Integral100 [61-70]

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

$\begin{aligned} &\int \sqrt{x^2 + 4x}dx = \int \sqrt{x^2 + 4x+4-4}dx \\ &=\int \sqrt{(x+2)^2-4}dx (x+2=2sec\theta, dx=2sec\theta tan \theta d\theta)\\ &=\int \sqrt{(4sec^2\theta-4) } 2sec\theta tan \theta d\theta\\ &=\int 4sec\theta tan^2\theta d\theta =4 \int sec \theta (sec^2\theta -1) d\theta \\ &=4\int sec^3 \theta d\theta -4\int sec\theta d\theta \end{aligned}$

$\int sec^3 x dx = \int sec x sec^2x dx=secxtanx-\int sec x tan^2 x dx\\ =sec x tan x-\int sec x (sec^2x-1)dx\\ =sec x tan x-\int sec^3x dx +\int sec x dx\\ 2\int sec^3x dx = secx tanx+ln|sec x+tan x|\\ \int sec^3 x dx = \frac{1}{2}(sec x tan x + ln |sec x + tan x|)+C$

$=4\int sec^3 \theta d\theta -4\int sec\theta d\theta\\ =2(sec \theta tan \theta + ln |sec \theta + tan \theta|)-4ln |sec \theta + tan \theta|\\ =2sec \theta tan \theta - 2ln|sec \theta + tan \theta|+C\\ (sec \theta = \frac{x+2}{2}, R.T. angle=\theta, a=2, h=x+2, o=\sqrt{x^2+4x})\\ =2 \frac{x+2}{2}\frac{\sqrt{x^2+4}}{2}-2ln|\frac{x+2+\sqrt{x^2+4}}{2}|+C\\ =\frac{(x+2)\sqrt{x^2+4}}{2}-2ln|\frac{x+2+\sqrt{x^2+4}}{2}|+C\\ =\frac{(x+2)\sqrt{x^2+4}}{2}-2ln|x+2+\sqrt{x^2+4}|+C_2\\$

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62. $\int (x^2)e^{x^3}dx$

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

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63. $\int (x^3)e^{x^2}dx$

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

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64. $\int tan(x)ln(cos(x))dx$

$\begin{aligned} &\int tan(x)ln(cos(x)) dx = \int \frac{sin(x)}{cos(x)}ln(cos(x)) dx \\ &(u=cos(x), du=-sin(x)dx)\\ &=-\int ln(u)\frac{1}{u} du = -(ln(u)ln(u)-\int \frac{ln(u)}{u} du)\\ &[ 2\int \frac{ln(u)}{u} du = (ln(u))^2 ]\\ &=-\frac{1}{2}(ln(\cos x))^2+C \end{aligned}\\$
Alt.
$(u=ln(cos(x), du = -\frac{sin x}{cos x}dx=-tan x dx)\\ \int tan(x)ln(cos(x)) dx = -\int u du \\ =-\frac{1}{2}(ln(\cos x))^2+C$

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65. $\int 1/(x^3 - 4x^2)dx$

$\begin{aligned} &\int \frac{1}{(x^3 - 4x^2)}dx \\ &=\int \frac{1}{x^2(x-4)}dx\\ &(\frac{ax+b}{x^2}+\frac{c}{x-4}=\frac{(c+a)x^2+(b-4a)x-4b}{x^2(x-4)})\\ &(b=-\frac{1}{4}, a=b/4=-\frac{1}{16}, c=-a=\frac{1}{16})\\ &=\int \frac{-\frac{1}{16}x-\frac{1}{4}}{x^2}dx+\int \frac{\frac{1}{16}}{x-4}dx\\ &=-\frac{1}{16} (\int \frac{x+4}{x^2}dx-\int \frac{1}{x-4}dx)\\ &=-\frac{1}{16} (\int \frac{1}{x}dx+4\int x^{-2}dx-ln|x-4|)\\ &=-\frac{1}{16} (ln|x|-\frac{4}{x}-ln|x-4|)+C \end{aligned}$

