Question

Solve:

$\stackrel{\pi /2}{\underset{-\pi /2}{\int }}(x^3+xcosx+{\tan }^{5}x+1)dx=$

• A. $0$
• B. $2$
• C. $\pi$
• D. $1$

Answer 1

The correct option is C $\pi$

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

$\Rightarrow {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$\Rightarrow a+b-x=\pi /2-\pi /2-x=-x$

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

$\Rightarrow I+I={\int }_{-\pi /2}^{\pi /2}0+0+0+2dx$

$\Rightarrow 2I=[2x{]}_{-\pi /2}^{\pi /2}=2(\pi /2+\pi /2)=2\pi$

$I=\pi$.