크레이지J의 탐구생활

31. $\int \frac{1}{\sqrt{x-x^{3/2}}}dx$

$\begin{aligned} &\int \frac{1}{\sqrt{x-x^{3/2}}}dx\\ &=\int \frac{1}{\sqrt{x}\sqrt{1-x^{1/2}}}dx\\ &=\int \frac{x^{-1/2}}{\sqrt{1-x^{1/2}}}dx\\ &(u=1-x^{1/2}, du=-\frac{1}{2}x^{-1/2} dx) \\ &=-2\int \frac{1}{u^{1/2}} du=-2\int u^{-1/2} du\\ &=-2 (2)u^{1/2}=-4\sqrt{u}\\ &=-4\sqrt{1-\sqrt{x}}+C \end{aligned}$

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

$\begin{aligned} &\int \frac{1}{\sqrt{x-x^2}}dx \\ &=\int \frac{1}{x\sqrt{x^{-1}-1}} dx =\int \frac{x^{-1}}{\sqrt{x^{-1}-1}} dx\\ &(u=x^{-1}-1, du=-x^{-2}dx)\\ &=-\int \frac{x^{-1}}{\sqrt{u}}x^2du \end{aligned}$

$\begin{aligned} &\int \frac{1}{\sqrt{x-x^2}}dx \\ &=\int \frac{1}{\sqrt{x}\sqrt{1-x}} dx \\ &(u=\sqrt{x}, du=\frac{1}{2\sqrt{x}}dx)\\ &=\int \frac{1}{u\sqrt{1-u^2}} 2udu\\ &=2\int \frac{1}{\sqrt{1-u^2}}du \\ \end{aligned}$
Right Triangle : h=1, o=u, a=sqrt(1-u^2) ,sin$\theta$ = u
$=2\int \frac{1}{\cos \theta} \cos \theta d\theta =2\theta = 2 \arcsin{u}+C\\ =2\arcsin{\sqrt{x}}+C$

33. $\int e^{2lnx} dx$

$\begin{aligned} &\int e^{2lnx} dx\\ &=\int e^{lnx}e^{lnx} dx =\int (e^{lnx})^2 dx =\int x^2 dx\\ or&=\int e^{lnx^2} dx =\int x^2 dx\\ &=\frac{1}{3}x^3+C \end{aligned}$

34. $\int lnx/sqrt x dx$

$\begin{aligned} &\int \frac{\ln x}{\sqrt x}dx \\ &(u=\sqrt x , du = \frac{1}{2\sqrt x}dx)\\ &=2\int ln u^2 du=4\int ln u du \\ &=4 (uln |u| -u ) \\ &=4 (\sqrt x ln |\sqrt x| - \sqrt x ) + C\\ &=2\sqrt x ln (x) - 4\sqrt x + C\\ \end{aligned}$

$\int ln x dx = (ln x) x - \int 1/x * x dx = x(lnx)-x$

35. $\int \frac{1}{e^x+e^{-x}} dx$

$\begin{aligned} &\int \frac{1}{e^x+e^{-x}} dx \\ & \text{we know} \; cosh x=\frac{e^x+e^{-x}}{2}\\ &=\frac{1}{2}\int \frac{1}{cosh x} dx \\ &=\int \frac{e^x}{e^{2x}+1} dx (u=e^x, du=e^xdx)\\ &=\int \frac{du}{u^2+1} =\arctan {u}\\ &=\arctan{e^x}+C \end{aligned}$

36. $\int log_2 x dx$

$\begin{aligned} &\int log_2 x dx =\int \frac{ln x}{ln 2} dx\\ &=\frac{1}{ln 2}\int ln x dx\\ &=\frac{1}{ln 2}(x ln x - x)+C\\ &=x log_2x - \frac{x}{ln 2} + C\\ \end{aligned}$

37. $\int x^3*sin(2x) dx$

$\begin{aligned} &\int x^3\sin{2x} dx \\ &=x^3(-\frac{1}{2}cos2x)-3x^2(-\frac{1}{4}sin2x)+6x(\frac{1}{8}cos2x)-6(\frac{1}{16}sin2x)\\ &=cos 2x (-\frac{1}{2}x^3+\frac{3}{4}x)+sin2x(\frac{3}{4}x^2-\frac{3}{8})+C \end{aligned}$

