크레이지J의 탐구생활

71. $\int 1/(cbrt(x)+1)dx$

$\begin{aligned} &\int \frac{1}{\sqrt[3]{x}+1}dx \\ &(u=\sqrt[3]{x}, du=\frac{1}{3\sqrt[3]{x^2}}dx=\frac{1}{3}x^{-\frac{2}{3}}dx=\frac{1}{3}u^{-2}dx)\\ &=3\int \frac{u^2}{u+1}du=3\int \frac{u^2-1+1}{u+1}du\\ &=3\int \frac{(u-1)(u+1)}{u+1}du+3\int\frac{1}{u+1}du \\ &=3(\frac{1}{2}u^2-u)+3ln|u+1|\\ &=\frac{3}{2}u^2-3u+3ln|u+1|+C\\ &=\frac{3}{2}\sqrt[3]{x}^2-3\sqrt[3]{x}+3ln|\sqrt[3]{x}+1|+C\\ \end{aligned}$
Alt.
$(u=\sqrt[3]{x}+1, x=(u-1)^3, dx=3(u-1)^2du)\\ =\int \frac{1}{u} 3(u-1)^2du\\ =3\int \frac{u^2-2u+1}{u} du=3\int u-2+(1/u) du\\ =3(\frac{1}{2}u^2-2u+ln|u|)\\ =\frac{3}{2}(\sqrt[3]{x}+1)^2-6(\sqrt[3]{x}+1)+3ln|\sqrt[3]{x}+1|+C$

72. $\int \frac{1}{cbrt(x + 1)}dx$

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

73. $\int (sin(x)+cos(x))^2 dx$

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

74. $\int 2xln(1+x) dx$

$\begin{aligned} &\int 2xln(1+x) dx =2\int xln(1+x) dx\\ & \text{(ln쪽을 미분하는 방향으로 부분적분)}\\ &=2( ln(1+x)\frac{1}{2}x^2-\int \frac{1}{1+x}\frac{1}{2}x^2dx )\\ &= ln(1+x)x^2-\int \frac{x^2}{1+x}dx = ln(1+x)x^2-\int \frac{x^2-1+1}{1+x}dx\\ &= ln(1+x)x^2-\int x-1+\frac{1}{1+x}dx\\ &= ln(1+x)x^2-\frac{1}{2}x^2+x-ln(1+x)+C\\ \end{aligned}$

75. $\int 1/(x(1+sin^2ln(x)))dx$

$\begin{aligned} &\int \frac{1}{x(1+sin^2ln(x))} dx \\ &( u=ln(x), du=\frac{1}{x}dx : 1/x과 ln(x)에서 치환 착안.)\\ &=\int \frac{du}{1+sin^2u} : (sin에 대해 치환, 변환(반각)해보고, cos으로 나눠도 본다.\\ & sec, tan을 만들 수 있다는 것을 착안.)\\ &=\int \frac{sec^2u}{sec^2u+tan^2u}du\\ &=\int \frac{sec^2u}{1+2tan^2u}du, (t=tanu, dt=sec^2udu)\\ &=\int \frac{dt}{1+2t^2}, (t=tanu, dt=sec^2udu)\\ &=\frac{1}{\sqrt{2}}arctan {\sqrt{2}t}=\frac{1}{\sqrt{2}}arctan({\sqrt{2} tan(ln(x))})+C \end{aligned}$

