(a) Evaluate ∫_-1^1| x^4- x | d x.

$\displaystyle \mathrm{I}=\int_{-1}^{1}\left|\mathrm{x}^{4}-\mathrm{x}\right| \mathrm{dx}=\int_{-1}^{0}\left(\mathrm{x}^{4}-\mathrm{x}\right) \mathrm{dx}-\int_{0}^{1}\left(\mathrm{x}^{4}-\mathrm{x}\right) \mathrm{dx}$

$$\begin{aligned} & \left.\left.=\left(\frac{\mathrm{x}^{$\displaystyle 5$}}{$\displaystyle 5$}-\frac{\mathrm{x}^{$\displaystyle 2$}}{$\displaystyle 3$}\right)\right]_{-$\displaystyle 1$}^{$\displaystyle 0$}-\left(\frac{\mathrm{x}^{$\displaystyle 5$}}{$\displaystyle 5$}-\frac{\mathrm{x}^{$\displaystyle 2$}}{$\displaystyle 2$}\right)\right]_{$\displaystyle 0$}^{$\displaystyle 1$} & =\frac{7}{10}+\frac{3}{10}=\mathrm{I} \end{aligned}

$$Or (b) $\displaystyle \int \frac{\sin ^{-1} \mathrm{x}}{\left(1-\mathrm{x}^{2}\right)^{3 / 2}} \mathrm{d} \mathrm{x}=\int \operatorname{tsec}^{2} t \mathrm{d} t,\left(\right.$ Putting $\displaystyle \sin ^{-1} \mathrm{x}=t, \mathrm{x}=\sin t$, also $\displaystyle \left.\frac{1}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{d} \mathrm{x}=\mathrm{d} t\right)$

$$\begin{aligned} & =t \cdot \tan t+\log |\cos t|+c & =\frac{\mathrm{x} \sin ^{-$\displaystyle 1$} \mathrm{x}}{\sqrt{$\displaystyle 1$-\mathrm{x}^{$\displaystyle 2$}}}+\frac{1}{2} \log \left|$\displaystyle 1$-\mathrm{x}^{$\displaystyle 2$}\right|+c \end{aligned}