[연습문제] 부정적분

1. 다음 적분을 계산하여라.

\((1)\ \displaystyle\int\sqrt{3x+1}dx={2\over9}(3x+1)^{3\over2}\)

\((2)\ \displaystyle\int x(x^2-2)^2dx=\frac{(x^2-2)^3}{6}\)

\((3)\ \displaystyle\int(x^2-2)^2dx=\int(x^4-2x^2+4)dx={x^5\over5}-{2\over3}x^3+4x\)

\((4)\ \displaystyle\int\frac{dx}{5-2x}=-\frac{\ln|5-2x|}{2}\)

\((5)\ \displaystyle\int\frac{\sqrt{x}}{1+x\sqrt{x}}dx={2\over3}\int\frac{dt}{1+t}={2\over3}\ln(1+t)={2\over3}\ln\left(1+x^{3\over2}\right)\)

\((6)\ \displaystyle\int\left(1-{1\over z}\right)^2\frac{dz}{z^2}=-\int(1-t)^2dt=\frac{(1-t)^3}{3}={1\over3}\left(1-{1\over z}\right)^3\)

2. 다음 적분을 계산하여라.

\((1)\ \displaystyle\int\frac{\cos3x}{\sin3x}dx={1\over3}\int\frac{d(\sin3x)}{\sin3x}=\frac{\ln|\sin3x|}{3}\)

\((2)\ \displaystyle\int2\sqrt{7t-13}dt={4\over21}(7t-13)^{3/2}\)

\((3)\ \displaystyle\int(\ln{x}+1)e^{x\ln{x}}dx=\int(\ln{x}+1)x^xdx=\int dt=t=x^x\)

\((4)\ \displaystyle\int\frac{5x-1}{5x^2-2x+1}dx={1\over2}\int\frac{(5x^2-2x+1)'}{5x^2-2x+1}dx=\frac{\ln(5x^2-2x+1)}{2}\)

\((5)\ \displaystyle\int\frac{\tan^4x}{\cos^2x}dx=\int\tan^4xd(\tan{x})=\frac{\tan^5x}{5}\)

\((6)\ \displaystyle\int3^{\sin{x}}\cos{x}dx=\int3^{\sin{x}}d(\sin{x})=\frac{3^{\sin{x}}}{\ln3}\)

\((7)\ \displaystyle\int\frac{e^{\sqrt{x}}}{\sqrt{x}}dx=2\int e^{\sqrt{x}}d(\sqrt{x})=2e^{\sqrt{x}}\)

\((8)\ \displaystyle\int\frac{15x^4-2x^2-55x}{3x^2+11}dx=\int\left(5x^2-19-\frac{55x-209}{3x^2+11}\right)dx\)

\(\displaystyle={5\over3}x^3-19x-{55\over6}\ln(3x^2+11)+19\sqrt{11\over3}{\rm Tan}^{-1}\left(\sqrt{3\over11}x\right)\)

\((9)\ \displaystyle\int\frac{e^x\csc^2e^x}{\cot{e^x}-1}dx=\int\frac{\csc^2t}{\cot{t}-1}dt=-\int\frac{ds}{s-1}=-\ln|s-1|=-\ln|\cot{t}-1|\)

\(=-\ln|\cot{e^x}-1|\)

\((10)\ \displaystyle\int\frac{\sec{x}\tan{x}}{\cos^2x}dx=-\int\frac{d(\cos{x})}{\cos^4x}={1\over3\cos^3x}\)

3. 다음 적분을, 부분적분법을 이용하여 계산하여라.

