Question

(i) $\underset{-1}{\overset{1}{\int }}\log \left(\frac{2-x}{2+x}\right)dx$

(ii) $\underset{-\mathrm{\pi }}{\overset{\mathrm{\pi }}{\int }}\left(1-x^2\right)\sin x{\cos }^{2}xdx$

Answer 1

(ii) $\underset{-\mathrm{\pi }}{\overset{\mathrm{\pi }}{\int }}\left(1-x^2\right)\sin x{\cos }^{2}xdx$


$$\begin{gathered} \mathrm{Let}I=\underset{-\mathrm{\pi }}{\overset{\mathrm{\pi }}{\int }}\left(1-x^2\right)\sin x{\cos }^{2}xdx \\ \mathrm{We}\mathrm{know}, \\ \underset{-a}{\overset{a}{\int }}f\left(x\right)dx=\left\{\begin{array}{l}0,\mathrm{if}f\left(-x\right)=-f\left(x\right)\\ 2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\mathrm{if}f\left(-x\right)=f\left(x\right)\end{array}\right. \\ \mathrm{Here}, \\ f\left(x\right)=\left(1-x^2\right)\sin x{\cos }^{2}x \\ f\left(-x\right)=\left(1-{\left(-x\right)}^2\right)\sin \left(-x\right){\cos }^2\left(-x\right) \\ =\left(1-x^2\right)\sin \left(-x\right){\cos }^{2}x \\ =-\left(1-x^2\right)\sin x{\cos }^{2}x \\ =-f\left(x\right) \\ \mathrm{Thus}, \\ \underset{-\mathrm{\pi }}{\overset{\mathrm{\pi }}{\int }}\left(1-x^2\right)\sin x{\cos }^{2}xdx=0 \end{gathered}$$


Hence, $\underset{-\mathrm{\pi }}{\overset{\mathrm{\pi }}{\int }}\left(1-x^2\right)\sin x{\cos }^{2}xdx$ = 0.