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

Evaluate the following integrals:

Question 15. \int\limits_{- \pi/2}^{\pi/2} \left\{ \sin \left| x \right| + \cos \left| x \right| \right\} dx

Solution:

We have,

I = \int\limits_{- \pi/2}^{\pi/2} \left\{ \sin \left| x \right| + \cos \left| x \right| \right\} dx

Since f(- x) = sin|- x| + cos|- x|

= sin |x| + cos |x|

= f(x)

So, f(x) is an even function.

Therefore, we get

I = \int\limits_{- \pi/2}^{\pi/2} \left\{ \sin \left| x \right| + \cos \left| x \right| \right\} dx

I = 2 \int_0^\frac{\pi}{2} \left( \sin x + \cos x \right) dx

I = 2 \left[ - \cos x + \sin x \right]_0^\frac{\pi}{2}

I = 2 (0 + 1 + 1 – 0)

I = 4

Question 16. \int\limits_0^4 \left| x - 1 \right| dx

Solution:

We have,

I = \int\limits_0^4 \left| x - 1 \right| dx

We know,

\left| x - 1 \right| = \begin{cases} - \left( x - 1 \right) &,& 0 \leq x \leq 1\\x - 1&,& 1 < x \leq 4\end{cases}

So we get,

I = \int_0^4 \left| x - 1 \right| d x

I = \int_0^1 - \left( x - 1 \right) dx + \int_1^4 \left( x - 1 \right) dx

I = \left[ - \frac{x^2}{2} + x \right]_0^1 + \left[ \frac{x^2}{2} - x \right]_1^4

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

I = 5

Question 17. \int\limits_1^4 \left\{ \left| x - 1 \right| + \left| x - 2 \right| + \left| x - 4 \right| \right\} dx

Solution:

We have,

I = \int\limits_1^4 \left\{ \left| x - 1 \right| + \left| x - 2 \right| + \left| x - 4 \right| \right\} dx

We know,

\left| x - 1 \right| = \begin{cases} - \left( x - 1 \right) &,& x \leq 1\\x - 1&,& 1 < x \leq 4\end{cases}

\left| x - 2 \right| = \begin{cases} - \left( x - 2 \right) &,& 1 \leq x \leq 2\\x - 2&,& 2 < x \leq 4\end{cases}

\left| x - 4 \right| = \begin{cases} - \left( x - 4 \right) &,& 1 \leq x \leq 4\\x - 4&,& x > 4\end{cases}

So we get,

I = \int_1^4 \left( x - 1 \right) d x - \int_1^2 \left( x - 2 \right) d x + \int_2^4 \left( x - 2 \right) d x - \int_1^4 \left( x - 4 \right) d x

I = \left[ \frac{x^2}{2} - x \right]_1^4 - \left[ \frac{x^2}{2} - 2x \right]_1^2 + \left[ \frac{x^2}{2} - 2x \right]_2^4 - \left[ \frac{x^2}{2} - 4x \right]_1^4

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. \int_{- 5}^0 \left\{ \left| x \right| + \left| x + 2 \right| + \left| x + 5 \right| \right\} dx

Solution:

We have,

I = \int_{- 5}^0 \left\{ \left| x \right| + \left| x + 2 \right| + \left| x + 5 \right| \right\} dx

We know,

\left| x \right| = \begin{cases} - x &,& - 5 \leq x \leq 0\\x&,& x > 0\end{cases}

\left| x + 2 \right| = \begin{cases} - \left( x + 2 \right) &,& - 5 \leq x \leq - 2\\x + 2&,& - 2 < x \leq 0\end{cases}

\left| x + 5 \right| = \begin{cases} - \left( x + 5 \right) &,& - 5 \leq x \leq 0\\x + 5&,& x > - 5\end{cases}

So we get,

I = - \int_{- 5}^0 x d x - \int_{- 5}^{- 2} \left( x + 2 \right) d x + \int_{- 2}^0 \left( x + 2 \right) d x + \int_{- 5}^0 \left( x + 5 \right) d x

I = - \left[ \frac{x^2}{2} \right]_{- 5}^0 - \left[ \frac{x^2}{2} + 2x \right]_{- 5}^{- 2} + \left[ \frac{x^2}{2} + 2x \right]_{- 2}^0 + \left[ \frac{x^2}{2} + 5x \right]_{- 5}^0

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

I = 63/2

Question 19. \int\limits_0^4 \left( \left| x \right| + \left| x - 2 \right| + \left| x - 4 \right| \right) dx

Solution:

We have,

I = \int\limits_0^4 \left( \left| x \right| + \left| x - 2 \right| + \left| x - 4 \right| \right) dx

We know,

\left| x \right| = \begin{cases} - x &,& - 5 \leq x \leq 0\\x&,& x > 0\end{cases}

