Integrals — Class 12 Maths Solution

Option c is correct

$I = \int\limits_{ - \pi /2}^{\pi /2} {\left( {{x^3} + x\cos x + {{\tan }^5}x + 1} \right)dx}$

$\Rightarrow$ $I = \int\limits_{ - \pi /2}^{\pi /2} {\left( {{x^3} + x\cos x + {{\tan }^5}x + 1} \right)dx} + \int\limits_{ - \pi /2}^{\pi /2} {1\,dx}$

$I = {I_1} + \left[ x \right]_{ - \pi /2}^{\pi /2} = {I_1} + \cfrac{\pi }{2} + \cfrac{\pi }{2}$

Where ${I_1} = \int\limits_{ - \cfrac{\pi }{2}}^{\cfrac{\pi }{2}} {\left( {{x^3} + x\cos x + {{\tan }^5}x} \right)dx}$ $\Rightarrow$ $I = {I_1} + \pi$

Now, for ${I_1}$ , let $f\left( x \right) = {x^3} + x\cos x + {\tan ^5}x$

$\therefore$ $f\left( { - x} \right) = {\left( { - x} \right)^3} + \left( { - x} \right)\cos \left( { - x} \right) + {\tan ^5}\left( { - x} \right)$

$= \left( { - {x^3} - x\cos x - {{\tan }^5}x} \right) = - f\left( x \right)$
$\therefore$ $f\left( x \right)$ is an odd function

Thus, ${I_1} = 0$ ,

Hence, $I = \pi$