Integral100 [11-20]

 

11. $\int \frac{\sin x}{{\sec }^{2019}x}dx$

$$\begin{array}{rl}& \int \frac{\sin x}{{\sec }^{2019}x}dx\\ & =\int \sin x{\cos }^{2019}xdx\\ & (u=\cos x,du=-\sin xdx)\\ & =-\int u^{2019}du\\ & =-\frac{1}{2020}{u}^{2020}+C\\ & =-\frac{1}{2020}{\cos }^{2020}x+C\end{array}$$

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12. $\int \frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}dx$

$$\begin{array}{rl}& \int \frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}dx\\ & (x=sin\theta ,dx=\cos \theta d\theta )\\ & (cos\theta =\sqrt{1-{\sin }^2\theta }=\sqrt{1-x^2})\\ & =\int \frac{\sin \theta {\sin }^{-1}\sin \theta }{\sqrt{1-{\sin }^2\theta }}\cos \theta d\theta \\ & =\int \theta \sin \theta d\theta \\ & =\theta (-\cos \theta )-(-\sin \theta )+C\\ & =\sin \theta -\theta \cos \theta +C\\ & =x-{\sin }^{-1}(x)\sqrt{1-x^2}+C\end{array}$$

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13. $\int \frac{2\sin x}{\sin 2x}dx$

$$\begin{array}{rl}& \int \frac{2\sin x}{\sin 2x}dx\\ & =\int \frac{2\sin x}{2\sin x\cos x}dx\\ & =\int \frac{1}{\cos x}dx\end{array}$$
$=\int \sec xdx$
$$\begin{array}{rl}& =\int \frac{\cos x}{{\cos }^{2}x}dx=\int \frac{\cos x}{1-{\sin }^{2}x}dx\\ & (u=\sin x,du=\cos xdx)\\ & =\int \frac{1}{1-u^2}du=\int \frac{1}{(1-u)(1+u)}u\\ & =\int \frac{1}{2}\left(\frac{1}{1-u}+\frac{1}{1+u}\right)du\\ & =\frac{1}{2}(-\ln |1-u|+ln|1+u|)+C\\ & =\frac{1}{2}\ln |\frac{1+u}{1-u}|+C\\ & =\frac{1}{2}\ln \left|\frac{1+\sin x}{1-\sin x}\right|+C\end{array}$$
Alternatives…
$$\begin{gathered} \ln (\sin \frac{x}{2}+cos\frac{x}{2})-\ln (cos\frac{x}{2}-sin\frac{x}{2}) \\ =\ln \left|\frac{\sin \frac{x}{2}+cos\frac{x}{2}}{cos\frac{x}{2}-sin\frac{x}{2}}\right| \\ =\ln \left|\frac{(\sin \frac{x}{2}+cos\frac{x}{2}{)}^2}{co{s}^2\frac{x}{2}-si{n}^2\frac{x}{2}}\right| \\ =\ln \left|\frac{1+2sin\frac{x}{2}cos\frac{x}{2}}{co{s}^2\frac{x}{2}-si{n}^2\frac{x}{2}}\right| \\ =\ln \left|\frac{1+\sin x}{\cos x}\right| \end{gathered}$$
$$\begin{gathered} \frac{1}{2}\ln \left|\frac{1+\sin x}{1-\sin x}\right| \\ =\ln \left|\frac{\sqrt{1+\sin x}}{\sqrt{1-\sin x}}\right| \\ =\ln \left|\frac{\sqrt{1+\sin x}\sqrt{1+\sin x}}{\sqrt{1-\sin x}\sqrt{1+\sin x}}\right| \\ =\ln \left|\frac{1+\sin x}{\sqrt{1-{\sin }^{2}x}}\right| \\ =\ln \left|\frac{1+\sin x}{\cos x}\right| \\ =\ln \left|\sec x+\tan x\right| \end{gathered}$$

