(a)

Evaluate : \[\int_{0}^{3 / 2}|x \cos \pi x| d x \]

(b)

Find : \[\int \frac{d x}{\sin x+\sin 2 x} \]

\[\begin{aligned} I & =\int_{0}^{3 / 2}|x \cos \pi x| d x \\ & =\int_{0}^{1 / 2} x \cos \pi x d x-\int_{1 / 2}^{3 / 2} x \cos \pi x d x \end{aligned} \] Consider $\displaystyle \int x \cos \pi x d x$ \[=\frac{x \sin \pi x}{\pi}-\int \frac{\sin \pi x}{\pi} d x \] using ($\displaystyle 2$)in($\displaystyle 1$), \[\begin{aligned} & \left.\left.\frac{x \sin \pi x}{\pi}+\frac{\cos \pi x}{\pi^{2}}\right]_{0}^{1 / 2}-\frac{x \sin \pi x}{\pi}+\frac{\cos \pi x}{\pi^{2}}\right]_{1 / 2}^{3 / 2} \\ & =\left(\frac{1}{2 \pi}-\frac{1}{\pi^{2}}\right)-\left(-\frac{3}{2 \pi}-\frac{1}{2 \pi}\right) \\ & =\frac{5}{2 \pi}-\frac{1}{\pi^{2}} \end{aligned} \] \[\begin{aligned} I & =\int \frac{d x}{\sin x+\sin 2 x} \\ & =\int \frac{d x}{\sin x(1+2 \cos x)} \\ & =\int \frac{\sin x}{\sin ^{2} x(1+2 \cos x)} d x \\ & =\int \frac{\sin x}{(1-\cos x)(1+\cos x)(1+2 \cos x)} d x \end{aligned} \] Put $\displaystyle \cos x=t \Rightarrow \sin x d x=d t$ \[\begin{aligned} I & =-\int \frac{d t}{(1-t)(1+t)(1+2 t)} \\ & =-\frac{1}{6} \int \frac{d t}{1-t}+\frac{1}{2} \int \frac{d t}{1+t}-\frac{4}{3} \int \frac{d t}{1+2 t} \\ & =\frac{1}{6} \log |1-t|+\frac{1}{2} \log |1+t|-\frac{2}{3} \log |1+2 t|+C \\ & =\frac{1}{6} \log |1-\cos x|+\frac{1}{2} \log |1+\cos x|-\frac{2}{3} \log |1+2 \cos x|+C \end{aligned} \]