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# Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.1 | Set 2

• Last Updated : 26 May, 2021

### Question 23.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I = [(a2 + b2)/2][Ï€/2]

I = Ï€(a2 + b2)/4

Therefore, the value of is Ï€(a2 + b2)/4.

### Question 24.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I = 2[sinÏ€/4 – cosÏ€/4 – 0 + 1]

I = 2[1/âˆš2 – 1/âˆš2 – 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 = 2âˆš2[sinÏ€/4 – sin0]

I = 2âˆš2[1/âˆš2- sin0]

I = 2âˆš2[1/âˆš2]

I = 2

Therefore, the value of is 2.

### Question 26.

Solution:

We have,

I =

By using integration by parts, we get,

I = x âˆ«sinxdx – âˆ«(âˆ«sin x (1)dx)dx

I = -xcosx – âˆ«(âˆ«sin xdx)dx

I = -xcosx + âˆ«cosxdx

I = -xcosx + sinx

So we get,

I =

I = [-Ï€/2cosÏ€/2 + sinÏ€/2 + 0 – 0]

I = 0 + 1 + 0 â€“ 0

I = 1

Therefore, the value of is 1.

### Question 27.

Solution:

We have,

I =

By using integration by parts, we get,

I = xâˆ«cosxdx – âˆ«(âˆ«cos x (1)dx)dx

I = xsinx – âˆ«(âˆ«cosxdx)dx

I = xsinx – âˆ«sinxdx

I = x sin x + cos x

So we get,

I =

I = [Ï€/2sinÏ€/2 + cosÏ€/2 – 0 – cos0]

I = Ï€/2 + 0 – 0 – 1

I = Ï€/2 – 1

Therefore, the value of is Ï€/2 – 1.

### Question 28.

Solution:

We have,

I =

By using integration by parts, we get,

I = x2sinx – âˆ«(2xâˆ«(cosx)dx)dx

I = x2sinx – âˆ«(2xsinx)dx

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

I = x2sinx – 2[-xcosx + âˆ«sinxdx]

I = x2sinx – 2[-xcosx + sinx]

I = x2sinx + 2xcosx – 2sinx

So we get,

I =

I = [(Ï€/2)2sinÏ€/2 + 2(Ï€/2)cosÏ€/2 – 2sinÏ€/2 – 0 – 0 + sin0]

I = [Ï€2/4 + 0 – 2 – 0 – 0 + 0]

I = Ï€2/4 – 2

Therefore, the value of  is Ï€2/4 – 2.

### Question 29.

Solution:

We have,

I =

By using integration by parts, we get,

I = -x2cosx – âˆ«(2xâˆ«sinxdx)dx

I = -x2cosx + âˆ«(2xcosx)dx

I = -x2cosx + 2[xsinx – âˆ«(âˆ«cosxdx)dx]

I = -x2cosx + 2[xsinx – âˆ«sinxdx]

I = -x2cosx + 2[xsinx + cosx]

I = -x2cosx + 2xsinx + 2cosx

So we get,

I =

I = -(Ï€/4)2cosÏ€/4 + 2Ï€/4sinÏ€/4 + 2cosÏ€/4 + 0 – 0 – 2

I = –Ï€2/16(1/ âˆš2) + Ï€/2(1/âˆš2) + 2(1/âˆš2) + 0 – 0 – 2

I = –Ï€2/16âˆš2 + Ï€/2âˆš2 + âˆš2 –  2

Therefore, the value of is -Ï€2/16âˆš2 + Ï€/2âˆš2 + âˆš2 –  2.

### Question 30.

Solution:

We have,

I =

By using integration by parts, we get,

I = 1/2x2sin2x – âˆ«(2xâˆ«cos2xdx)dx

I = 1/2x2sin2x – âˆ«(xsin2x)dx

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

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

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

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

So we get,

I =

I = [1/2(Ï€2/4)sinÏ€ + 1/2(Ï€/2)cosÏ€ – 0 – 0 – 0 + 0]

I = -Ï€/4

Therefore, the value of is -Ï€/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 âˆ«sin2x – âˆ«(âˆ«sin2xdx)dx]

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

So we get,

I =

I = [1/6[Ï€3/8] + 0 + 0 – Ï€/8]

I = Ï€3/48 – Ï€/8

Therefore, the value of  is Ï€3/48 – Ï€/8.

### Question 32.

Solution:

We have,

I =

By using integration by parts, we get,

I =

I = xlogx – âˆ«1dx

I = xlogx – x

So we get,

I =

I = 2log2 – 2 – log1 + 1

I = 2 log 2 â€“ 1

Therefore, the value of is 2 log 2 â€“ 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) â€“ 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/âˆš5tan-1(âˆš5) – 1/5log1 – 3/âˆš5tan-1(0)]

I = [1/5 log6 + 3âˆš5 tan-1(âˆš5) – 0 – 0]

I = 1/5 log6 + 3âˆš5 tan-1(âˆš5)

Therefore, the value of  is 1/5 log6 + 3âˆš5 tan-1(âˆš5).

### Question 39.

Solution:

We have,

I =

I =

I =

I =

I =

I =

Let x â€“ 1/2 = t, so we have,

=> dx = dt

Now, the lower limit is, x = 0

=> t = x â€“ 1/2

=> t = 0 â€“ 1/2

=> t = 1/2

Also, the upper limit is, x = 2

=> t = x â€“ 1/2

=> t = 2 â€“ 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 = 4/2âˆš7[tan-1(5/âˆš7) – tan-1(1/âˆš7)]

I = 2/âˆš7[tan-1(5/âˆš7) – tan-1(1/âˆš7)]

Therefore, the value of is 2/âˆš7[tan-1(5/âˆš7) – tan-1(1/âˆš7)].

### 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 = Ï€/2

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I = 1/4[Ï€/2 – 0] – 1/16[sin2Ï€ – 0]

I = 1/4[Ï€/2] – 1/16[0 – 0 ]

I = Ï€/8

Therefore, the value of is Ï€/8.

### Question 42.

Solution:

We have,

I =

I =

I =

I =

I =

I = [sin-1(1/2) – sin-1(-1/2)]

I = Ï€/6 -(-Ï€/6)

I = Ï€/6 + Ï€/6

I = Ï€/3

Therefore, the value of  is Ï€/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 = Ï€/2 – (-Ï€/2)

I = Ï€/2 + Ï€/2

I = Ï€

Therefore, the value of  is Ï€.

### Question 44.

Solution:

We have,

I =

I =

I =

Let x + 1 = t, so we have,

=> dx = dt

Now, the lower limit is, x = â€“1

=> t = x + 1

=> t = â€“ 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(Ï€/4) – 0

I = Ï€/8

Therefore, the value of is Ï€/8.

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