# Choose the correct answer in the value of $\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2}(x^3+x\cos x+\tan^5x+1)dx$ is

$(a)\;0\qquad(b)\;2\qquad(c)\;\pi\qquad(d)\;1$

Toolbox:
• (i)$\int \limits_a^bf(x)dx=F(b)-F(a)$
• (ii)If $f(-x)=-f(x)$ it is an odd function then $\int \limits_0^a f(x)dx=0$
• (iii)If $f(-x)=-f(x)$ it is an odd function then $\int \limits_0^a f(x)dx=2 \int \limits_0^a f(x)dx$
Given $I=\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2}(x^3+x\cos x+\tan^5x+1)dx$

On seperating the term

$\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2}(x^3dx+\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2}x\cos x+\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2}\tan^5x+\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2}1dx$

Consider $x^3;(-x)^3=-x^3$

Hence it is an odd function

Therefore $\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2} x^3dx=0$

Consider $xcosx\;-x\cos(-x)=-x\cos x$

Hence it is an odd function

Therefore $\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2} x \cos x=0$

Consider $\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2} \tan ^5 x.\tan ^5(-x)=-\tan ^5x$

It is an odd function

Hence $\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2} \tan ^5x=0$

Therefore $I=0+0+0+0+2\int \limits_0^{\frac{\pi}{2}}dx$

On Integrating

$I=[2x]_0^{\pi/2}$

On applying limits

$I.2.\frac{\pi}{2}=\pi$

Correct answer is C

answered Feb 14, 2013 by