76. $\int sqrt((1-x)/(1+x))dx$

Try
$\begin{aligned} &\int \sqrt{\frac{1-x}{1+x}} dx \\ &(\text{전체치환? 일부치환. 분자분모곱, pm 등})\\ &( u = \sqrt{\frac{1-x}{1+x}}, du=\frac{1}{2}\sqrt{\frac{1+x}{1-x}}\frac{1-x+1+x}{1-2x+x^2}dx)\\ &( du = \frac{1}{2}\sqrt{\frac{1+x}{1-x}}\frac{2}{(1-x)^2}dx=\frac{1}{u(1-x)^2}dx)\\ &(u^2=\frac{1-x}{1+x}, xu^2+x=1-u^2, x=\frac{1-u^2}{1+u^2} )\\ &=\int u^2(1-x)^2 du =\int u^2(1-\frac{1-u^2}{1+u^2})^2du\\ &=\int u^2(\frac{2u^2}{1+u^2})^2 du = \int 4u^6(\frac{1}{1+u^2})^2du\\ \end{aligned}$
Try
$\begin{aligned} &\int \sqrt{\frac{1-x}{1+x}} dx \\ &(u=\frac{1-x}{1+x}, du=\frac{-1-x-(1-x)}{1+2x+x^2}dx=\frac{-2}{(1+x)^2}dx)\\ &( xu+u=1-x, x(u+1)=1-u, x=\frac{1-u}{1+u})\\ &=-\frac{1}{2}\int \sqrt{u} (1+x^2) du = -\frac{1}{2}\int \sqrt{u} (1+(\frac{1-2u+u^2}{1+2u+u^2})^2) du \\ &=-\frac{1}{2}\int \sqrt{u} (\frac{2+2u^2}{1+2u+u^2})^2 du \\ &=-2\int\sqrt{u}\frac{(1+u^2)^2}{(1+u)^4}du \end{aligned}$
Solve
$\begin{aligned} &\int \sqrt{\frac{1-x}{1+x}} dx =\int \sqrt{\frac{(1-x)(1-x)}{(1+x)(1-x)}} dx\\ &=\int \sqrt{\frac{(1-x)^2}{1-x^2}} dx =\int \frac{1-x}{\sqrt{1-x^2}}dx\\ &=\int \frac{1}{\sqrt{1-x^2}}dx-\int \frac{x}{\sqrt{1-x^2}}dx\\ & (u=1-x^2, du=-2xdx)\\ &=\sin^{-1}x+\frac{1}{2}\int\frac{1}{\sqrt{u}}du=\sin^{-1}x+\frac{1}{2}\int u^{-\frac{1}{2}}du\\ &=\sin^{-1}x+\sqrt{1-x^2}+C \end{aligned}$

77. $\int x^{x/ln(x)}dx$

Interesting…
$\begin{aligned} &\int x^{\frac{x}{ln(x)}} dx\\ &( (\frac{x}{ln(x)})'=\frac{ln(x)-1}{(ln(x))^2})\\ &(y=x^{\frac{x}{ln(x)}}, ln(y)=\frac{x}{ln(x)}ln(x))\\ &( \frac{1}{y}y'= \frac{ln(x)-1}{(ln(x))^2}ln(x)+\frac{x}{ln(x)}\frac{1}{x})\\ &( \frac{1}{y}y'= \frac{ln(x)-1}{ln(x)}+\frac{1}{ln(x)}=1)\\ &y'=y\\ &=x^{\frac{x}{ln(x)}}+C \end{aligned}$
Alt.
$\begin{aligned} &\int x^{\frac{x}{ln(x)}} dx = \int (e^{ln x})^{\frac{x}{ln(x)}}dx=\int e^x dx\\ &= e^x+C\\ &(x^{\frac{x}{ln(x)}}=e^x) \end{aligned}$

78. $\int arcsin(sqrt(x))dx$

$\begin{aligned} &\int arcsin(\sqrt x) dx \\ &( \text{ 루트부분을 치환. 부분적분} )\\ &( u = \sqrt x, u^2=x, 2udu=dx)\\ &=2\int u sin^{-1}(u) du= (arcsin은 미분가능)\\ &=2( sin^{-1}u \frac{1}{2}u^2-\int \frac{1}{\sqrt{1-u^2}}\frac{1}{2}u^2 du )\\ &=u^2sin^{-1}u-\int \frac{u^2}{\sqrt{1-u^2}}du\\ &=u^2sin^{-1}u+\int \frac{-1+1-u^2}{\sqrt{1-u^2}}du\\ &=x\sin^{-1}\sqrt{x}-\int \frac{1}{\sqrt{1-u^2}}du+\int \sqrt{1-u^2}du\\ &=x\sin^{-1}\sqrt{x}-\sin^{-1}\sqrt{x}+\int \sqrt{1-u^2}du\\ \end{aligned}$