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14. $\int {\cos }^{2}2xdx$

$$\begin{array}{rl}& \int {\cos }^{2}2xdx\\ & =\int \frac{1+\cos 4x}{2}dx\\ & =\frac{1}{2}x+\frac{1}{2}\int \cos 4xdx\\ & =\frac{1}{2}x+\frac{1}{2}\frac{1}{4}\sin 4x+C\\ & =\frac{1}{2}x+\frac{1}{8}\sin 4x+C\end{array}$$
Check…
$$\begin{gathered} \frac{d}{dx}[\frac{1}{2}x+\frac{1}{8}\sin 4x] \\ =\frac{1}{2}+\frac{1}{2}\cos 4x=\frac{1+\cos 4x}{2} \\ ={\cos }^{2}2x \end{gathered}$$

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15. $\int \frac{1}{x^3+1}dx$

$x^3+1=(x+1)(x^2-x+1)=x^3-x^2+x+x^2-x+1$
$\frac{ax+b}{x^2-x+1}+\frac{c}{x+1},a+c=0,b+a-c=0,b+c=1$
$c=-a,b+2a=0,b=-2a,-2a+-a=1$
$a=-\frac{1}{3},c=\frac{1}{3},b=\frac{2}{3}$

$$\begin{array}{rl}& \int \frac{1}{x^3+1}dx\\ & =\int \frac{1}{(x+1)(x^2-x+1)}dx\\ & =\int \frac{-\frac{1}{3}x+\frac{2}{3}}{x^2-x+1}dx+\int \frac{\frac{1}{3}}{x+1}dx\\ & =-\frac{1}{3}\int \frac{x-2}{x^2-x+1}dx+\frac{1}{3}\ln |x+1|+C\end{array}$$
$$\begin{gathered} \int \frac{x-2}{x^2-x+1}dx \\ =\int \frac{x-2}{(x-\frac{1}{2}{)}^2+\frac{3}{4}}dx \\ (u=x-\frac{1}{2}) \\ =\int \frac{u-\frac{3}{2}}{u^2+\frac{3}{4}}du \\ =\int \frac{u}{u^2+\frac{3}{4}}du-\frac{3}{2}\int \frac{1}{u^2+\frac{3}{4}}du \end{gathered}$$
Tip. 원래는 이렇게 하는 것이 더 낫다. 분모의 미분형태(2x-1)를 분자에서 파생.
$$\begin{gathered} \int \frac{x-2}{x^2-x+1}dx \\ =\frac{1}{2}\int \frac{(2x-1)-3}{x^2-x+1}dx \\ =\frac{1}{2}\int \frac{2x-1}{x^2-x+1}dx-\frac{3}{2}\int \frac{1}{(x-\frac{1}{2}{)}^2+\frac{3}{4}}dx \end{gathered}$$

$$\begin{gathered} A.\int \frac{u}{u^2+\frac{3}{4}}du \\ (t=u^2+\frac{3}{4},dt=2udu) \\ =\frac{1}{2}\int \frac{1}{t}dt=\frac{1}{2}\ln |t|=\frac{1}{2}\ln |x^2-x+1| \\ B.\int \frac{1}{u^2+\frac{3}{4}}du \\ =\int \frac{1}{u^2+(\frac{\sqrt{3}}{2}{)}^2}du \\ =\frac{2}{\sqrt{3}}\arctan (\frac{2}{\sqrt{3}}u)=\frac{2}{\sqrt{3}}\arctan (\frac{2}{\sqrt{3}}(x-\frac{1}{2})) \\ \therefore \int \frac{x-2}{x^2-x+1}dx=\int \frac{u}{u^2+\frac{3}{4}}du-\frac{3}{2}\int \frac{1}{u^2+\frac{3}{4}}du \\ =\frac{1}{2}\ln |x^2-x+1|-\frac{3}{2}\frac{2}{\sqrt{3}}\arctan (\frac{2}{\sqrt{3}}(x-\frac{1}{2})) \\ =\frac{1}{2}\ln |x^2-x+1|-\sqrt{3}\arctan (\frac{2x-1}{\sqrt{3}}) \end{gathered}$$

$$\begin{array}{rl}\therefore & \int \frac{1}{x^3+1}dx\\ & =-\frac{1}{3}\int \frac{x-2}{x^2-x+1}dx+\frac{1}{3}\ln |x+1|+C\\ & =-\frac{1}{6}\ln |x^2-x+1|+\frac{\sqrt{3}}{3}\arctan (\frac{2x-1}{\sqrt{3}})+\frac{1}{3}\ln |x+1|+C\end{array}$$