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66. $\int sin(x)cos(2x)dx$

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

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67. $\int 2^{ln(x)} dx$

$\begin{aligned} &\int 2^{ln(x)} dx \\ &(y=2^{ln(x)}, log_2y=\frac{ln y}{ln 2}=ln(x))\\ &(\frac{1}{yln 2}dy=\frac{1}{x}dx, \frac{dy}{dx}=\frac{yln 2}{x})\\ &=\int y \frac{x}{yln 2}dy=\frac{1}{ln 2}\int x dy\\ &=\frac{1}{ln 2}\int e^{log_2y} dy\\ &=\frac{1}{ln 2}\int y^{log_2e} dy\\ &=\frac{1}{ln 2}\int y^{\frac{1}{ln 2}} dy\\ &(\frac{1}{ln 2}+1 =\frac{1+ln 2}{ln 2} )\\ &=\frac{1}{ln 2}\frac{ln2}{(1+ln 2)} y^{(\frac{1}{ln 2}+1)}\\ &=\frac{1}{(1+ln 2)} y y^{1/ln 2}\\ &=\frac{1}{(1+ln2)} 2^{ln x}2^{(ln x)(1/ln 2)}+C\\ &=\frac{1}{(1+ln2)} 2^{ln x}x^{(ln 2)(1/ln 2)}+C\\ &=\frac{x 2^{lnx}}{1+ln2}+C \end{aligned}$
Alt.
$\begin{aligned} &\int 2^{ln(x)} dx = \int x^{ln(2)} dx\\ &=\frac{1}{1+ln2}x^{(1+ln2)}+C\\ &=\frac{x x^{ln2}}{1+ln2}+C=\frac{x 2^{lnx}}{1+ln2}+C\\ \end{aligned}$

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68. $\int sqrt(1+cos(2x)) dx$

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

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69. $\int 1/(1+tan(x)) x$

$\begin{aligned} &\int \frac{1}{1+tan(x)} dx \\ &=\int \frac{1-tan(x)}{1-tan^2(x)} dx=\int \frac{1+2tan(x)-3tan(x)}{1-tan^2(x)} dx\\ &=\int tan(2x) dx + \int \frac{1-3tan(x)}{1-tan^2(x)}dx\\ &=-\frac{1}{2}ln|cos 2x|+ \int \frac{\frac{cosx-3sinx}{cosx}}{\frac{cos^2x-sin^2x}{cos^2x}}dx \\ &=-\frac{1}{2}ln|cos 2x|+ \int \frac{cos^2x-3cos x sinx}{cos^2x-sin^2x}dx \\ &=\int \frac{cos x}{cos x+sin x} dx (u=sinx, du=cosxdx)\\ &=\int \frac{1}{\sqrt{1-u^2} + u} du\\ &=\int \frac{1+tan(x)}{1+tan^2(x)+2tan(x)} dx\\ &\int \frac{}{1+tan(x)} dx \\ &tan(2x)=\frac{2tan(x)}{1-tan^2(x)} \end{aligned}$
first!..
$\int \frac{1-tan(x)}{1+tan(x)} dx =\int \frac{cos-sin}{cos+sin} dx (u=cos+sin, du=-sin+cos dx)\\ =\int \frac{1}{u} du = ln|u|=ln|cos(x)+sin(x)|+C$
$\int \frac{1}{1+tan(x)} dx =\frac{1}{2}\int \frac{1-tan(x)+1+tan(x)}{1+tan(x)} dx\\ =\frac{1}{2} \int \frac{1-tan(x)}{1+tan(x)} dx +\frac{1}{2}x\\ =\frac{1}{2}ln|cos(x)+sin(x)|+\frac{1}{2}x+C$

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70. $\int_{1/e}^e sqrt(1- ln(x)^2)/x dx$

$\begin{aligned} &\int_{1/e}^e \frac{\sqrt{1- ln(x)^2}}{x} dx (u=ln(x), du=\frac{1}{x} dx)\\ &=\int_{-1}^{1} \sqrt{1-u^2} du =2\int_{0}^{1} \sqrt{1-u^2} du \\ &= \text{area half circle, radius 1} \\ &=\frac{\pi}{2}\\ & (u=sin \theta, du=cos\theta d\theta)\\ &=2\int_0^{\pi/2} cos^2\theta d\theta=\int_0^{\pi/2} 1+cos2\theta d\theta\\ &=\bigg[ \theta +\frac{1}{2}sin 2\theta \bigg]_0^{\pi/2}\\ &=\frac{\pi}{2} \end{aligned}$

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