Question

The value of ${\int }_{-\pi /2}^{\pi /2}(x^3+xcosx+ta{n}^{5}x+1)dx$ is equal to

• A. $0$
• B. $2$
• C. $\pi$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C $\pi$

${\int }_{-\pi /2}^{\pi /2}(x^3+x\cos x+{\tan }^{5}x+1)dx$

$={\int }_{-\pi /2}^{\pi /2}(x^3+x\cos x+{\tan }^{5}x)dx+{\int }_{-\pi /2}^{\pi /2}dx=0+\left[x{]}_{-\pi /2}^{\pi /2}\right[\because (x^3+x\cos x+{\tan }^{5}x)\text{is an odd function}]$

$=[\frac{\pi }{2}-(-\frac{\pi }{2})]=\pi$