Question

Evaluate: $\int x\cdotp \sin^{2} x \ dx$

Answer 1

We know that $sin^{2}x=\frac{ 1-cos x}{2}$

So replace in the integration

$\int x sin^{2}x dx= \frac{1}{2}\int x(1-cos x) dx\\\\=\frac{1}{2}\int x dx-\frac{1}{2}\int x\:cos x dx$

Here we integrate both terms separately and put in the equation at last

$\frac{1}{2}\int x dx=\frac{x^{2}}{4}$

$\frac{1}{2}\int x\:cos x dx = \frac{x}{2}\int cos x dx-\frac{1}{2}\int (\frac{dx}{dx}\int cos x dx)dx\\\\=\frac{x.sin x}{2}-\frac{1}{2}\int sin x dx\\\\=\frac{x.sin x}{2}+\frac{cos x}{2}$

Now put both integration in the eq

$\int x sin^{2}x dx=\frac{x^{2}}{4}- \frac{x.sin x}{2}+\frac{cos x}{2}+C$