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• RD Sharma Class 12 Solutions for Maths

# Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.1 | Set 3

### 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

=> t2 = 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 = ex cos(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(e+ 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 = log25 – 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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