Question

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

• A. $2\pi$
• B. $6\pi$
• C. $3\pi$
• D. $4\pi$

Answer 1

The correct option is A $2\pi$

$I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}(x^3+x\cos x+{\tan }^{8}x+1)dx$

$=\left[{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{x}^3+{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}x\cos x+{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\tan }^{5}x+{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}1.\right]$

Let $I_1={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{x}^3={\left[\frac{x^4}{4}\right]}_{-\pi /2}^{\pi /2}=0$

$I_2={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}x\cos x$

By Integration by parts method

$=x\int \cos dx-\int 1.\int \cos xdx.dx$

$=x\sin x-\int \sin xdx$

$=[x\sin x+\cos x{]}_{-\pi /2}^{\pi /2}=\frac{\pi }{2}(1)+0-\frac{\pi }{2}(-1)$

$I_3={\int }_{-\pi /2}^{\pi /2}{\tan }^{5}xdx={\int }_{-\pi /2}^{\pi /2}{\tan }^{3}x.{\tan }^{2}xdx$

$={\int }_{-\pi /2}^{\pi /2}{\tan }^{3}x({\sec }^{2}x-1)dx={\int }_{-\pi /2}^{\pi /2}{\tan }^{3}x{\sec }^{2}xdx-{\int }_{-\pi /2}^{\pi /2}{\tan }^{3}xdx$

$={\int }_{-\pi /2}^{\pi /2}{\tan }^{3}x{\sec }^{3}xdx-{\int }_{-\pi /2}^{\pi /2}\tan x.{\tan }^{2}xdx$

$={\int }_{-\pi /2}^{\pi /2}{\tan }^{3}x{\sec }^{2}xdx-{\int }_{-\pi /2}^{\pi /2}{\int }_{-\pi /2}^{\pi /2}\tan x.({\sec }^{2}x-1)dx$

$={\int }_{-\pi /2}^{\pi /2}{\tan }^{3}x\sin xdx-{\int }_{-\pi /2}^{\pi /2}\tan x{\sec }^{2}xdx+{\int }_{-\pi /2}^{\pi /2}\tan xdx$

Let $\tan x=t\therefore dt={\sec }^{2}xdx$

$=\int t^{3}dt-\int tdt+\ln |\sec x|$

$={[\frac{\tan 4x}{4}-\frac{{\tan }^{2}x}{2}+\ln |\sec x|]}_{-\pi /2}^{\pi /2}$

$=0$

$I_4={\int }_{-\pi /2}^{\pi /2}1.dx=[x{]}_{-\pi /2}^{\pi /2}=\pi$

$I=I_1+I_2+I_3+I_4$

$=0+\pi +0+\pi$

$=2\pi$