problem1
\[\int\frac{dx}{\sin 2x+2\sin x} \]
solution1.1
\[\int\frac{dx}{\sin 2x+2\sin x}=\int\frac{dx}{2sinx(1+cosx)}=\frac{1}{4}\int\frac{d(\frac{x}{2})}{sin\frac{x}{2}cos^3\frac{x}{2}}=\frac{1}{4}\int\frac{d(tan\frac{x}{2})}{tan\frac{x}{2}cos^2\frac{x}{2}}= \frac{1}{4}\int\frac{1+tan^2\frac{x}{2}}{tan\frac{x}{2}}d(tan\frac{x}{2})=\frac{1}{8}tan^2\frac{x}{2}+\frac{1}{4}ln\vert tan\frac{x}{2} \vert +C \]
solution1.2
\[\int\frac{dx}{sin2x+2sinx}=\int\frac{dx}{2sinx(1+cosx)}=\int\frac{sinxdx}{2(1-cos^2x)(1+cosx)}\\=-\frac{1}{8}\int[\frac{1}{1-cosx}+\frac{3+cosx}{(1+cosx)^2}]dcosx=\frac{1}{8}ln\frac{1-cosx}{1+cosx}+\frac{1}{4(1+cosx)}+C \]
problem2
\[\int(arcsinx)^2dx \]
solution2.1
\[\int(arcsinx)^2dx=x(arcsinx)^2-\int\frac{2xarcsinx}{\sqrt{1-x^2}}dx\\=x(arcsinx)^2-\int2arcsinxd{\sqrt{1-x^2}}\\=x(arcsinx)^2-2\sqrt{1-x^2}arcsinx-2x+C \]
solution2.2
\[u=arcsinx,x=sinu,dx=cosudu \]
\[\int(arcsinx)^2dx=\int u^2cosudu=\int u^2dsinu=u^2sinu-\int 2usinudu\\=u^2sinu+2ucosu-2sinu+C=x(arcsinx)^2+2\sqrt{1-x^2}arcsinx-2x+C \]
problem3
\[\int\frac{xe^{arctanx}}{(1+x^2)^{\frac{3}{2}}}dx \]
solution3.1
\[\int\frac{xe^{arctanx}}{(1+x^2)^{\frac{3}{2}}}dx=\int\frac{x}{(1+x^2)^{\frac{1}{2}}}de^{arctanx}\\=\frac{xe^{arctanx}}{(1+x^2)^{\frac{1}{2}}}-\int\frac{1}{(1+x^2)^{\frac{1}{2}}}de^{arctanx}\\=\frac{xe^{arctanx}}{(1+x^2)^{\frac{1}{2}}}-\frac{e^{arctanx}}{(1+x^2)^{\frac{1}{2}}}-\int\frac{xe^{arctanx}}{(1+x^2)^{\frac{3}{2}}}dx \]
小小的移向整理一哈
\[\int\frac{xe^{arctanx}}{(1+x^2)^{\frac{3}{2}}}dx=\frac{xe^{arctanx}}{2(1+x^2)^{\frac{1}{2}}}-\frac{e^{arctanx}}{2(1+x^2)^{\frac{1}{2}}}+C \]
solution3.2
令
\[x=tant \]
\[\int\frac{xe^{arctanx}}{(1+x^2)^{\frac{3}{2}}}dx=\int e^tsintdt \]
\[\int e^tsintdt=-e^tcost-\int e^tcostdt=-e^tcost+e^tsint-\int e^tsintdt \]
\[\int\frac{xe^{arctanx}}{(1+x^2)^{\frac{3}{2}}}dx=\frac{xe^{arctanx}}{2(1+x^2)^{\frac{1}{2}}}-\frac{e^{arctanx}}{2(1+x^2)^{\frac{1}{2}}}+C \]
problem4
\[\int x\tan ^2x\rm dx \]
solution4
\[\int x\tan ^2x\rm dx=\int x\rm d (tanx-x)=x\tan x-x^2-\int (tanx-x)\rm d x=x\tan x-\frac{x^2}{2}+ln |\cos x|+C \]
problem5
\[\int\frac{\sin 10x}{\sin x}\rm dx \]
solution5
考慮
\[\frac{\sin 2nx}{\sin x}=2\sum_{k=1}^{n}{\cos (2k-1)x} \]
\[\int\frac{\sin 10x}{\sin x}\rm dx=2(\frac{1}{9}\sin 9x+\frac{1}{7}\sin 7x+\frac{1}{5}\sin 5x+\frac{1}{3}\sin 3x+\sin x)+C \]
problem6
\[\int\frac{\sin 9x}{\sin x}\rm dx \]
solution6
考慮
\[\frac{\sin (2n+1)x}{\sin x}=2\sum_{k=1}^{n}{\cos 2kx}+1 \]
\[\int\frac{\sin 9x}{\sin x}\rm dx=2(\frac{1}{8}\sin 8x+\frac{1}{6}\sin 6x+\frac{1}{4}\sin 4x+\frac{1}{2}\sin 2x)+x+C \]
problem7
\[\int\frac{\rm dx}{\sin (x+a)\sin (x+b)} \]
solution7
Note that trigonometric identity gives
\[\begin{aligned} \sin (a-b) &=\sin ((x+a)-(x+b)) \\ &=\sin (a+x) \cos (b+x)-\sin (b+x) \cos (a+x) \end{aligned} \]
Using this technique, we can split the integrand into two parts and the answer follows directly. However we need to consider two different cases: the first case is \(\sin (a-b)=0\) and the other is \(\sin (a-b) \neq 0\).