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16. $\int x{\sin }^{2}xdx$

$$\begin{array}{rl}& \int x{\sin }^{2}xdx\\ & =\int x\frac{1-\cos 2x}{2}dx\\ & =\frac{1}{2}\int xdx-\frac{1}{2}\int x\cos 2xdx\end{array}$$

$$\begin{gathered} \int x\cos 2xdx=x(\frac{1}{2}\sin 2x)-\frac{1}{2}\int \sin 2xdx \\ =\frac{x\sin 2x}{2}-\frac{1}{2}\frac{1}{2}(-\cos 2x) \\ =\frac{x\sin 2x}{2}+\frac{\cos 2x}{4} \end{gathered}$$

$$\begin{array}{rl}& =\frac{1}{2}\int xdx-\frac{1}{2}\int x\cos 2xdx\\ & =\frac{1}{4}{x}^2-\frac{x\sin 2x}{4}-\frac{\cos 2x}{8}+C\\ & =\frac{x^2-x\sin 2x}{4}-\frac{\cos 2x}{8}+C\\ & =\frac{2x^2-2x\sin 2x-\cos 2x}{8}+C\end{array}$$

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17. $\int (x+\frac{1}{x}{)}^{2}dx$

$$\begin{array}{rl}& \int (x+\frac{1}{x}{)}^{2}dx\\ & =\int x^2+2+\frac{1}{x^2}dx\\ & =\frac{1}{3}{x}^3+2x-\frac{1}{x}+C\end{array}$$

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18 $\int \frac{3}{x^2+4x+29}dx$

$$\begin{array}{rl}& \int \frac{3}{x^2+4x+29}dx\\ & =3\int \frac{1}{(x+2{)}^2+5^2}dx\\ & =\frac{3}{5}\arctan \frac{x+2}{5}+C\end{array}$$

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19 $\int co{t}^5(x)dx$

$$\begin{array}{rl}& \int {\cot }^{5}xdx\\ & =\int \frac{co{s}^{5}x}{si{n}^{5}x}dx\\ & =\int \frac{co{s}^{4}x\cos x}{si{n}^{4}x\sin x}dx\\ & =\int \frac{(co{s}^{2}x{)}^2\cos x}{(si{n}^{2}x{)}^2\sin x}dx\\ & =\int \frac{(1-si{n}^{2}x{)}^2\cos x}{(si{n}^{2}x{)}^2\sin x}dx\\ & =\int \frac{(1-2si{n}^{2}x+si{n}^{4}x)\cos x}{(si{n}^{2}x{)}^2\sin x}dx\\ & =\int \frac{cosx}{si{n}^{5}x}dx-2\int \frac{cosx}{si{n}^{3}x}dx+\int \frac{cosx}{sinx}dx\\ & (u=sinx,du=cosxdx)\\ & =\int u^{-5}du-2\int u^{-3}du+\int \frac{1}{u}dx\\ & =-\frac{1}{4u^4}+\frac{1}{u^2}+ln|u|+C\\ & =-\frac{1}{4si{n}^{4}x}+\frac{1}{si{n}^{2}x}+ln|sinx|+C\\ & =-\frac{1}{4}cs{c}^{4}x+{\csc }^{2}x+ln|sinx|+C\end{array}$$

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20. ${\int }_{-1}^1\frac{tanx}{x^4+x^2+1}dx$

$$\begin{array}{rl}& {\int }_{-1}^1\frac{tanx}{x^4+x^2+1}dx\\ & =0\\ & (oddfunction;x->even.sin/cos->odd)\end{array}$$

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