기본적인 적분

다음 공식은 자주 이용되는 것이다.

\[\int f(ax+b)dx={1\over a}{\rm F}(ax+b)\qquad(a\ne0)\]

여기서 \(\int f(x)dx={\rm F}(x)\) 이다.

<증명> 먼저 \(\int f(x)dx={\rm F}(x)\) 이므로 \({\rm F'}(x)=f(x)\) 이다. 그러므로 다음과 같이 미분계산을 할 수 있다.

\[\left({1\over a}{\rm F}(ax+b)\right)={\rm F'}(ax+b)=f(ax+b)\]

따라서 위의 공식이 증명되었다.

위의 공식을 이용해서, 부정적분의 기본공식을 생각하면, 다음 공식을 얻는다. 단, \(a\ne0\)이라 한다.

\[\begin{align}&\int(ax+b)^\alpha dx={1\over a(\alpha+1)}(ax+b)^{\alpha+1}\qquad(\alpha\ne-1)\\&\int\frac{dx}{ax+b}=\frac{\ln|ax+b|}{a}\\&\int e^{ax}dx=\frac{e^{ax}}{a}\\&\int\cos(ax+b)dx=\ \ \,\frac{\sin(ax+b)}{a}\\&\int\sin(ax+b)dx=-\frac{\cos(ax+b)}{a}\end{align}\]

[예제 1] 다음 함수를 적분하여라.

\((1)\ (2x+1)^3\qquad\quad\ (2)\ \dfrac{1}{(2x+1)^3}\qquad(3)\ \dfrac{1}{2x+1}\)

\((4)\ \sqrt{\dfrac{x}{2}+3}\qquad\quad\ \,(5)\ \dfrac{1}{\sqrt{x-1}}\qquad\ \ \ (6)\ \dfrac{1}{\sqrt[3]{1-3x}}\)

\((7)\ \cos(2x+3)\qquad(8)\ \cos^2x\qquad\quad\ \ \ (9)\ (e^{2x}+e^{-2x})^2\)

<풀이>

\((1)\ \displaystyle\int(2x+3)^3dx={1\over2(3+1)}(2x+1)^{3+1}={1\over8}(2x+1)^4\)

\((2)\ \displaystyle\int\frac{dx}{(2x+1)^3}={1\over2(-3+1)}(2x+1)^{-3+1}=-\frac{1}{4(2x+1)^2}\)

\((3)\ \displaystyle\int\frac{dx}{2x+1}=\frac{\ln|2x+1|}{2}\)

\((4)\ \displaystyle\int\sqrt{{x\over2}+3}dx={1\over{1\over2}\left({1\over2}+1\right)}\left({x\over2}+2\right)^{{1\over2}+1}={4\over3}\left({x\over2}+1\right)\sqrt{{x\over2}+1}\)

\((5)\ \displaystyle\int\frac{dx}{x-1}={1\over-{1\over2}+1}(x-1)^{{1\over2}+1}=2\sqrt{x-1}\)

\((6)\ \displaystyle\int\frac{dx}{\sqrt[3]{1-3x}}=-{1\over3\left(-{1\over3}+1\right)}(1-3x)^{-{1\over3}+1}=-{1\over2}\sqrt[3]{(1-3x)^2}\)

\((7)\ \displaystyle\int\cos(2x+3)dx={1\over2}\sin(2x+3)\)

\((8)\ \displaystyle\int\cos^2xdx=\int\frac{\cos2x+1}{2}dx=\frac{\sin2x}{4}+{x\over2}\)

\((9)\ \displaystyle\int(e^{2x}+e^{-2x})^2dx=\int(e^{4x}+2+e^{-4x})={e^{4x}\over4}+2x-{e^{-4x}\over4}\)

한편, 지금까지 얻은 공식들을 이용해서 다음 공식들을 도입할 수 있다.

