(a) Find: ∫ (cos x)/(sin 3 x) d x; OR (b) Find: ∫ x^2 log ( x^2+1) d x

$\displaystyle \mathrm{I}=\int \frac{\cos \mathrm{x}}{3 \sin \mathrm{x}-4 \sin ^{3} \mathrm{x}} \mathrm{dx}$

Let $\displaystyle \sin \mathrm{x}=\mathrm{t} \Rightarrow \cos \mathrm{x} \mathrm{dx}=\mathrm{dt}$

$$\begin{aligned} \mathrm{I} & =\int \frac{d \mathrm{t}}{$\displaystyle 3$ \mathrm{t}-$\displaystyle 4$ \mathrm{t}^{$\displaystyle 3$}} & =\int \frac{1}{\mathrm{t}^{$\displaystyle 3$}\left(\dfrac{3}{\mathrm{t}^{$\displaystyle 2$}}-$\displaystyle 4$\right)} d \mathrm{t} \end{aligned}

$$Let $\displaystyle \frac{3}{\mathrm{t}^{2}}-4=\mathrm{z} \Rightarrow-\frac{6}{\mathrm{t}^{3}} \mathrm{dt}=\mathrm{dz}$

$$\begin{aligned} & \mathrm{I}=-\frac{1}{6} \int \frac{d \mathrm{z}}{\mathrm{z}} & =-\frac{1}{6} \log |\mathrm{z}|+C & =-\frac{1}{6} \log \left|$\displaystyle 3$ \operatorname{cosec}^{$\displaystyle 2$} \mathrm{x}-$\displaystyle 4$\right|+C \end{aligned}

$$(b)Let $\displaystyle \mathrm{I}=\int \mathrm{x}^{2} \log \left(\mathrm{x}^{2}+1\right) \mathrm{dx}$

$$\begin{aligned} & =\log \left(\mathrm{x}^{$\displaystyle 2$}+$\displaystyle 1$\right) \cdot \frac{\mathrm{x}^{$\displaystyle 3$}}{$\displaystyle 3$}-\int \frac{$\displaystyle 2$ \mathrm{x}}{\mathrm{x}^{$\displaystyle 2$}+$\displaystyle 1$} \cdot \frac{\mathrm{x}^{$\displaystyle 3$}}{$\displaystyle 3$} \mathrm{dx} & =\frac{\mathrm{x}^{$\displaystyle 3$}}{$\displaystyle 3$} \log \left(\mathrm{x}^{$\displaystyle 2$}+$\displaystyle 1$\right)-\frac{2}{3} \int \frac{\mathrm{x}^{$\displaystyle 4$}}{\mathrm{x}^{$\displaystyle 2$}+$\displaystyle 1$} \mathrm{dx} \end{aligned}