# 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 = âˆš3

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 – (âˆš3)^{5}/5 + âˆš3]I = 1/4[243/5 – 3 – 9âˆš3/5 + âˆš3]

I = 1/4((243 – 15 – 9âˆš3 + 5âˆš3)/5)

I = 1/4[(228 – 4âˆš3)/5]

I = 1/4[4(57 – âˆš3)/5]

I = (57 – âˆš3)/5

Therefore, the value ofis (57 – âˆš3)/5.

### Question 46.

**Solution:**

We have,

I =

By using binomial theorem in the expansion of (1 â€“ 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/Ï€[1 – 0]I = xe

^{2x}/2 – e^{2x}/4 + 2/Ï€So we get,

I =

I = [e

^{2}/2 + e^{2}/4 – 0 + 1/4] + 2/Ï€I = e

^{2}/4 + 1/4 + 2/Ï€

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

### 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)] + 2âˆš2/Ï€I = [0 – (-1)] + 2âˆš2/Ï€

I = 1 + 2âˆš2/Ï€

Therefore, the value ofis 1 + 2âˆš2/Ï€.

### Question 50.

**Solution:**

We have,

I =

I =

I =

I =

I = -e

^{Ï€}cotÏ€/2 + e^{Ï€/2 }cotÏ€/4I = 0 + e

^{Ï€/2}(1)I = e

^{Ï€/2}

Therefore, the value ofis e^{Ï€/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/âˆš2[sinÏ€(2e

^{Ï€}) – 0]I = 1/âˆš2[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 + Ï€/4) + 1/2âˆ«e^{x}sin(x/2 + Ï€/4)I = e

^{x }cos(x/2 + Ï€/4) + 1/2[ e^{x}sin(x/2 + Ï€/4) – 1/2 âˆ«e^{x}cos(x/2 + Ï€/4)dx]I = e

^{x}cos(x/2 + Ï€/4) + 1/2e^{x}sin(x/2 + Ï€/4) – 1/4I5I/4 = -3/ 2âˆš2(e

^{2Ï€ }+ 1)I = -3âˆš2/5(e

^{2Ï€}+ 1)

Therefore, the value ofis -3âˆš2/5(e^{2Ï€}+ 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,

=> â€“ 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 = Ï€/2

=> t = cos x

=> t = cos Ï€/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 = -sinÏ€ + 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 = Ï€/2 – (-Ï€/2)

I = Ï€/2 + Ï€/2

I = Ï€

Therefore, the value ofis Ï€.

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

**Solution:**

We have,

=>

=>

=>

=>

=> tan

^{-1}2k/4 – tan^{-1}0 = Ï€/16=> tan

^{-1}2k/4 – 0 = Ï€/16=> tan

^{-1}2k/4 = Ï€/16=> tan

^{-1}2k = Ï€/4=> 2k = tanÏ€

/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}â€“ 0 = 8=> a

^{3}= 8=> a = 2

Therefore, the value of a is 2.

### Question 61.

**Solution:**

We have,

I =

I =

I =

I =

I = -[âˆš2cos3

Ï€/2 – âˆš2cosÏ€]I = -(-âˆš2 – 0)

I = âˆš2

Therefore, the value ofis âˆš2.

### Question 62.

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I =

I = [-4cosÏ€/2 + 4cos0] + [4sinÏ€/2 – 4sin0]

I = 0 + 4 + 4 â€“ 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 = (

Ï€/8 – 1/4) – (3/4(Ï€/8 – 1/4) – 1/16)I =

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

I = Ï€/8 – 1/4 – 3Ï€/32 + 1/4

I = Ï€/8 – 3Ï€/32

I = (4Ï€ – 3Ï€)/32

I = Ï€/32

Therefore, the value ofisÏ€/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 = [tanÏ€/3 – tanÏ€/6] + [-cotÏ€/3 + cotÏ€/6]

I = [âˆš3 – 1/âˆš3] + [- 1/âˆš3 – âˆš3]

I = 2[âˆš3 – 1/âˆš3]

I = 4/âˆš3

Therefore, the value ofis 4/âˆš3.

### 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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