Integral100 [1-10]

Integral problems

1. $\int \tan^5x \sec^3x dx$

cf) $\frac{d}{dx} \sec{x}=\sec{x} \tan{x}$
$1+\tan^2{x} = \sec^2{x}=\frac{d}{dx} \tan{x}$
use $\sec{x} \tan{x}$ part.
$\begin{aligned} &\int \tan^5x \sec^3x dx\\ &=\int tan^4xsec^2x \tan{x} \sec{x} dx\\ &(u = \sec{x} , du = \sec{x}\tan{x}dx)\\ &=\int tan^4xsec^2x du\\ &=\int (\sec^2{x}-1)^2sec^2{x} du \\ &=\int (u^2-1)^2u^2du\\ &=\int (u^4-2u^2+1)u^2du\\ &=\int u^6-2u^4+u^2du\\ &=\frac{1}{7}u^7-\frac{2}{5}u^5+\frac{1}{3}u^3+C\\ &=\frac{1}{7}{\sec^7{x}}-\frac{2}{5}{\sec^5{x}}+\frac{1}{3}{\sec^3{x}}+C \end{aligned}$

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2. $\int \frac{\cos{2x}}{\sin{x}+\cos{x}}dx$

cf) $\cos{2x}=\cos^2{x}-\sin^2{x}=1-2\sin^2{x}=2cos^2{x}-1$

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

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3. $\int\frac{x^2+1}{x^4-x^2+1} dx$

cf)
$(x^4-x^2+1)(x^4+x^2+1)=\\ x^8-x^6+x^4+ x^6-x^4+x^2 +x^4-x^2+1\\ =x^8+x^4+1$
부분 분수로 나눠보자.
$\frac{x^2+1}{x^4-x^2+1}\\ =\frac{c}{x^2+ax+1}+\frac{d}{x^2+bx+1}\\ (미지수를 구한다)\\ =\frac{\frac{1}{2}}{x^2+\sqrt{3}x+1}+\frac{\frac{1}{2}}{x^2-\sqrt{3}x+1}$
따라서 적분을 취하면.
$\int\frac{x^2+1}{x^4-x^2+1} dx\\ =\frac{1}{2}\int \frac{1}{x^2+\sqrt{3}x+1}dx+\frac{1}{2}\int \frac{1}{x^2-\sqrt{3}x+1}dx\\ =\frac{1}{2}\int \frac{1}{ (x+\frac {\sqrt{3}}{2})^2+\frac{1}{4}}dx+\frac{1}{2}\int \frac{1}{ (x-\frac {\sqrt{3}}{2})^2+\frac{1}{4}}dx$
먼저 왼쪽 부분을 계산해 보자. 제곱의 형태를 삼각치환해 보자. (1+tan^x 꼴로 만든다.)
$\frac{1}{2}\tan\theta=x+\frac{\sqrt{3}}{2}$
$dx = \frac{1}{2}\sec^2\theta d\theta$
We know $\int \frac{1}{1+x^2}dx = \tan^{-1}x+C$.
$\frac{1}{2}\int \frac{1}{ (x+\frac {\sqrt{3}}{2})^2+\frac{1}{4}}dx\\ =\frac{1}{2}\int \frac{1}{\frac{1}{4}\tan^2\theta+\frac{1}{4}}dx\\ =\frac{1}{2}\int \frac{1}{\frac{1}{4}\sec^2\theta} \frac{1}{2}\sec^2\theta d\theta \\ =\theta+C = \tan^{-1}(2x+\sqrt{3})+C$
오른쪽 부분도 같은 방식으로 계산하면 된다.
$\frac{1}{2}\int \frac{1}{ (x-\frac {\sqrt{3}}{2})^2+\frac{1}{4}}dx\\ = \tan^{-1}(2x-\sqrt{3})+C$
$\therefore \int\frac{x^2+1}{x^4-x^2+1} dx\\ =\tan^{-1}(2x+\sqrt{3})+\tan^{-1}(2x-\sqrt{3})+C$

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Check!!!

$\begin{aligned} &\frac{d}{dx} \tan^{-1}(2x+\sqrt{3})+\tan^{-1}(2x-\sqrt{3})+C\\ &= \frac{1}{1+(2x+\sqrt{3})^2}2+\frac{1}{1+(2x-\sqrt{3})^2}2\\ &= \frac{2}{4+4\sqrt{3}x+4x^2}+\frac{2}{4-4\sqrt{3}x+4x^2}\\ &= \frac{1}{2+2\sqrt{3}x+2x^2}+\frac{1}{2-2\sqrt{3}x+2x^2}\\ &= \frac{4x^2+4}{4x^4-4x^2+4}\\ &= \frac{x^2+1}{x^4-x^2+1} \end{aligned}$