\left| x - 2 \right| = \begin{cases} - \left( x - 2 \right) &,& 0 \leq x \leq 2\\x - 2&,& 2 < x \leq 4\end{cases}

\left| x - 4 \right| = \begin{cases} - \left( x - 4 \right) &,& 0 \leq x \leq 4\\x - 4&,& x > 4\end{cases}

So we get,

I = \int_0^4 x d x - \int_0^2 \left( x - 2 \right) d x + \int_2^4 \left( x - 2 \right) d x - \int_0^4 \left( x - 4 \right) d x

I = \left[ \frac{x^2}{2} \right]_0^4 - \left[ \frac{x^2}{2} - 2x \right]_0^2 + \left[ \frac{x^2}{2} - 2x \right]_2^4 - \left[ \frac{x^2}{2} - 4x \right]_0^4

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

I = 20

Question 20. \int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx

Solution:

We have,

I = \int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx

We know,

\left| x + 1 \right| = \begin{cases}x + 1, & \text{if }x + 1 \geq 0 \\ - \left( x + 1 \right), & \text{if }x + 1 < 0\end{cases}

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 = \int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx

I = \int_{- 1}^0 \left( 2 - x \right)dx + \int_0^1 \left( x + 2 \right)dx + \int_1^2 3xdx

I = \left.\frac{\left( 2 - x \right)^2}{2 \times \left( - 1 \right)}\right|_{- 1}^0 + \left.\frac{\left( x + 2 \right)^2}{2}\right|_0^1 + \left.3 \times \frac{x^2}{2}\right|_1^2

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. \int_{- 2}^2 x e^{\left| x \right|} dx

Solution:

We have,

I = \int_{- 2}^2 x e^{\left| x \right|} dx

Now here,

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

= - x e^|x|

= – f(x)

So, f(x) is an odd function.

Therefore we get,

I = \int_{- 2}^2 x e^{\left| x \right|} dx

I = 0

Question 22. \int_{- \frac{\pi}{4}}^\frac{\pi}{2} \sin x\left| \sin x \right|dx

Solution:

We have,

I = \int_{- \frac{\pi}{4}}^\frac{\pi}{2} \sin x\left| \sin x \right|dx

I = \int_{- \frac{\pi}{4}}^0 \sin x\left| \sin x \right|dx + \int_0^\frac{\pi}{2} \sin x\left| \sin x \right|dx

As we know, \left| \sin x \right| = \begin{cases}\sin x, & 0 \leq x \leq \frac{\pi}{2} \\ - \sin x, & - \frac{\pi}{4} \leq x \leq 0\end{cases}

I = \int_{- \frac{\pi}{4}}^0 \sin x\left( - \sin x \right)dx + \int_0^\frac{\pi}{2} \sin x\sin xdx

I = - \int_{- \frac{\pi}{4}}^0 \sin^2 xdx + \int_0^\frac{\pi}{2} \sin^2 xdx

I = - \int_{- \frac{\pi}{4}}^0 \frac{1 - \cos2x}{2}dx + \int_0^\frac{\pi}{2} \frac{1 - \cos2x}{2}dx

I = - \frac{1}{2} \int_{- \frac{\pi}{4}}^0 dx + \frac{1}{2} \int_{- \frac{\pi}{4}}^0 \cos2xdx + \frac{1}{2} \int_0^\frac{\pi}{2} dx - \frac{1}{2} \int_0^\frac{\pi}{2} \cos2xdx

I = \left.- \frac{1}{2} \times x\right|_{- \frac{\pi}{4}}^0 +\left. \frac{1}{2} \times \frac{\sin2x}{2}\right|_{- \frac{\pi}{4}}^0 + \left.\frac{1}{2} \times x\right|_0^\frac{\pi}{2} - \left.\frac{1}{2} \times \frac{\sin2x}{2}\right|_0^\frac{\pi}{2}

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. \int_0^\pi \cos x\left| \cos x \right|dx

Solution:

We have,

I = \int_0^\pi \cos x\left| \cos x \right|dx

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 = \int_0^\pi \cos x\left| \cos x \right|dx

I = 0

Question 24. \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx

Solution:

We have,

I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx

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 = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx

I = 2 \int_0^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx

As we know, \left| x \right| = \begin{cases}x, & \text{if }x \geq 0 \\ - x, & \text{if }x < 0\end{cases}

I = 2 \int_0^\frac{\pi}{2} \left( 2\sin x + \cos x \right)dx

I = 4 \int_0^\frac{\pi}{2} \sin x\ dx + 2 \int_0^\frac{\pi}{2} \cos x\ dx

I = \left.4 \times \left( - \cos x \right)\right|_0^\frac{\pi}{2} + \left.2 \times \sin x\right|_0^\frac{\pi}{2}