38. $\int x^2[1+x^3]^{1/3} dx$

$\begin{aligned} &\int x^2 \sqrt[3]{1+x^3} dx \\ &(u=1+x^3, du=3x^2dx) \\ &=\frac{1}{3}\int \sqrt[3]u du = \frac{1}{3}\int u^{\frac{1}{3}} du \\ &=\frac{1}{3} \frac{3}{4}u^{1+\frac{1}{3}}=\frac{1}{4}u\sqrt[3]{u}\\ &=\frac{1}{4}(1+x^3)\sqrt[3]{1+x^3}+C \end{aligned}$

39. $\int 1/(x^2 + 4)^2 dx$

$\begin{aligned} &\int \frac{1}{(x^2 + 4)^2} dx \\ &(x=2tany, dx=2sec^2ydy, y=\arctan{\frac{x}{2}}) \\ &=\int \frac{2sec^2y}{(4(tan^2y+1))^2}dy=\int \frac{sec^2y}{8sec^4y}dy\\ &=\frac{1}{8}\int cos^2y dy =\frac{1}{16}\int 1+\cos{2y}dy\\ &=\frac{y}{16}+\frac{1}{32}\sin{2y}=\frac{1}{16}arctan{\frac{x}{2}}+\frac{1}{16}sin y cos y\\ &(right triangle angle=y, h=sqrt(x^2+4) a=2, o=x)\\ &=\frac{1}{16}arctan{\frac{x}{2}}+\frac{1}{16}\frac{x}{\sqrt{x^2+4}} \frac{2}{\sqrt{x^2+4}}\\ &=\frac{1}{16}arctan{\frac{x}{2}}+\frac{x}{8(x^2+4)}+C \end{aligned}$

40. $\int_1^2 sqrt(x^2-1) dx$

$\begin{aligned} &\int_1^2 \sqrt{x^2-1} dx , (x=sec(y), dx=sec(y) tan(y) dy)\\ &\int \sqrt{\tan^2y} \sec y \tan y dy\\ &=\int \sec y \tan^2 y dy (sec y tan y -I-> sec y) \\ &=\tan y \sec y -\int \sec^3 y dy\\ \end{aligned}$

$\int \sec^3 x dx = \int \sec x \sec^2 x dx (sec^2x-I->tan x)\\ =\sec x \tan x - \int \sec x \tan x \tan x dx\\ =\sec x \tan x - \int \sec x (\sec^2 x -1 ) dx \\ =\sec x \tan x - \int \sec^3 x dx +\int \sec x dx \\ =\sec x \tan x + ln |sec x + tan x|-\int \sec^3x dx\\ = \frac{1}{2}(\sec x \tan x + ln |sec x + tan x|)$

x=sec(y), y=arcsec x, RT. angle=y, h=x, a=1, o=sqrt(x^2-1)
$=\tan y \sec y -\int \sec^3 y dy\\ =\tan y \sec y -\frac{1}{2}(\sec y \tan y + ln |sec y + tan y|)\\ =\frac{1}{2}x\sqrt{x^2-1}-\frac{1}{2}ln| \sqrt{x^2-1}+x|+C\\ \int_1^2 \sqrt{x^2-1} dx=\left[\frac{1}{2}x\sqrt{x^2-1}-\frac{1}{2}ln| \sqrt{x^2-1}+x|\right]_1^2\\ =\sqrt{3}-\frac{1}{2}ln(\sqrt{3}+2) \\$

Author: crazyj7@gmail.com

21. $\int \sin^3{x} \cos^2{x}dx$

$\begin{aligned} &\int \sin^3{x} \cos^2{x}dx\\ &=\int \sin{x}\sin^2{x}\cos^{2}x dx\\ &=\int \sin{x}(1-\cos^2{x})\cos^{2}x dx\\ &(u=cosx, du=-sinx dx) \\ &=-\int (1-u^2)u^{2} du = \int u^4-u^2dx\\ &=\frac{1}{5}cos^5x-\frac{1}{3}cos^3x+C\\ \end{aligned}$

22. $\int \frac{1}{x^2\sqrt{x^2+1}} dx$

$\begin{aligned} &\int \frac{1}{x^2\sqrt{x^2+1}} dx\\ &=\int \frac{1}{tan^2\theta sec\theta} sec^2\theta d\theta (x=\tan{\theta}, dx=sec^2\theta d\theta) \\ &=\int sec\theta cot^2\theta d\theta =\int \frac{cos^2\theta}{cos\theta sin^2\theta} d\theta \\ &=\int \frac{cos \theta}{sin^2 \theta} d\theta (t=sin\theta, dt=cos\theta d\theta)\\ &=\int \frac{dt}{t^2} = -\frac{1}{t}=-\frac{1}{sin\theta}=-csc\theta+C\\ &=-\csc({\arctan{x}}) + C \\ &=-\frac{\sqrt{x^2+1}}{x} + C \\ \end{aligned}$
Alternative

