Question

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

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

Answer 1

The correct option is C $\pi$

We know that for an odd function $f\left(x\right)$:-

${\int }_{-a}^{a}f\left(x\right)dx=0$

And for even function $g\left(x\right)$ :-

${\int }_{-a}^{a}g\left(x\right)dx=2{\int }_0^{a}g\left(x\right)dx$

Now in the given integration- $x^5,x^3\sec x$ and ${\tan }^{-1}x$ are odd function.And hence their integration is $0$.

Now, integration is left as:-

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

$=\pi$

Hence, answer is option-(C).