\((1)\ \displaystyle\int e^x\sin{x}dx=e^x\sin{x}-\int e^x\cos{x}dx=e^x\sin{x}-e^x\cos{x}-\int e^x\sin{x}dx\)

\(\displaystyle={1\over2}e^x(\sin{x}-\cos{x})\)

\((2)\ \displaystyle\int ye^ydy=ye^y-\int e^ydy=e^y(y-1)\)

\((3)\ \displaystyle\int\ln{x}dx=x\ln{x}-\int dx=x(\ln{x}-1)\)

\((4)\ \displaystyle\int t\ln{t}dt={t^2\over2}\ln{t}-\int{t\over2}dt={t^2\over4}(2\ln{t}-1)\)

\((5)\ \displaystyle\int x\sec{x}\tan{x}dx=x\sec{x}-\int\sec{x}dx=x\sec{x}-\ln\left|\tan\left({x\over2}+{\pi\over4}\right)\right|\)

\((6)\ \displaystyle\int t^2\ln{t}dt={t^3\over3}\ln{t}-\int{t^3\over3}dt={t^3\over9}(3\ln{t}-1)\)

\((7)\ \displaystyle\int e^{ax}\sin{bx}dx=\frac{e^{ax}}{a}\sin{bx}-\int\frac{b}{a}e^{ax}\cos{bx}dx\)

\(\displaystyle=\frac{e^{ax}}{a}\sin{bx}-\frac{b}{a^2}e^{ax}\cos{bx}-\frac{b^2}{a^2}\int e^{ax}\sin{bx}dx=\frac{e^{ax}(a\sin{bx}-b\sin{bx})}{a^2+b^2}\)

\((8)\ \displaystyle\int\sec^3wdw=\tan{w}\sec{w}-\int\tan^2w\sec{w}dw=\tan{w}\sec{w}-\int\sec^3wdw+\int\sec{w}dw\)

\(\displaystyle={1\over2}\left\{\tan{w}\sec{w}+\ln\left|\tan\left({w\over2}+{\pi\over4}\right)\right|\right\}\)

4. 다음 적분을 구하여라.

\((1)\ \displaystyle\int\sin(5x+9)=-\frac{\cos(5x+9)}{5}\)

\((2)\ \displaystyle\int\frac{\cos\sqrt{x}}{\sqrt{x}}dx=2\int\cos{t}dt=2\sin{t}=2\sin\sqrt{x}\)

\((3)\ \displaystyle\int\frac{\sin(\ln{x^2})}{x}dx=\int\sin2tdt=-\frac{\cos2t}{2}=-\frac{\cos(2\ln{x})}{2}\)

\((4)\ \displaystyle\int e^{2x}\cos(2x-1)dx={e^{2x}\over2}\cos(2x-1)+\int e^{2x}\sin(2x-1)dx\)

\(\displaystyle={e^{2x}\over2}\cos(2x-1)+{e^{2x}\over2}\sin(2x-1)-\int e^{2x}\cos(2x-1)dx={e^{2x}\over4}\{\sin(2x-1)+\cos(2x-1)\}\)

\((5)\ \displaystyle\int x^4\sec(x^5-3)\tan(x^5-3)dx={1\over5}\int\sec{t}\tan{t}dt={1\over5}\sec{t}={1\over5}\sec(x^5-3)\)

\((6)\ \displaystyle\int\theta\sec\theta^2d\theta={1\over2}\int\sec{t}dt={1\over2}\ln\left|\tan\left({t\over2}+{\pi\over4}\right)\right|={1\over2}\ln\left|\tan\left({\theta^2\over2}+{\pi\over4}\right)\right|\)

5. 다음 적분을 계산하여라. (점화식 참조)

\((1)\ \displaystyle\int\cos^42xdx={1\over2}\int\cos^4tdt={1\over8}\left(\sin{t}\cos^3t+3\int\cos^2tdt\right)={\sin4t\over64}+{\sin2t\over8}+{3\over16}t\)

\(\displaystyle={\sin8x\over64}+{\sin4x\over8}+{3\over8}x\)

\((2)\ \displaystyle\int\sin^33xdx={1\over3}\int\sin^3tdt={1\over3}\int(\cos^2t-1)d(\cos{t})={\cos^3t\over9}-{\cos{t}\over3}\)

\(\displaystyle={\cos^33x\over9}-{\cos3x\over3}\)

\((3)\ \displaystyle\int\sin^6xdx=-{\sin^5x\cos{x}\over6}+{5\over6}\int\sin^4xdx\)

