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$\begin{align}2.\int x\sin^2 xdx&=\int(\frac{1}{2}x-\frac{1}{2}x\cos 2x)dx\\&=\frac{1}{4}x^2-\frac{1}{2}\int x\cos 2xdx \\&=\frac{1}{4}x^2-\frac{1}{2}(\frac{1}{2}x\sin 2x-\int\frac{1}{2}\sin 2xdx)\\&=\frac{1}{4}x^2-\frac{1}{4}x\sin 2x-\frac{1}{8}\cos 2x+C\end{align}$

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$\begin{array}{l} \int {\frac{1}{{\sin 2x\cos x}}dx} \\ = \int {\frac{{{{\sin }^2}x + {{\cos }^2}x}}{{2\sin x{{\cos }^2}x}}dx} \\ = \frac{1}{2}\int {\frac{{\sin x}}{{{{\cos }^2}x}}dx + \frac{1}{2}\int {\frac{1}{{\sin x}}dx} } \\ = \frac{{ - 1}}{2}\int {\frac{1}{{{{\cos }^2}x}}d\cos x + \frac{1}{2}\int {\frac{{\sin x}}{{{{\sin }^2}x}}dx} } \\ = \frac{1}{{2\cos x}} - \frac{1}{2}\int {\frac{1}{{1 - {{\cos }^2}x}}d\cos x} \\ = \frac{1}{{2\cos x}} - \frac{1}{2}\int {\frac{1}{{(1 - \cos x)(1 + \cos x)}}d\cos x} \\ = \frac{1}{{2\cos x}} - \frac{1}{4}\int {\frac{1}{{1 + \cos x}} + \frac{1}{{1 - \cos x}}d\cos x} \\ = \frac{1}{{2\cos x}} - \frac{1}{4}\ln \frac{{1 + \cos x}}{{1 - \cos x}} + C \end{array}$

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$\begin{aligned} \int \frac{d x}{1+\sin ^{2} x} &=\int\frac{sin^2x+cos^2x}{2sin^2x+cos^2x}dx\xlongequal{上下除以sin^2x}\int\frac{1+tan^2x}{2tanx^2+1}dx\\&=\int\frac{sec^2x}{2tanx^2+1}dx=\int \frac{\operatorname{dtan} x}{1+2 \tan ^{2} x} \\ &=\frac{1}{\sqrt{2}} \int \frac{d(\sqrt{2} \tan x)}{1+(\sqrt{2} \tan x)^{2}}=\frac{1}{\sqrt{2}} \arctan (\sqrt{2} \tan x)+C \end{aligned}$

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$\eqalign{ & \int {{x^2}\cos x{\kern 1pt} {\text{d}}x} \cr & u = {x^2},v' = \cos x, \cr & u' = 2x,v = \sin x, \cr & = {x^2}\sin x - \int {2x\sin x{\kern 1pt} {\text{d}}x} \cr & u = {\text{x,v'}} = \sin x, \cr & u' = 1,v = - \cos x, \cr & = {x^2}\sin x - 2\left( { - x\cos x - \int { - \cos x{\kern 1pt} {\text{d}}x} } \right) \cr & = {x^2}\sin x + 2x\cos x - 2\sin x + C \cr & = \left( {{x^2} - 2} \right)\sin x + 2x\cos x + C \cr}$

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