Question

The value of $\underset{-\pi /2}{\overset{\pi /2}{\int }}\left(x^3+x\cos x+\tan 5x+1\right)dx$ is

• A. 0
• B. 2
• C. $\pi$
• D. 1

Answer 1

Answer: (C)

Let $I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\left(x^3+x\cos x+{\tan }^{5}x+1\right)dx$
$\Rightarrow I=\underset{-\pi /2}{\overset{\pi /2}{\int }}{x}^{3}dx+\underset{-\pi /2}{\overset{\pi /2}{\int }}x\cos xdx$
$+\underset{-\pi /2}{\overset{\pi /2}{\int }}{\tan }^{5}xdx+\underset{-\pi /2}{\overset{\pi /2}{\int }}1dx$
We know that,
$\underset{-a}{\overset{a}{\int }}f(x)dx=\{\begin{array}{ll}2\underset{0}{\overset{a}{\int }}f(x)dx,& \text{ if }f(x)\text{ is even }\\ 0,& \text{ if }f(x)\text{ is odd }\end{array}$
$\therefore I=0+0+0+2\underset{0}{\overset{\pi /2}{\int }}1dx.$
$[\because x^3,x\cos x\text{ and }{\tan }^5(x)\text{ are odd functions] }$
$\therefore I=2[x{]}_0^{\pi /2}=\frac{2\pi }{2}=\pi$