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• 다른 솔루션. (divide by x^2)
  $\begin{aligned} &\frac{x^2+1}{x^4-x^2+1}= \frac{1+\frac{1}{x^2} }{x^2-1+\frac{1}{x^2}}\\ &=\frac{1+\frac{1}{x^2} }{x^2-2+\frac{1}{x^2}+1}\\ &=\frac{1+\frac{1}{x^2} }{ (x-\frac{1}{x})^2+1}\\ \end{aligned}$
  $u=x-\frac{1}{x} \quad du=(1+\frac{1}{x^2}) dx$

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

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

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

5. $\int \csc^3{x} \sec{x} dx$

$\begin{aligned} &\int \csc^3{x} \sec{x} dx \\ &= \int \frac{1}{\sin^3{x}\cos{x}} dx\\ &= \int \frac{\cos^2x+\sin^2x}{\sin^3{x}\cos{x}} dx\\ &= \int \frac{\cot^2x+1}{\sin{x}\cos{x}} dx\\ &=\int \frac{\cot^2x}{\sin{x}\cos{x}}dx+\int \frac{1}{\sin{x}\cos{x}} dx\\ &= \int \frac{1}{\sin{x}\cos{x}} dx + \int \frac{\cos^2{x}}{\sin^2{x}\sin{x}\cos{x}} dx\\ \end{aligned}$
(이하는 아래 계산 과정과 동일)

$1+\tan^2{x}=sec^2{x}$
$1+\cot^2{x}=csc^2{x}$

$\begin{aligned} &\int \csc^3{x} \sec{x} dx \\ &= \int (1+\cot^2{x})\csc{x} \sec{x} dx\\ &= \int \csc{x} \sec{x} + \cot^2{x}\csc{x} \sec{x} dx\\ &= \int \csc{x} \sec{x} dx + \int \cot^2{x}\csc{x} \sec{x} dx\\ &= \int \frac{1}{\sin{x}\cos{x}} dx + \int \frac{\cos^2{x}}{\sin^2{x}\sin{x}\cos{x}} dx\\ &= \int \frac{\cos{x}}{\sin{x}}+\frac{\sin{x}}{\cos{x}} dx + \int \frac{\cos{x}}{\sin^3{x}} dx\\ &=\int \cot{x} dx + \int \tan{x} dx +\int \frac{\cos{x}}{\sin^3{x}} dx\\ &=\ln|\sin{x}| - \ln |\cos{x}|+\int \frac{\cos{x}}{\sin^3{x}} dx\\ \end{aligned}$

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

So,
$=\ln|\sin{x}| - \ln |\cos{x}|+\int \frac{\cos{x}}{\sin^3{x}} dx\\ =\ln|\sin{x}| - \ln |\cos{x}|-\frac{1}{2\sin^2{x}}+C\\ =\ln|\sin{x}| - \ln |\cos{x}|-\frac{1}{2}\csc^2{x}+C\\ =-\frac{1}{2}\csc^2{x}+\ln|\sin{x}| - \ln |\cos{x}|+C\\ =-\frac{1}{2}\csc^2{x}+\ln|\tan{x}| +C$

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6. $\int\frac{\cos{x}}{\sin^2{x}-5\sin{x}-6}dx$

$\begin{aligned} &\int\frac{\cos{x}}{\sin^2{x}-5\sin{x}-6}dx\\ &=\int \frac{\cos{x}}{(\sin{x}-6)(\sin{x}+1)}dx\\ & (u = \sin{x} \quad du = \cos{x} dx) \\ &=\int \frac {1} {(u-6)(u+1)} du\\ & \frac {1} {(u-6)(u+1)} = \frac{a}{u-6}+\frac{b}{u+1}\\ &a+b=0, a-6b=1, 7a=1, a=\frac{1}{7}, b=-\frac{1}{7}\\ &=\frac{1}{7}\int \frac{1}{u-6}dx -\frac{1}{7}\int \frac{1}{u+1} dx \\ &=\frac{1}{7} \ln|u-6|-\frac{1}{7}\ln|u+1|+C\\ &=\frac{1}{7} \ln |\sin{x}-6|-\frac{1}{7}\ln|\sin{x}+1|+C\\ &=\frac{1}{7} \ln \big | \frac{\sin{x}-6}{\sin{x}+1} \big |+C \end{aligned}$

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7. $\int \frac{1}{\sqrt{e^x}} dx$

$\begin{aligned} &\int \frac{1}{\sqrt{e^x}} dx\\ &= \int e^{-\frac{x}{2}} dx \\ &= \frac{1}{-\frac{1}{2}} e^{-\frac{x}{2}}\\ &=-2e^{-\frac{x}{2}}\\ &= -\frac{2}{\sqrt{e^x}}+C \end{aligned}$

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8. $\int \frac{e^x \sqrt{e^x-1}}{e^x+3} dx$

