Also, the upper limit is, x = 1

=> sin2 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 of is 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 of  is π/3.

Question 43.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I = [sin-1(2/2) – sin-1(-2/2)]

I = sin-11 – sin-1(-1)

I = π/2 – (-π/2)

I = π/2 + π/2

I = π

Therefore, the value of  is 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-12/2 – 1/2tan-10/2

I = 1/2tan-11 – 1/2tan-10

I = 1/2(π/4) – 0

I = π/8

Therefore, the value of is π/8.

### Question 45.

Solution:

We have,

I =

Let 2x + 1 = t2, so we have,

=> 2 dx = 2t dt

=> dx = t dt

Now, the lower limit is, x = 1

=> t2 = 2x + 1

=> t2 = 2(1) + 1

=> t2 = 3

=> t = √3

Also, the upper limit is, x = 4

=> t= 2x + 1

=> t2 = 2(4) + 1

=> t2 = 9

=> t = 3

So, the equation becomes,

I =

I =

I =

I =

I =

I = 1/4 [35/5 – 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 of is (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 of  is 1/42.

### Question 47.

Solution:

We have,

I =

I =

I =

I =

By using integration by parts, we get,

I =

I =

I = ex/x

So we get,

I =

I = e2/2 – e1/1

I = e2/2 – e

Therefore, the value of is e2/2 – e.

### Question 48.

Solution:

We have,

I =

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

I =

I = xe2x/2 – (1/2)(e2x/2) + 2/π[1 – 0]

I = xe2x/2 – e2x/4 + 2/π

So we get,

I =

I = [e2/2 + e2/4 – 0 + 1/4] + 2/π

I = e2/4 + 1/4 + 2/π

Therefore, the value of  is e2/4 + 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 = [e1(1 – 1) – e0(0 – 1)] + 2√2/π

I = [0 – (-1)] + 2√2/π

I = 1 + 2√2/π

Therefore, the value of  is 1 + 2√2/π.

### Question 50.

Solution:

We have,

I =

I =

I =

I =

I = -eπ cotπ/2 + eπ/2 cotπ/4

I = 0 + eπ/2(1)

I = eπ/2

Therefore, the value of is eπ/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(2eπ) – 0]

I = 1/√2[0 – 0]

I = 0

Therefore, the value of is 0.

### Question 52.

Solution:

We have,

I =

By using integration by parts, we get,

I = excos(x/2 + π/4) + 1/2∫exsin(x/2 + π/4)

I = ecos(x/2 + π/4) + 1/2[ exsin(x/2 + π/4) – 1/2 ∫excos(x/2 + π/4)dx]

I = excos(x/2 + π/4) + 1/2exsin(x/2 + π/4) – 1/4I

5I/4 = -3/ 2√2(e2π + 1)

I = -3√2/5(e + 1)

Therefore, the value of is -3√2/5(e + 1).

### Question 53.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I = 2/3[23/2 – 1] + 2/3[1 – 0]

I =

I = 25/2/3

Therefore, the value of is 25/2/3.

### 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– log33

I = log32 – log27

I = log32/27

Therefore, the value of is 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 of  is 2/3.

### Question 56.

Solution:

We have,

I =

I =

I =

I =

I = -sinπ + sin0

I = 0

Therefore, the value of  is 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 = e4/4 – e2/2

Therefore, the value of  is e4/4 – e2/2.

### 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 of  is π.

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

Solution:

We have,

=>

=>

=>

=>

=> tan-12k/4 – tan-10 = π/16

=> tan-12k/4 – 0 = π/16

=> tan-12k/4 = π/16

=> tan-12k = 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,

=>

=>

=>

=>

=> a3 – 0 = 8

=> a3 = 8

=> a = 2

Therefore, the value of a is 2.

### Question 61.

Solution:

We have,

I =

I =

I =

I =

I = -[√2cos3π/2 – √2cosπ]

I = -(-√2 – 0)

I = √2

Therefore, the value of  is √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 of  is 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 of  is π/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 of  is 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 of  is 4/√3.

### Question 66.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

### 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 of  is log2/4 + 1/4.

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