Case \(1 \sin (a-b)=0 .\)
It is evident that
\[\sin (a-b)=0 \Rightarrow a-b=k \pi, k \in \mathbb{Z} \Rightarrow \sin (x+a)=(-1)^{k} \sin (x+b) \]
so we have
\[\begin{aligned} I:=\int \frac{\mathrm{d} x}{\sin (a+x) \sin (b+x)} &=(-1)^{k} \int \frac{\mathrm{d} x}{\sin ^{2}(b+x)} \\ &=(-1)^{k} \int \csc ^{2}(b+x) \mathrm{d} x \\ &=(-1)^{k+1} \cot ^{2}(b+x)+C \end{aligned} \]
Case \(\mathbf{2} \sin (a-b) \neq 0\)
By using the relation Eq. (8), it follows that
\[\begin{aligned} I &=\int \frac{\mathrm{d} x}{\sin (a+x) \sin (b+x)} \\ &=\frac{1}{\sin (a-b)} \int \frac{\sin (a-b)}{\sin (a+x) \sin (b+x)} \mathrm{d} x \\ &=\frac{1}{\sin (a-b)} \int \frac{\sin (a+x) \cos (b+x)-\sin (b+x) \cos (a+x)}{\sin (a+x) \sin (b+x)} \mathrm{d} x \\ &=\frac{1}{\sin (a-b)} \int(\cot (b+x)-\cot (a+x)) \mathrm{d} x \\ &=\frac{1}{\sin (a-b)} \ln \left|\frac{\sin (b+x)}{\sin (a+x)}\right|+C \end{aligned} \]
Finally, the answer reads
\[I= \begin{cases}(-1)^{k+1} \cot ^{2}(b+x)+C, & \text { if } \sin (a-b)=0 \\ \frac{1}{\sin (a-b)} \ln \left|\frac{\sin (b+x)}{\sin (a+x)}\right|+C, & \text { if } \sin (a-b) \neq 0\end{cases} \]
problem8
\[\int\frac{3\cos x-\sin x}{\cos x+\sin x}\rm dx \]
solution8
對於形如\(\int\frac{a\cos x+b\sin x}{c\cos x+d\sin x}\rm dx\)
一般有一個\({a\cos x+b\sin x}=A{(c\cos x+d\sin x)}+B{(c\cos x+d\sin x)^{(1)}}\)
\[\int\frac{3\cos x-\sin x}{\cos x+\sin x}\rm dx=\int\rm dx+2\int\frac{d(\cos x+\sin x)}{\cos x+\sin x}=x+2ln|\cos x+\sin x|+C \]
problem9
\[I_1=\int\frac{\sin x}{a\cos x+b\sin x}\rm dx \]
\[I_2=\int\frac{\cos x}{a\cos x+b\sin x}\rm dx \]
求解\(I_1,I_2\)
solution9
\[\begin{cases} aI_2+bI_1=\int\frac{a\cos x+b\sin x}{a\cos x+b\sin x}\rm dx=x+C_1\\ bI_2-aI_1=\int\frac{b\cos x-a\sin x}{a\cos x+b\sin x}\rm dx=ln\left|a\cos x+b\sin x\right|+C_2 \end{cases} \]
\[\begin{cases} I_1=\frac{1}{a^2+b^2}({bx-aln\left|a\cos x+b\sin x\right|})+C\\ I_2=\frac{1}{a^2+b^2}({ax+bln\left|a\cos x+b\sin x\right|})+C \end{cases} \]
problem10
\[\int\frac{x+\sin x}{1+\cos x}\rm dx \]
solution10
\[\int\frac{x+\sin x}{1+\cos x}\rm dx=\int(x+\sin x)d(\tan \frac{x}{2})=x\tan \frac{x}{2}+C \]
problem11
\[\int \frac{\cos ^{2} x-\sin x}{\cos x\left(1+\cos x e^{\sin x}\right)} \mathrm{d} x \]
solution11
Note that
\[\mathrm{d}\left(\cos x e^{\sin x}\right)=\left(\cos ^{2} x-\sin x\right) e^{\sin x} \mathrm{~d} x \]
it follows
\[\begin{aligned} \int \frac{\cos ^{2} x-\sin x}{\cos x\left(1+\cos x e^{\sin x}\right)} \mathrm{d} x &=\int \frac{e^{\sin x}\left(\cos ^{2} x-\sin x\right)}{e^{\sin x} \cos x\left(1+\cos x e^{\sin x}\right)} \mathrm{d} x \\ &=\int \frac{\mathrm{d}\left(e^{\sin x} \cos x\right)}{e^{\sin x} \cos x\left(1+\cos x e^{\sin x}\right)} \\ &=\int \frac{\mathrm{d} u}{u(1+u)}\left(u=\cos x e^{\sin x}\right) \\ &=\ln \left|\frac{u}{1+u}\right|+C \\ &=\ln \left|\frac{e^{\sin x} \cos x}{1+e^{\sin x} \cos x}\right|+C \end{aligned} \]