\(\displaystyle=-{\sin^5x\cos{x}\over6}-{5\over24}\sin^3x\cos{x}-{5\over16}\sin{x}\cos{x}+{5\over16}x\)

\(\displaystyle=-{\sin6x\over192}+{9\over192}\sin4x-{15\over64}\sin2x-{5\over16}x\)

\((4)\ \displaystyle\int\sec^43xdx={1\over3}\int\sec^4tdt={1\over3}\int\sec^2t(\tan^2t+1)dt={1\over3}\left({\tan^3t\over3}+\tan{t}\right)\)

\(\displaystyle={\tan^33x\over9}+{\tan3x\over3}\)

\((5)\ \displaystyle\int\cot^6tdt=\int\cot^4t(\csc^2t-1)dt=-{\cot^5t\over5}-\int\cot^2t(\csc^2t-1)dt\)

\((6)\ \displaystyle\int\csc^42xdx={1\over2}\int\csc^4tdt={1\over2}\int\csc^2t(\cot^2t+1)dt={1\over2}\left(-{\cot^3t\over3}-\cot{t}\right)\)

\(\displaystyle=-{\cot^32x\over6}-{\cot2x\over2}\)

\((7)\ \displaystyle\int\sin^52t\cos^22tdt={1\over2}\int\sin^5x\cos^2xdx={1\over2}\left(-{\sin^4x\cos^3x\over7}+{4\over7}\int\sin^3x\cos^2xdx\right)\)

\(\displaystyle={1\over2}\left(-{\cos^7x\over7}+{2\over5}\cos^5x-{\cos^3x\over3}\right)=-{\cos^22t\over14}+{\cos^52t\over5}-{\cos^32t\over6}\)

\((8)\ \displaystyle\int\sin^4\left(\theta\over2\right)\cos^2\left(\theta\over2\right)d\theta=2\int\sin^4x\cos^2xdx=2\left(-{\sin^3x\cos^3x\over6}+{1\over2}\int\sin^2x\cos^2xdx\right)\)

\(\displaystyle=2\left\{-{\sin^3x\cos^3x\over6}+{1\over2}\left(-{\sin4x\over32}+{x\over8}\right)\right\}={\sin6x\over96}-{\sin4x\over32}-{\sin2x\over32}+{x\over8}\)

\(\displaystyle={\sin3\theta\over96}-{\sin2\theta\over32}-{\sin\theta\over32}+{\theta\over16}\)

6. 다음 적분을 계산하여라.

\((1)\ \displaystyle\int2x\sqrt{3x-1}dx={2\over9}\int\left(t\sqrt{t}+\sqrt{t}\right)dt={6\over165}t\sqrt{t}(3t+5)={6\over135}(3x-1)\sqrt{3x-1}(9x+2)\)

\((2)\ \displaystyle\int\frac{\cos{x}}{4+\sin^2x}dx=\int\frac{dt}{4+t^2}={1\over2}{\rm Tan}^{-1}{t\over2}={1\over2}{\rm Tan}^{-1}{\sin{x}\over2}\)

\((3)\ \displaystyle\int\frac{x}{\sqrt[3]{5x-6}}dx={1\over25}\int\frac{t+6}{\sqrt[3]{t}}dt={3\over125}\sqrt[3]{t^2}(t+15)={3\over125}\sqrt[3]{(5x-6)^2}(5x+9)\)

\((4)\ \displaystyle\int\frac{e^{7x}}{\sqrt{1-e^{14x}}}dx={1\over7}\int\frac{dt}{\sqrt{1-t^2}}={1\over7}{\rm Sin}^{-1}t={1\over7}{\rm Sin}^{-1}e^{7x}\)

\((5)\ \displaystyle\int\frac{x^2}{\sqrt{ax+b}}dx={1\over a^2}\int\left(t\sqrt{t}-2b\sqrt{t}+{b^2\over\sqrt{t}}\right)dt={2\sqrt{t}\over15a^2}(3t^2-10bt+15b^2)\)

\(=\dfrac{2\sqrt{ax+b}}{15a^2}(3a^2x^2-4abx+8b^2)\)

\((6)\ \displaystyle\int\frac{dx}{x^2+4x+9}=\int\frac{dx}{(x+2)^2+(\sqrt{5})^2}={1\over\sqrt{5}}{\rm Tan}^{-1}{x+2\over\sqrt{5}}\)

7. 다음 적분을 계산하여라.

