Question

# The value of $\underset{-\mathrm{\pi }/2}{\overset{\mathrm{\pi }/2}{\int }}\left({x}^{3}+x\mathrm{cos}x+{\mathrm{tan}}^{5}x+1\right)dx,\phantom{\rule{0ex}{0ex}}$ is (a) 0 (b) 2 (c) π (d) 1

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Solution

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

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