$\begin{aligned} &\int \frac{e^x \sqrt{e^x-1}}{e^x+3} dx\\ &u=\sqrt{e^x-1}, \quad du=\frac{1}{2} \frac{1}{\sqrt{e^x-1} } e^x dx\\ Q&=\int \frac{u}{u^2+4} 2 \sqrt{e^x-1} du\\ &=2\int \frac{u^2}{u^2+4} du\\ &=2\int \frac{u^2+4-4}{u^2+4} du\\ &=2\int 1-\frac{4}{u^2+4} du \\ &=2 \left [ u-4\int \frac{1}{u^2+2^2} du \right ] +C\\ &=2 \left [ u-4 \frac{1}{2} \arctan \frac{u}{2} \right ] +C\\ &=2u-4\arctan \frac{u}{2} +C \\ &=2\sqrt{e^x-1}-4\arctan \frac{\sqrt{e^x-1}}{2} +C \\ \end{aligned}$
Again…
$\begin{aligned} &\int \frac{e^x \sqrt{e^x-1}}{e^x+3} dx\\ & (u=e^x-1, \quad du=e^xdx) \\ Q&=\int \frac{\sqrt{u}}{u+4} du\\ & (t = \sqrt{u} , \quad dt = \frac{1}{2\sqrt{u}} du, du=2\sqrt{u}dt) \\ &=2\int \frac{t^2}{t^2+4} dt =2\int \frac{1}{1+\frac{4}{t^2}} dt=2\int \frac{1}{1+(\frac{2}{t})^2} dt\\ &(s=\frac{2}{t}, t=\frac{2}{s} , ds =-\frac{2}{t^2}dt )\\ &=2\int \frac{1}{1+s^2} (-\frac{t^2}{2}) ds \\ &=-\int \frac{t^2}{1+s^2} ds =-\int \frac{\frac{4}{s^2}}{1+s^2} ds\\ &=-4\int \frac{1}{s^2(1+s^2)}ds =-4\int \frac{1}{s^2}-\frac{1}{1+s^2} ds \\ &=-4\int\frac{1}{s^2} ds+4 \int \frac{1}{1+s^2} ds \\ &=-4(-1)\frac{1}{s}+4 \arctan{s} +C \\ &=\frac{4}{\frac{2}{t}} + 4 \arctan{ \frac{2}{t} } +C =2t + 4 \arctan \frac{2}{t} +C \\ &= 2 \sqrt{u} + 4 \arctan \frac{2}{\sqrt{u}} +C\\ &= 2 \sqrt{e^x-1} + 4 \arctan \frac{2}{\sqrt{e^x-1}} +C\\ \end{aligned} \\Fail\\ \text{Where is Incorrect?} \\$

Where is incorrect?? … No. It’s all right.
$\arctan{\frac{1}{x}} = \frac{\pi}{2}-\arctan{x} ,(x>0)$
So, Integration constant is ignored.

$\begin{aligned} &= 2 \sqrt{e^x-1} + 4 \arctan \frac{2}{\sqrt{e^x-1}} +C\\ &= 2 \sqrt{e^x-1} + 4 ( \frac{\pi}{2}-\arctan \frac{\sqrt{e^x-1}}{2}) +C\\ &= 2 \sqrt{e^x-1} - 4\arctan \frac{\sqrt{e^x-1}}{2}+2\pi +C\\ &= 2 \sqrt{e^x-1} - 4\arctan \frac{\sqrt{e^x-1}}{2} +C_2\\ \end{aligned}$

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9. $\int \frac{1}{x+\sqrt{x}} dx$

$\begin{aligned} &\int \frac{1}{x+\sqrt{x}} dx\\ & (t=\sqrt{x}, t^2=x, dt=\frac{1}{2\sqrt{x}}dx)\\ &=\int \frac{1}{t^2+t}{2\sqrt{x}}dt=\int \frac{2t}{t^2+t}dt\\ &=2\int \frac{1}{1+t}dt=2\ln|1+t|+C\\ &=2\ln|1+\sqrt{x}|+C \end{aligned}$

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10. $\int_{-1}^{5}|x-3| dx$

$\begin{aligned} &\int_{-1}^{5}|x-3| dx \\ &=\int_{-1}^{3} |x-3|dx + \int_{3}^{5} |x-3| dx\\ &=\int_{-1}^{3} 3-x dx + \int_{3}^{5} x-3 dx\\ &=\left[ 3x-\frac{x^2}{2} \right]_{-1}^{3} + \left[ \frac{x^2}{2}-3x \right]_{3}^{5} \\ &=(9-\frac{9}{2})-(-3-\frac{1}{2})+(\frac{25}{2}-15)-(\frac{9}{2}-9)\\ &=\frac{9}{2}+\frac{7}{2}-\frac{5}{2}+\frac{9}{2}\\ &=10 \end{aligned}$

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