Question

Choose the correct answer. The value of
${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}(x^3+xcosx+ta{n}^{5}x+1)dx$ is
$\text{(a) zero (b) 2 (c)}\pi \text{(d) 1}$

Answer 1

Let ${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}(x^3+xcosx+ta{n}^{5}x+1)dxis\Rightarrow I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{x}^{3}dx+{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}xcosxdx+{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}ta{n}^{5}xdx+{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}1dx$
We know that ${\int }_{-a}^{a}f\left(x\right)dx=\{\begin{array}{c}2{\int }_0^{a}f\left(x\right)dx,f\left(x\right)iseven\\ 0,iff\left(x\right)isodd\end{array}\therefore I=0+0+0+2{\int }_0^{\frac{\pi }{2}}1dx.\therefore I=2[x{]}_0^{\frac{\pi }{2}}=\frac{2\pi }{2}=\pi$
$\text{Hence the correct option is (c)}$