# Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.1 | Set 1

### Question 1.

Solution:

We have,

I =

I =

I =

I =

I = 2[âˆš9 – âˆš4 ]

I = 2 (3 âˆ’ 2)

I = 2 (1)

I = 2

Therefore, the value of is 2.

### Question 2.

Solution:

We have,

I =

I =

I = log (3 + 7) âˆ’ log (âˆ’2 + 7)

I = log 10 âˆ’ 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 = Ï€/6

So, the equation becomes,

I =

I =

I =

I =

I =

I =  Ï€/6 – 0

I = Ï€/6â€‹

Therefore, the value of is Ï€/6.

### Question 4.

Solution:

We have,

I =

I =

I =

I =

I = Ï€/4

Therefore, the value of is Ï€/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)2 + 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 = logâˆš2

Therefore, the value of is logâˆš2.

### Question 6.

Solution:

We have,

I =

I =

I =

I =

I =

I = 1/ab[tan-1âˆž – tan-10]

I = 1/ab[Ï€/2 – 0]

I = 1/ab[Ï€/2]

I = Ï€/2ab

Therefore, the value of is Ï€/2ab.

### Question 7.

Solution:

We have,

I =

I =

I = [tan-11 – tan-1(-1)]

I = [Ï€/4 – (-Ï€/4)]

I = [Ï€/4 + Ï€/4]

I = 2Ï€/4

I = Ï€/2

Therefore, the value of is Ï€/2.

### Question 8.

Solution:

We have,

I =

I =

I = -eâˆž – (-e0)

I = âˆ’ 0 + 1

I = 1

Therefore, the value of is 1.

### Question 9.

Solution:

We have,

I =

I =

I =

I =

I =

I = [1 âˆ’ 0] âˆ’ [log(1 + 1) âˆ’ log(0 + 1)]

I = 1 âˆ’ [log2 âˆ’ log1]

I = 1 – log2/1

I = 1 âˆ’ log 2

I = log e âˆ’ log 2

I = loge/2

Therefore, the value of  is loge/2.

### Question 10.

Solution:

We have,

I =

I =

I =

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

I = [âˆ’0 + 1] + 1

I = 1 + 1

I = 2

Therefore, the value of is 2.

### Question 11.

Solution:

We have,

I =

I =

I = log(sinÏ€/2) – log(sinÏ€/4)

I = log1 – log1/âˆš2

I =

I = logâˆš2

Therefore, the value of is logâˆš2.

### Question 12.

Solution:

We have,

I =

I =

I = log(secÏ€/4 + tanÏ€/4 – log(sec0 + tan0)

I = log(âˆš2 + 1) – log(1 + 0)

I =

I = log(âˆš2 + 1)

Therefore, the value of is log(âˆš2 + 1).

### Question 13.

Solution:

We have,

I =

I =

I = [log|cosecÏ€/4 – cotÏ€/4|] – [log|cosecÏ€/6 – cotÏ€/6|]

I = [log|âˆš2 – 1|] – [log|2 – âˆš3|]

I =

Therefore, the value of is .

### Question 14.

Solution:

We have,

I =

Let x = cos 2t, so we have,

=> dx = â€“2 sin 2t dt

Now, the lower limit is,

=> x = 0

=> cos 2t = 0

=> 2t = Ï€/2

=> t = Ï€/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,

=> â€“ sin t dt = dz

=> sin t dt = â€“ dz

Now, the lower limit is,

=> t = 0

=> z = cos t

=> z = cos 0

=> z = 1

Also, the upper limit is,

=> t = Ï€/4

=> z = cos t

=> z = cos Ï€/4

=> z = 1/âˆš2

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I = -4[(log1/âˆš2 – 1/2(2)) – (log1 – 1/2)]

I = -4[(log1/âˆš2 – 1/4) – (0 – 1/2)]

I = -4[log1/âˆš2 – 1/4 – 0 + 1/2]

I = -4[-logâˆš2 + 1/4]

I = 4logâˆš2 – 1

I = 4 Ã— 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 Ï€ â€“ tan0] â€“ [sec Ï€ â€“ sec 0]

I = [0 â€“ 0] â€“ [â€“1 â€“ 1]

I = 0 â€“ (â€“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 Ï€/4 â€“ tan(â€“Ï€/4)] â€“ [sec Ï€/4 â€“ sec (â€“Ï€/4)]

I = [1 â€“ (â€“1)] â€“ [sec Ï€/4 â€“ sec (Ï€/4)]

I = 2 â€“ 0

I = 2

Therefore, the value of is 2.

### Question 17.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I = 1/2[Ï€/2 – 0] + 1/4[sinÏ€ – sin0]

I = 1/2[Ï€/2] + 1/4[0 – 0]

I = Ï€/4

Therefore, the value of  is Ï€/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[sinÏ€/2 – sin0] + 1/2[sinÏ€/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[sinÏ€/2 – sin0] – 1/6[sin3Ï€/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[-cotÏ€/2 + cot2Ï€/3]

I = 2[-1/âˆš3 – 0]

I = -2/âˆš3

Therefore, the value of  is -2/âˆš3.

### Question 22.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I =

I = 1/4[Ï€/2 + Ï€/4 + 0 + 0 – 0 – 0 – 0 – 0]

I = 1/4[3Ï€/4]

I = 3Ï€/16

Therefore, the value of is 3Ï€/16.

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