problem12
\[\int \frac{\arctan \frac{1}{x}}{1+x^{2}} \mathrm{~d} x \]
solution12
Set \(u=\frac{1}{x} \Rightarrow \mathrm{d} x=-\frac{\mathrm{d} u}{u^{2}}\), and inserting this into the original integral yields
\[\begin{aligned} \int \frac{\arctan \frac{1}{x}}{1+x^{2}} \mathrm{~d} x &=-\int \frac{\arctan u}{1+u^{2}} \mathrm{~d} u \\ &=-\frac{1}{2} \int \mathrm{d}\left(\arctan ^{2} u\right) \\ &=-\frac{1}{2} \arctan ^{2} u+C \\ &=-\frac{1}{2} \arctan ^{2} \frac{1}{x}+C \end{aligned} \]
problem13
\[\int \frac{e^{\sin 2 x} \sin ^{2} x}{e^{2 x}} \mathrm{~d} x \]
solution13
Using the relation
\[\sin ^{2} x=\frac{1-\cos 2 x}{2} \]
find that
\[\begin{aligned} \int \frac{e^{\sin 2 x} \sin ^{2} x}{e^{2 x}} \mathrm{~d} x &=\int \frac{1-\cos 2 x}{2} e^{\sin 2 x-2 x} \mathrm{~d} x \\ &=-\frac{1}{4} \int e^{\sin 2 x-2 x} \mathrm{~d}(\sin 2 x-2 x) \\ &=-\frac{e^{\sin 2 x-2 x}}{4}+C \end{aligned} \]
problem14
\[\int \frac{\mathrm{d} x}{\sin ^{4} x+\cos ^{4} x} \]
solution14
Note that
\[\sin ^{4} x=\left(\sin ^{2} x\right)^{2}=\frac{(1-\cos 2 x)^{2}}{4}, \quad \cos ^{4} x=\left(\cos ^{2} x\right)^{2}=\frac{(\cos 2 x+1)^{2}}{4} \]
taking those relations into the integral indicates
\[\begin{aligned} \int \frac{\mathrm{d} x}{\sin ^{4} x+\cos ^{4} x} &=\int \frac{4 \mathrm{~d} x}{(1-\cos 2 x)^{2}+(1+\cos 2 x)^{2}} \\ &=\int \frac{2 \mathrm{~d} x}{1+\cos ^{2} 2 x} \\ &=\int \frac{2 \mathrm{~d} x}{2-\sin ^{2} 2 x} \\ &=\int \frac{2 \sec ^{2} 2 x \mathrm{~d} x}{2 \sec ^{2} 2 x-\tan ^{2} 2 x} \\ &=\int \frac{\mathrm{d}(\tan 2 x)}{2+\tan ^{2} 2 x} \\ &=\frac{1}{\sqrt{2}} \arctan \left(\frac{\tan 2 x}{\sqrt{2}}\right)+C \end{aligned} \]
problem15
\[\int \frac{x \cos x}{\sin ^{3} x} \mathrm{~d} x \]
solution15
Using integration by parts yields
\[\begin{aligned} \int \frac{x \cos x}{\sin ^{3} x} \mathrm{~d} x &=-\frac{1}{2} \int x \mathrm{~d}\left(\sin ^{-2} x\right) \\ &=-\frac{1}{2} x \csc ^{2} x+\frac{1}{2} \int \csc ^{2} x \mathrm{~d} x \\ &=-\frac{1}{2} x \csc ^{2} x-\frac{1}{2} \cot x+C \end{aligned} \]
problem16
\[\int(\tan x+1)^{2} e^{2 x} \mathrm{~d} x \]
solution16
Observing that
\[\mathrm{d}(\tan x)=\sec ^{2} x \mathrm{~d} x \]
and expand the integrand gives
\[\begin{aligned} & \int(\tan x+1)^{2} e^{2 x} \mathrm{~d} x \\ =& \int\left(2 \tan x+\sec ^{2} x\right) e^{2 x} \mathrm{~d} x \\ =& \int \tan x \mathrm{~d}\left(e^{2 x}\right)+\int e^{2 x} \sec ^{2} x \mathrm{~d} x \\ =& e^{2 x} \tan x+C \end{aligned} \]