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

### 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) – 1/4I5I/4 = -3/ 2√2(e

^{2π }+ 1)I = -3√2/5(e

^{2π}+ 1)

Therefore, the value ofis -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.