Question:
Evaluate the following integrals:
$\int x \cos ^{3} x^{2} \sin x^{2} d x$
Solution:
Let $\cos x^{2}=t$
Then $d\left(\cos x^{2}\right)=d t$
Since $d\left(x^{n}\right)=n x^{n-1}$ and $d(\cos x)=-\sin x d x$
$d t=2 x\left(-\sin x^{2}\right)=-2 x \sin x^{2} d x$
$x \sin x^{2} d x=-\frac{d t}{2}$
hence $\int x \cos ^{3} x^{2} \sin x^{2} d x=\int t^{3} x-\frac{d t}{2}$
$=-\frac{1}{2} \int \mathrm{t}^{3} \mathrm{dt}$
$=-\frac{1}{2} \times \frac{\mathrm{t}^{4}}{4}+\mathrm{c}$
$=-\frac{1}{8} \cos ^{4} \mathrm{x}^{2}+\mathrm{c}$
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