\((1)\ \displaystyle\int\frac{x^2}{\sqrt{a^2-x^2}}=a^2\int\sin^2tdt=a^2\left(-{\sin{t}\cos{t}\over2}+{t\over2}\right)=-\frac{x\sqrt{a^2-x^2}}{2}+{a^2\over2}{\rm Sin}^{-1}{x\over a}\)

\((2)\ \displaystyle\int\frac{dx}{(x^2+9)^{3\over2}}={1\over9}\int\cos{t}dt={1\over9}\sin{t}=\frac{dx}{9\sqrt{x^2+9}}\)

\((3)\ \displaystyle\int\frac{\sqrt{x^2-4}}{x^2}dx=-\frac{\sqrt{x^2-4}}{x}+\int\frac{dx}{\sqrt{x^2-4}}=-\frac{\sqrt{x^2-4}}{x}+\ln\left|x+\sqrt{x^2-4}\right|\)

\((4)\ \displaystyle\int\frac{dx}{x^2\sqrt{x^2-a^2}}={1\over a^2}\int\cos{t}dt={\sin{t}\over a^2}=\frac{\sqrt{x^2-a^2}}{a^x}\)

8. 다음 적분을 계산하여라.

\((1)\ \displaystyle\int\frac{dx}{x^2-x}=\int\left({1\over x-1}-{1\over x}\right)dx=\ln\left|\frac{x-1}{x}\right|\)

\((2)\ \displaystyle\int\frac{4x-19}{2x^2+7x-15}dx=\int\left({3\over x+5}-{2\over2x-3}\right)dx=3\ln|x+5|-\ln|2x-3|\)

\((3)\ \displaystyle\int\frac{2x^2+x-3}{x^2-2x}dx=\int\left\{2+{3\over2x}-{7\over2(x-2)}\right\}dx=2x+{3\over2}\ln|x|-{7\over2}\ln|x-2|\)

\((4)\ \displaystyle\int\frac{4x^2+9x-1}{x^3+2x^2-x-2}dx=\int\left({3\over x+1}+{2\over x-1}+{1\over x+2}\right)dx\)

\(=3\ln|x+1|+2\ln|x-1|-\ln|x+2|\)

9. 다음 적분을 계산하여라.

\((1)\ \displaystyle\int\frac{dx}{2+\cos{x}}=2\int\frac{dt}{(\sqrt{3})^2+t^2}={2\over\sqrt{3}}{\rm Tan}^{-1}{t\over\sqrt{3}}={2\over\sqrt{3}}{\rm Tan}^{-1}{\tan{x\over2}\over\sqrt{3}}\)

\((2)\ \displaystyle\int\frac{\sin{x}}{2+\cos{x}}dx=-\int\frac{dt}{t+2}=-\ln|t+2|=-\ln|\cos{x}+2|\)

\((3)\ \displaystyle\int\frac{dx}{4\sin{x}+3\cos{x}}=-2\int\frac{dt}{3t^2-8t-3}=-2\int\frac{dt}{(3t+1)(t-3)}\)

\(\displaystyle=-{1\over5}\int\left({1\over t-3}-{3\over3t+1}\right)dt={\ln|3t+1|\over5}-{\ln|t-3|\over5}={\ln\left|3\tan{x\over2}+2\right|\over5}-{\ln\left|\tan{x\over2}-3\right|\over5}\)

\((4)\ \displaystyle\int\frac{\cot{x}}{1+\cos{x}}dx={1\over2}\int\left({1\over t}-t\right)dt={\ln|t|\over2}-{t^2\over4}={\ln\left|\tan{x\over2}\right|\over2}-{\tan^2{x\over2}\over4}\)