$\int \sqrt{1-u^2}du, (u=sin\theta, du=cos\theta d\theta)\\ =\int cos^2\theta d\theta=\frac{1}{2}\int1+cos2\theta d\theta=\frac{1}{2}\theta+\frac{1}{2}\int cos 2\theta d\theta\\ =\frac{1}{2}\theta+\frac{1}{4} sin 2\theta =\frac{1}{2}sin^{-1}u+\frac{1}{2} sin \theta cos \theta\\ =\frac{1}{2}sin^{-1}u+\frac{1}{2} u \sqrt{1-u^2}$

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

79. $\int arctan(x) dx$

$\begin{aligned} &\int arctan(x) dx \\ &(y=arctan(x), x=tan(y), dx=sec^2ydy)\\ &=\int y sec^2y dy (부분적분)\\ &=y \tan(y)-\int tan(y) dy=x\arctan{x}+ln|cos y|+C\\ &=x\arctan{x}+ln|\frac{1}{\sqrt{1+x^2}}|+C\\ \end{aligned}$
Alt.
$\begin{aligned} &\int arctan(x) dx \\ &(\text{we know} \int \frac{1}{1+x^2} dx=arctan(x))\\ &=arctan(x) x -\int \frac{x}{1+x^2} dx (x^2을 치환)\\ &=x \arctan{x}-\frac{1}{2}ln|1+x^2|+C \end{aligned}$

80. $\int_0^5 f(x) dx$

if x<=2, f(x)=10
if x>=2, f(x)=$3x^2-2$
$\begin{aligned} &=\int_0^2 f(x)dx+ \int_2^5 f(x) dx \\ &=\int_0^2 10 dx+ \int_2^5 3x^2-2 dx \\ &=20+\bigg[x^3-2x \bigg ]_2^5=20+(115-4)\\ &=131 \end{aligned}$

Author: crazyj7@gmail.com

51. $\int \sec^6x dx$

$\begin{aligned} &\int \sec^6x dx =\int \sec^2x sec^4 x dx\\ &(D( tanx ) \rightarrow sec^2x) \\ &=sec^4x tanx-\int 4sec^3xsecxtanxtanxdx\\ &=sec^4xtanx-4\int sec^4xtan^2x dx\\ &=sec^4xtanx-4\int sec^4xsec^2x-sec^4x dx\\ &=sec^4xtanx-4\int sec^6x dx+4\int sec^4x dx\\ &5\int sec^6x dx = sec^4xtanx+4\int sec^4 dx\\ &\int sec^4 dx=\int \sec^2x (1+\tan^2x) dx=\int sec^2x dx+\int sec^2xtan^2xdx\\ &=\tan{x}+\int sec^2xtan^2x dx\\ &\int sec^2xtan^2x dx (u=tan x, du= sec^2x dx)\\ &=\int u^2 du = \frac{1}{3}u^3=\frac{1}{3}\tan^3{x}\\ &5\int sec^6x dx = sec^4xtanx+4\int sec^4 dx\\ &=sec^4xtanx+4(tan x+\frac{1}{3}\tan^3{x})\\ &\therefore \int sec^6x dx=\frac{1}{5}sec^4xtanx +\frac{4}{5}tanx+\frac{4}{15}tan^3x +C\\ &=\frac{1}{5}(1+2tan^2x+tan^4x)tanx +\frac{4}{5}tanx+\frac{4}{15}tan^3x +C\\ &=\frac{1}{5}tan^5x+\frac{5}{5}tanx+\frac{10}{15}tan^3x +C\\ &=\frac{1}{5}tan^5x+\frac{2}{3}tan^3x+tan {x} +C\\ \end{aligned}$
Alternative

$\begin{aligned} &\int \sec^6x dx =\int (\sec^2x)^2 sec^2 x dx=\int (1+\tan^2x)^2 sec^2 x dx\\ &(u=tanx, du=sec^2x dx)\\ &=\int (1+u^2)^2 du = \int u^4+2u^2+1du=\frac{1}{5}u^5+\frac{2}{3}u^3+u\\ &=\frac{1}{5}tan^5x+\frac{2}{3}tan^3x+tan {x} +C\\ \end{aligned}$

