크레이지J의 탐구생활

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