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. \int_{- \frac{\pi}{2}}^\pi \sin^{- 1} \left( \sin x \right)dx

Solution:

We have,

I = \int_{- \frac{\pi}{2}}^\pi \sin^{- 1} \left( \sin x \right)dx

I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \sin^{- 1} \left( \sin x \right)dx + \int_\frac{\pi}{2}^\pi \sin^{- 1} \left( \sin x \right)dx

I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} xdx + \int_\frac{\pi}{2}^\pi \left( \pi - x \right)dx

As π/2 ≤ x ≤ π, we get

=> –π ≤ –x ≤ –π/2

=> 0 ≤ π – x ≤ π/2

So, we get

I = \left.\frac{x^2}{2}\right|_{- \frac{\pi}{2}}^\frac{\pi}{2} + \left.\frac{\left( \pi - x \right)^2}{2 \times \left( - 1 \right)}\right|_\frac{\pi}{2}^\pi

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

I = 0 + π²/8

I = π²/8

Question 26. \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{- \frac{\pi}{2}}{\sqrt{\cos x \sin^2 x}}dx

Solution:

We have,

I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{- \frac{\pi}{2}}{\sqrt{\cos x \sin^2 x}}dx

I = - \frac{\pi}{2} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{1}{\sqrt{\cos x \sin^2 x}}dx

I = - \frac{\pi}{2} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{1}{\sqrt{\cos x}\left| \sin x \right|}dx

As we know, f\left( - x \right) = \sqrt{\cos\left( - x \right)}\left| \sin\left( - x \right) \right|

= √cos x|-sin x|

= √cos x|sin x|

= f(x)

So, f(x) is an odd function.

Therefore we get,

I = - \frac{\pi}{2} \times 2 \int_0^\frac{\pi}{2} \frac{1}{\sqrt{\cos x}\left| \sin x \right|}dx 

I = -\pi\int_0^\frac{\pi}{2} \frac{1}{\sqrt{\cos x}\left| \sin x \right|}dx

As we know, \left| \sin x \right| = \sin x, 0 \leq x \leq \frac{\pi}{2}   ,

I = - \pi \int_0^\frac{\pi}{2} \frac{1}{\sqrt{\cos x}\sin x}dx

I = - \pi \int_0^\frac{\pi}{2} \frac{\sin x}{\sqrt{\cos x}\left( 1 - \cos^2 x \right)}dx

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 = 2\pi \int_1^0 \frac{zdz}{z\left( 1 - z^4 \right)}

I = 2\pi \int_1^0 \frac{dz}{1 - z^4}

I = 2\pi \int_1^0 \frac{dz}{\left( 1 - z \right)\left( 1 + z \right)\left( 1 + z^2 \right)}

I = 2\pi \int_1^0 \frac{\frac{1}{4}}{1 - z}dz + 2\pi \int_1^0 \frac{\frac{1}{4}}{1 + z}dz + 2\pi \int_1^0 \frac{\frac{1}{2}}{1 + z^2}dz

I = \left.\frac{2\pi}{4} \times \frac{\log\left( 1 - z \right)}{- 1}\right|_1^0 + \left.\frac{2\pi}{4} \times \log\left( 1 + z \right)\right|_1^0 + \left.\frac{2\pi}{2} \times \tan^{- 1} z\right|_1^0

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. \int_0^2 2x\left[ x \right]dx

Solution:

We have,

I = \int_0^2 2x\left[ x \right]dx

I = \int_0^1 2x\left[ x \right]dx + \int_1^2 2x\left[ x \right]dx

As we know, \left[ x \right] = \begin{cases}0, & 0 \leq x < 1 \\ 1, & 1 \leq x < 2\end{cases}

I = \int_0^1 2x \times 0dx + \int_1^2 2x \times 1dx

I = 0 + 2 \int_1^2 xdx

I = \left.2 \times \frac{x^2}{2}\right|_1^2

I = 4 – 1

I = 3

Question 28. \int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx

Solution:

We have,

I = \int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx

I = \int_0^\pi \cos^{- 1} \left( \cos x \right)dx +\int_\pi^{2\pi} \cos^{- 1} \left( \cos x \right)dx

As we know, π ≤ x ≤ 2π

=> –2π ≤ –x ≤ –π

=> 0 ≤ 2π – x ≤ π

Therefore, we get

I = \int_0^\pi xdx + \int_\pi^{2\pi} \left( 2\pi - x \right)dx

I = \left.\frac{x^2}{2}\right|_0^\pi + \left.\frac{\left( 2\pi - x \right)^2}{2 \times \left( - 1 \right)}\right|_\pi^{2\pi}

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