52. $\int 1/(5x - 2)^4 dx$

$\begin{aligned} &\int \frac{1}{(5x - 2)^4}dx (u=5x-2, du=5dx)\\ &=\frac{1}{5}\int u^{-4}du=-\frac{1}{15}u^{-3}=-\frac{1}{15}(5x-2)^{-3}+C \end{aligned}$

53. $\int ln (1+x^2) dx$

$\begin{aligned} &\int ln (1+x^2)dx =\int 1*ln (1+x^2)dx \\ &=(ln|1+x^2|)(x)-\int \frac{2x^2}{1+x^2} dx\\ &=xln|1+x^2|-2\int \frac{x^2+1-1}{1+x^2} dx\\ &=xln|1+x^2|-2\int 1-\frac{1}{1+x^2} dx\\ &=xln|1+x^2|-2x+2\int\frac{1}{1+x^2} dx\\ &=xln|1+x^2|-2x+2 \arctan{x}+C\\ \end{aligned}$

54. $\int \frac{1}{x^4+x} dx$

$\begin{aligned} &\int \frac{1}{x^4+x} dx =\int \frac{1}{x(x^3+1)}dx \\ &=\int \frac{1}{x(x+1)(x^2-x+1)}dx \\ &=\int \frac{1}{x}+\frac{-\frac{1}{3}}{x+1}+\frac{-\frac{2}{3}x+\frac{1}{3}}{x^2-x+1}dx \\ &=ln|x|-\frac{1}{3}ln|x+1|-\frac{2}{3} \int \frac{x-\frac{1}{2}}{(x-\frac{1}{2})^2+\frac{3}{4}}dx\\ &=ln|x|-\frac{1}{3}ln|x+1|-\frac{1}{3} ln |x^2-x+1|+C\\ &=\frac{1}{3}ln|x^3|-\frac{1}{3}ln|x+1|-\frac{1}{3} ln |x^2-x+1|+C\\ &=\frac{1}{3}ln|x^3\frac{1}{(x+1)}\frac{1}{(x^2-x+1)}+C\\ &=\frac{1}{3}ln|\frac{x^3}{x^3+1}|+C=-\frac{1}{3}ln|\frac{x^3+1}{x^3}|+C\\ &=-\frac{1}{3}ln|1+x^{-3}|+C \end{aligned}$
Alternative (미분형태가 나오도록… 차수를 1차이나게…)
$\begin{aligned} &\int \frac{1}{x^4+x} dx =\int \frac{1}{x^4(1+x^{-3})}dx \\ &=\int \frac{x^{-4}}{1+x^{-3}}dx (u=1+x^{-3}, du=-3x^{-4}dx)\\ &=-\frac{1}{3}\int \frac{1}{u}dx=-\frac{1}{3}ln|u|=-\frac{1}{3}ln|1+x^{-3}|+C \end{aligned}$

55. $\int \frac{1-tan x }{1+tan x} dx$

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

$\frac{1}{2}ln|1+sin(2x)|=ln|\sqrt{1+2sinxcosx}|\\ =ln|\sqrt{cos^2x+sin^2x+2sinxcosx}|\\ =ln|cos x + sin x|$

56. $\int x·sec(x)·tan(x) dx$

$\begin{aligned} &\int x \sec{x} \tan{x} dx dx (D sec \rightarrow sec x tan x) \\ &=x sec x - \int sec x dx\\ &=x sec x - ln|sec x+tan x | + C\\ \end{aligned}$
Check…
$D(x \sec x - ln|\sec x+\tan x | ) \\ = \sec x+x(\sec x \tan x)-\frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x } \\ = \sec x + x \sec x \tan x - \sec x = x \sec x \tan x$

57. $\int arcsec(x) dx$

$\begin{aligned} &\int arcsec(x) dx \\ &( y=arcsec(x), x=sec(y), dx=sec(y)tan(y)dy )\\ &=\int y sec(y) tan (y) dy = y sec(y) - \int sec(y) dy \\ &=y sec(y) - ln | sec(y)+tan(y) | \\ &(R.T angle=y, a=1, h=x, o=sqrt(x^2-1) )\\ &=x arcsec(x)-ln |x+tan(arcsec(x)) |+C \\ &=x arcsec(x)-ln |x+\sqrt{x^2-1} |+C \\ \end{aligned}$
Alt.

