Question

The value of $\underset{-\mathrm{\pi }/2}{\overset{\mathrm{\pi }/2}{\int }}\left(x^3+x\cos x+{\tan }^{5}x+1\right)dx,$ is

(a) 0
(b) 2
(c) π
(d) 1

Answer 1

(c) $\mathrm{\pi }$

$$\begin{gathered} {\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\left(x^3+x\cos x+{\tan }^{5}x+1\right)dx \\ ={\left[\frac{x^4}{4}\right]}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}+{\left[x\sin x\right]}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}-{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\sin xdx+{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}{\tan }^{3}x\left(se{c}^{2}x-1\right)dx+{\left[x\right]}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \\ =\frac{{\mathrm{\pi }}^4}{64}-\frac{{\mathrm{\pi }}^4}{64}+\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{2}-{\left[-\cos x\right]}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}+{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}{\tan }^{3}xse{c}^{2}xdx-{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}{\tan }^{3}xdx+\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{2} \\ =\mathrm{\pi }+0+{\left[\frac{{\tan }^4\mathrm{x}}{4}\right]}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}-{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\mathrm{tanx}{\sec }^2\mathrm{x}\mathrm{dx}-{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\mathrm{tanx}\mathrm{dx} \\ =\mathrm{\pi }-{\left[\frac{{\tan }^2\mathrm{x}}{2}\right]}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}-{\left[-\log \left(\cos x\right)\right]}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \\ =\mathrm{\pi } \end{gathered}$$