Question
Evaluate the following integrals:
$\int\text{x}\cos^3\text{x}^2\sin\text{x}^2\text{ dx}$

Answer

$\int\text{x}\cos^3\text{x}^2\sin\text{x}^2\text{ dx}$
Let $\text{x}^2=\text{t}$
$2\text{x}\text{dx}=\text{dt}$
$\text{x}\text{ dx}=\frac{\text{dt}}{2}$
Now, $\int\text{x}\cos^3\text{x}^2\sin\text{x}^2\text{ dx}$
$=\frac{1}{2}\int\cos^3\text{t}\cdot\sin\text{t}\text{ dt}$
Again let $\cos\text{t}=\text{p}$
$-\sin\text{t}\text{ dt}=\text{dp}$
$\sin\text{t}\text{ dt}=-\text{dp}$
So, $\frac{1}{2}\int\cos^3\text{t}\cdot\sin\text{t}\text{ dt}$
$=-\frac{1}{2}\text{p}^3\text{ dp}$
$=-\frac{1}{2}\Big(\frac{\text{p}^4}{4}\Big)+\text{C}$
$=-\frac{\text{p}^4}{8}+\text{C}$
$=-\frac{\cos^4\text{t}}{8}+\text{C}$
$=-\frac{\cos^4\text{x}^2}{8}+\text{C}$

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