$D(\arcsin{x}) \\ ( y=arcsec(x), x=sec(y), dx=sec(y)tan(y)dy )\\ =\frac{dy}{dx} = \frac{1}{sec(y)tan(y)}=cos^2y \csc y\\ =(\frac{1}{x})^2 \frac{x}{\sqrt{x^2-1}}=\frac{1}{x\sqrt{x^2-1}}$

$\int arcsec(x) dx = arcsec(x)(x)-\int \frac{1}{x \sqrt{x^2-1}} x dx \\ =x arcsec(x)-\int \frac{1}{\sqrt{x^2-1}}dx$

$\int \frac{1}{\sqrt{x^2-1}}dx (x=sec\theta, dx=sec\theta tan\theta d \theta)\\ =\int \frac{1}{tan \theta} sec\theta tan\theta d \theta\\ =\int sec \theta d \theta =ln|sec \theta+tan \theta|$

$\int arcsec(x) dx =x arcsec(x)-ln | sec \theta+tan\theta| \\ (R.T angle=\theta, a=1, h=x, o=sqrt(x^2-1))\\ \int arcsec(x) dx =x arcsec(x)-ln | x+\sqrt{x^2-1}| +C$

58. $\int (1 - cos(x))/(1 + cos(x)) dx$

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

$\int \frac{1}{sin^2x} dx=\int \frac{cos^2x + sin^2x}{sin^2x}\\ =\int \frac{cos^2x}{sin^2x} dx +x = \int cot^2x dx +x\\ =\int csc^2x-1 dx+x =\int csc^2x dx = -cot x \\ \int \frac{cos x}{sin^2x} dx (u=sin x, du=cos x dx)\\ =\int \frac{du}{u^2} = -\frac{1}{u}=-csc x\\ \int \frac{cos^2x}{sin^2x} dx =\int cot^2x dx =\int csc^2x-1dx\\ =-cot x - x$

$=\int \frac{1}{sin^2x}dx-2\int \frac{cos x}{sin^2 x}dx+\int \frac{cos^2x}{sin^2x} dx \\ = -cot x +2 csc x -cot x - x\\ =2 csc x -2cot x -x +C =2 (csc x -cot x) -x +C\\ =2(\frac{1-cos x}{sin x})-x+C\\$

$=2(\frac{\frac{1-cosx}{2}}{\frac{sinx}{2}})-x+C\\ =2(\frac{sin^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}/2} )-x+C\\ =2tan(\frac{x}{2})-x+C$

Alt.
$\begin{aligned} &\int \frac{1 - cos(x)}{1 + cos(x)}dx \\ &( sin^2(\frac{x}{2})=\frac{1-cosx}{2} , cos^2(\frac{x}{2})=\frac{1+cosx}{2} )\\ &=\int \frac{sin^2(\frac{x}{2})}{cos^2(\frac{x}{2})} dx =\int \frac{1-cos^2(\frac{x}{2})}{cos^2(\frac{x}{2})} dx \\ &=\int sec^2{\frac{x}{2}} dx-x \\ &=2tan(\frac{x}{2})-x+C \end{aligned}$

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

$\begin{aligned} &\int x^2\sqrt{x+4}dx \\ &(u=\sqrt{x+4}, du=\frac{1}{2\sqrt{x+4}}dx)\\ &=\int(u^2-4)^2u 2u du=\int 2u^2(u^4-8u^2+16)du\\ &=\int 2u^6-16u^4+32u^2 du\\ &=\frac{2}{7}u^7-\frac{16}{5}u^5+\frac{32}{3}u^3+C\\ &=\frac{2}{7}(x+4)^\frac{7}{2}-\frac{16}{5}(x+4)^\frac{5}{2}+\frac{32}{3}(x+4)^\frac{3}{2}+C\\ \end{aligned}$