(a) Find: ∫ cos^3 x e^log sin x d x

$\displaystyle \int \cos ^{3} x \mathrm{e}^{\log \sin x} d x=\int \cos ^{3} x \cdot \sin x d x$, Assuming $\displaystyle \cos x=t$ and $\displaystyle \sin x d x=-d t$

$$\begin{aligned} & =-\int t^{$\displaystyle 3$} d t & =-\frac{t^{$\displaystyle 4$}}{$\displaystyle 4$}+C=-\frac{\cos ^{$\displaystyle 4$} x}{$\displaystyle 4$}+C \end{aligned}

$$Or (b) $\displaystyle \int \frac{1}{5+4 x-x^{2}} d x=\int \frac{1}{3^{2}-(x-2)^{2}} d x$

$$=\frac{1}{6} \log \left|\frac{1+x}{5-x}\right|+C