Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.3 | Set 2

Evaluate the following integrals:

Question 15. 

Solution:

We have,

I = 

Since f(- x) = sin|- x| + cos|- x|

= sin |x| + cos |x|

= f(x)

So, f(x) is an even function.

Therefore, we get

I = 

I = 

I = 

I = 2 (0 + 1 + 1 – 0)

I = 4

Question 16. 

Solution:

We have,

I = 

We know,

So we get,

I = 

I = 

I = 

I = -1/2 + 1 - 0 + 8 - 4 - 1/2 + 1

I = 5

Question 17. 

Solution:

We have,

I = 

We know,

So we get,

I = 

I = 

I = 8 - 4 - 1/2 + 1 - (2 - 4 - 1/2 + 2) + 8 - 8 - 2 + 4 - (8 - 16 - 1/2 + 4)

I = 23/2

Question 18. 

Solution:

We have,

I = 

We know,

So we get,

I = 

I = 

I = 25/2 - (2 - 4 - 25/2 + 10) - 2 + 4 + (-25/2 + 25)

I = 63/2

Question 19. 

Solution:

We have,

I = 

We know,

So we get,

I = 

I = 

I = 8 – (2 – 4) + 8 – 8 – 2 + 4 – (8 – 16)

I = 20

Question 20. 

Solution:

We have,

I = 

We know,

When –1 < x < 0,

|x + 1| + |x| + |x - 1| = x + 1 + (- x) + [-(x - 1)]

= 2 – x

And when 0 < x < 1,

|x + 1| + |x| + |x - 1| = x + 1 + x + [-(x - 1)]

= x + 2

And when 1 ≤ x ≤ 2,

|x + 1| + |x| + |x - 1| = x + 1 + x + x - 1

= 3x

So we get,

I = 

I = 

I = 

I = - 1/2(4 - 9) + 1/2( 9 - 4) + 3/2(4 - 1)

I = 5/2 + 5/2 + 9/2

I = 19/2

Question 21. 

Solution:

We have,

I = 

Now here,

f(- x) = (- x)e^{|- x|}

= - x e^|x|

= – f(x)

So, f(x) is an odd function.

Therefore we get,

I = 

I = 0

Question 22. 

Solution:

We have,

I = 

I = 

As we know, 

I = 

I = 

I = 

I = 

I = 

I = -1/2(0 + π/4) + 1/4(0 + sin π/2) + 1/2 ( π/2 - 0) - 1/4(sin π - 0)

I = - π/8 + 1/4 (0 + 1) + π/8 - 1/4 (0 - 0)

I = π/8 + 1/4

Question 23. 

Solution:

We have,

I = 

Now here,

f(π - x) = cos(π - x)|cos(π - x)|

= -cos x|-cos x| 

= - cos x|cos x| 

= – f(x)

So, f(x) is an odd function.

Therefore we get,

I = 

I = 0

Question 24. 

Solution:

We have,

I = 

Now here,

f(- x) = 2sin|- x| + cos|- x|

= 2sin|x| + cos|x| 

= f(x)

So, f(x) is an odd function.

Therefore we get,

I = 

I = 

As we know, 

I = 

I = 

I = 

I = - 4(cos π/2 - cos 0) + 2(sin π/2 - sin 0)

I = –4 ( 0 – 1) + 2 (1 – 0)

I = 4 + 2

I = 6

Question 25. 

Solution:

We have,

I = 

I = 

I = 

As π/2 ≤ x ≤ π, we get

=> –π ≤ –x ≤ –π/2

=> 0 ≤ π – x ≤ π/2

So, we get

I = 

I = 1/2 (π²/4 - π²/4) - 1/2( 0 - π²/4)

I = 0 + π²/8

I = π²/8

Question 26. 

Solution:

We have,

I = 

I = 

I = 

As we know, 

= √cos x|-sin x|

= √cos x|sin x|

= f(x)

So, f(x) is an odd function.

Therefore we get,

I = 

I = 

As we know, ,

I = 

I = 

Let cos x = z². So, we have

=> – sin x dx = 2z dz

Now, the lower limit is, x = 0

=> z² = cos x

=> z² = cos 0

=> z² = 1

=> z = 1

Also, the upper limit is, x = π/2

=> z² = cos x

=> z² = cos π/2

=> z² = 0

=> z = 0

So, the equation becomes,

I = 

I = 

I = 

I = 

I = 

I = - π/2(log1 - log0) + π/2(log1 - log2) + π(tan^-1 0 - tan^-1 1)

I = - π/2[0 - ∞] + π/2(0 - log2) + π(0 - π/4)

I = -∞ - π/2 log2 - π²/4

I = –∞

Question 27. 

Solution:

We have,

I = 

I = 

As we know, 

I = 

I = 

I = 

I = 4 – 1

I = 3

Question 28. 

Solution:

We have,

I = 

I = 

As we know, π ≤ x ≤ 2π

=> –2π ≤ –x ≤ –π

=> 0 ≤ 2π – x ≤ π

Therefore, we get

I = 

I = 

I = 1/2( π - 0) - 1/2(0 - π)

I = π²/2 + π²/2

I = π²

Summary

This exercise typically focuses on more advanced applications of definite integrals and special techniques. Key concepts often covered include:

• Evaluation of improper integrals
• Integrals involving special functions (e.g., exponential, logarithmic)
• Application of integration techniques to solve real-world problems
• Evaluating integrals with trigonometric substitutions
• Dealing with piecewise functions in definite integrals