$\begin{aligned} &\int \frac{1}{x^2\sqrt{x^2+1}} dx\\ &=\int \frac{1}{x^2 x \sqrt{1+x^{-2}}} dx\\ &=\int \frac{x^{-3}}{\sqrt{1+x^{-2}}} dx (u=1+x^{-2}, du=-2x^{-3}dx)\\ &=\int -\frac{1}{2}u^{-1/2}du = -\frac{1}{2}2u^{1/2}=-\sqrt{1+\frac{1}{x^2}}+C\\ \end{aligned}$

23. $\int \sin{x}\sec{x}\tan{x} dx$

$\begin{aligned} &\int \sin{x}\sec{x}\tan{x} dx = \int \tan^2x dx =\int sec^2x -1dx\\ &=\int \frac{1-cos^2x}{cos^2x} dx = \int sec^2x dx-x\\ &=\tan{x}-x+C\\ \end{aligned}$

24. $\int sec^3(x)dx$

$\begin{aligned} &\int sec^3(x)dx=\int sec(x)sec^2(x) dx\\ &= \sec x \tan x - \int \sec x \tan x \tan x dx\\ &= \sec x \tan x - \int \sec x \tan^2 x dx\\ &= \sec x \tan x - \int \sec x (sec^2x-1) dx\\ &= \sec x \tan x + \int sec x dx - \int sec^3x dx\\ &= \sec x \tan x + \ln | secx+tanx| - \int sec^3x dx \\ \end{aligned}$

$\begin{aligned} &2\int sec^3(x)dx=\sec x \tan x +\ln | secx+tanx| \\ &\therefore \int sec^3(x)dx=\frac{1}{2} (\sec x \tan x +\ln | secx+tanx|)+C \\ \end{aligned}$

25. $\int 1/(x*sqrt(9x^2-1)) dx$

$\begin{aligned} &\int \frac{1}{x\sqrt{9x^2-1}} dx \\ &(3x=\sec{y}, 3dx=\sec y \tan y dy)\\ &=\int \frac{3}{\sec{y} \tan{y} } \frac{\sec y \tan y }{3} dy \\ &= y =\sec^{-1} 3x +C \\ \end{aligned}$

26. $\int cos(sqrt(x)) dx$

$\begin{aligned} &\int \cos ({\sqrt{x}}) dx \\ &(u=\sqrt{x}, du=\frac{1}{2\sqrt{x}}dx )\\ &= \int \cos {u} 2 \sqrt{x} du\\ &= 2\int u\cos {u} du = 2( u sin u - (-cos u) ) \\ &= 2u sin u +2 cos u +C\\ &=2\sqrt{x}\sin{\sqrt{x}} + 2 \cos{\sqrt{x}} + C \\ \end{aligned}$

27. $\int \cosec{x}dx$

$\begin{aligned} &\int \cosec{x} dx \\ &=\int \frac{\cosec{x}(cosec{x}+cot{x}) }{ (cosec{x}+cot{x}) } dx \\ &( u = cosec{x}+cot{x} , du = (-\cosec{x}\cot{x}-\cosec^2{x}) dx )\\ &=-\int \frac{1}{u} du \\ &=-\ln|{\cosec{x}+\cot{x}}| + C \\ \end{aligned}$

