这是一份leeds利兹大学MATH021201作业代写的成功案例
Here $1=\int x \cos ^{3} x^{2} \sin x^{2} d x$.
First, putting $\mathrm{x}^{2}=\mathrm{t}$,
so that $2 \mathrm{x} \mathrm{dx}=\mathrm{dt}$, we have
$$
I=\frac{1}{2} \int \cos ^{3} t \sin t d t
$$
Now putting $\cos \mathrm{t}=\mathrm{u}$, so that $-\sin t \mathrm{dt}=\mathrm{du}$, we have
$$
\begin{aligned}
&1=-\frac{1}{2} \int u^{3} d u=-\frac{1}{2} \frac{u^{4}}{4}=-\frac{1}{8} u^{4} \
&=-\frac{1}{8} \cos ^{4} t=-\frac{1}{8} \cos ^{4} x^{2} .
\end{aligned}
$$
$\left[\because \mathrm{t}=\mathrm{x}^{2}\right]$
MATH021201 COURSE NOTES ：
Putting $x e^{x}=t$ so that $\left(e^{x}+x e^{x}\right) d x=d t$ or $\mathrm{e}^{\mathrm{y}}(1+\mathrm{x}) \mathrm{dx}=\mathrm{dt}$, we have
$$
I=\int \frac{d t}{\sin ^{2} t}=\int \operatorname{cosec}^{2} t d t=-\cot t=-\cot \left(x e^{x}\right) .
$$