28. $\int sqrt(x^2+4x+13) dx$

$\begin{aligned} &\int \sqrt{x^2+4x+13} dx \\ &=\int \sqrt{(x+2)^2+3^2} dx \\ & (u=\frac{x+2}{3} , 3 du = dx) \\ &=3\int \sqrt{3^2u^2+3^2} du \\ &=9\int \sqrt{u^2+1} du \\ &(u=tan {y}, du=sec^2ydy)\\ &=9\int \sec{y} \sec^2{y} dy = 9\int sec^3y dy \\ &=9\left[sec(y)tan(y)-\int sec(y)tan(y)tan(u) dy\right] \\ &=9sec(y)tan(y)-9\int sec(y)(sec^2y-1) dy \\ &=9sec(y)tan(y)-9\int sec^3(y)dy+9\int sec y dy \\ &18\int sec^3(y)dy = 9sec(y)tan(y)+9\ln|sec(y)+tan(y)|\\ &9\int sec^3(y)dy = \frac{9}{2}(sec(y)tan(y)+\ln|sec(y)+tan(y)|)\\ &= \frac{9}{2}sec(y)tan(y)+\frac{9}{2}\ln|sec(y)+tan(y)|\\ &(tan{y}=\frac{x+2}{3}, angle=y, h=\sqrt{x^2+4x+13} , adj=3, opposite=x+2)\\ &= \frac{9}{2}\frac{\sqrt{x^2+4x+13}}{3}\frac{x+2}{3}+\frac{9}{2}\ln{|\frac{\sqrt{x^2+4x+13}}{3}+\frac{x+2}{3}|} \\ &=\frac{(x+2)\sqrt{x^2+4x+13}}{2} +\frac{9}{2}\ln{|\frac{x+2+\sqrt{x^2+4x+13}}{3}|}+C\\ &=\frac{(x+2)\sqrt{x^2+4x+13}}{2} +\frac{9}{2}\ln{|x+2+\sqrt{x^2+4x+13}|}+C_2\\ \end{aligned}\\$

29. $\int e^{2x}*cosx dx$

$\begin{aligned} &\int e^{2x} \cos{x} dx \\ &=e^{2x} sin{x}-(2e^{2x})(-cos{x})+\int (4e^{2x})(-cos{x}) dx \\ &=e^{2x}sin{x}+2e^{2x}cos{x}-4\int e^{2x}cos{x}dx\\ &5\int e^{2x}\cos{x} dx= e^{2x}sin{x}+2e^{2x}cos{x} \\ &\int e^{2x} \cos{x} dx =\frac{1}{5}e^{2x}sin{x}+ \frac{2}{5}e^{2x}cos{x}+C\\ \end{aligned}$

30. $\int_3^5 (x-3)^9 dx$

$\begin{aligned} &\int_3^5 (x-3)^9 dx (u=x-3)\\ &=\int_0^2 u^9 du =\bigg[ \frac{u^{10}}{10} \bigg ]_0^2 \\ &=102.4 \end{aligned}$

Author: crazyj7@gmail.com

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

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

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

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

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…
$$\begin{gathered} \ln (\sin \frac{x}{2}+cos\frac{x}{2})-\ln (cos\frac{x}{2}-sin\frac{x}{2}) \\ =\ln \left|\frac{\sin \frac{x}{2}+cos\frac{x}{2}}{cos\frac{x}{2}-sin\frac{x}{2}}\right| \\ =\ln \left|\frac{(\sin \frac{x}{2}+cos\frac{x}{2}{)}^2}{co{s}^2\frac{x}{2}-si{n}^2\frac{x}{2}}\right| \\ =\ln \left|\frac{1+2sin\frac{x}{2}cos\frac{x}{2}}{co{s}^2\frac{x}{2}-si{n}^2\frac{x}{2}}\right| \\ =\ln \left|\frac{1+\sin x}{\cos x}\right| \end{gathered}$$
$$\begin{gathered} \frac{1}{2}\ln \left|\frac{1+\sin x}{1-\sin x}\right| \\ =\ln \left|\frac{\sqrt{1+\sin x}}{\sqrt{1-\sin x}}\right| \\ =\ln \left|\frac{\sqrt{1+\sin x}\sqrt{1+\sin x}}{\sqrt{1-\sin x}\sqrt{1+\sin x}}\right| \\ =\ln \left|\frac{1+\sin x}{\sqrt{1-{\sin }^{2}x}}\right| \\ =\ln \left|\frac{1+\sin x}{\cos x}\right| \\ =\ln \left|\sec x+\tan x\right| \end{gathered}$$

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…
$$\begin{gathered} \frac{d}{dx}[\frac{1}{2}x+\frac{1}{8}\sin 4x] \\ =\frac{1}{2}+\frac{1}{2}\cos 4x=\frac{1+\cos 4x}{2} \\ ={\cos }^{2}2x \end{gathered}$$

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.
$$\begin{gathered} \int \frac{x-2}{x^2-x+1}dx \\ =\frac{1}{2}\int \frac{(2x-1)-3}{x^2-x+1}dx \\ =\frac{1}{2}\int \frac{2x-1}{x^2-x+1}dx-\frac{3}{2}\int \frac{1}{(x-\frac{1}{2}{)}^2+\frac{3}{4}}dx \end{gathered}$$

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

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

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

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

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

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

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

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

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

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

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

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