\[\begin{split}&\int\frac{dx}{a^2+x^2}={1\over a}{\rm Tan}^{-1}{x\over a}\quad(a>0)\\&\int\frac{dx}{\sqrt{a^2-x^2}}={\rm Sin}^{-1}{x\over a}\quad(a>0)\\&\int\frac{dx}{x^2-a^2}={1\over2a}\ln\left|\frac{x-a}{x+a}\right|\\&\int\frac{dx}{\sqrt{x^2+\rm A}}=\ln|x+\sqrt{x^2+\rm A}|\\&\int\frac{dx}{(x-a)(x-b)}={1\over a-b}\ln\left|\frac{x-a}{x-b}\right|\quad(a\ne  b)\\&\int a^xdx=\frac{a^x}{\ln{a}}\quad(a\ne1,\,a>0)\end{split}\]

<증명>

\(\displaystyle\int\frac{dx}{a^2+x^2}={1\over a^2}\int\frac{dx}{1+\left({x\over a}\right)^2}={1\over a}={1\over a}{\rm Tan}^{-1}{x\over a}\)

\(\displaystyle\int \frac{dx}{\sqrt{a^2-x^2}}={1\over a}\int\frac{dx}{\sqrt{1-({x\over a})^2}}={\rm Sin}^{-1}{x\over a}\)

\(\displaystyle\int\frac{dx}{x^2-a^2}=\int{1\over2a}\left({1\over x-a}-{1\over x+a}\right)dx={1\over2a}(\ln|x-a|-\ln|x+a|)={1\over2a}\ln\left|\frac{x-a}{x+a}\right|\)

\(\displaystyle\left(\ln(x+\sqrt{x^2+{\rm A}})\right)'=\frac{1}{x+\sqrt{x^2+\rm A}}(x+\sqrt{x^2+\rm A})=\frac{1}{x+\sqrt{x^2+\rm A}}\left(1+\frac{x}{\sqrt{x^2+\rm A}}\right)\)

\(=\dfrac{1}{\sqrt{x^2+A}}\)

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

\(\displaystyle\int a^xdx=\int e^{x\ln{a}}dx=\frac{e^{x\ln{a}}}{\ln{a}}=\frac{a^x}{\ln{a}}\)

[예제 2] 다음 함수를 적분하여라.

\(\displaystyle(1)\ \frac{1}{\sqrt{4-x^2}}\qquad\qquad\ \ (2)\ \frac{1}{1+9x^2}\qquad\qquad\ \ (3)\ \frac{1}{4-x^2}\)

\(\displaystyle(4)\ \frac{1}{\sqrt{9x^2-5}}\qquad\qquad(5)\ \frac{1}{\sqrt{5+x^2}}\qquad\qquad\,(6)\ \frac{1}{x^2-4x+7}\)

\(\displaystyle(7)\ \frac{1}{x^2-4x+1}\qquad\quad(8)\ \frac{1}{\sqrt{4x-1-x^2}}\qquad(9)\ \frac{1}{\sqrt{x^2-4x+7}}\)

\(\displaystyle(10)\ \frac{1}{x^2-4x+3}\qquad\ \ (11)\ 3^x+\cos{x}\qquad\quad\ \ (12)\ (10^x+1)^2\)

<풀이>

\((1)\ \displaystyle\int\frac{dx}{\sqrt{4-x^2}}={\rm Sin}^{-1}{x\over2}\)

\((2)\ \displaystyle\int\frac{dx}{1+9x^2}={1\over9}\int\frac{dx}{\left({1\over3}\right)^2+x^2}={1\over3}{\rm Tan^{-1}}3x\)

\((3)\ \displaystyle\int\frac{dx}{4-x^2}=-\int\frac{dx}{x^2-2^2}={1\over4}\ln\left|\frac{x+2}{x-2}\right|\)

\((4)\ \displaystyle\int\frac{dx}{9x^2-5}={1\over3}\int\frac{dx}{x^2-{5\over9}}={1\over3}\ln\left|x+\sqrt{x^2-{5\over9}}\right|\)

\((5)\ \displaystyle\int\frac{dx}{\sqrt{5+x^2}}=\ln(x+\sqrt{5+x^2})\)

\((6)\ \displaystyle\int\frac{dx}{x^2-4x+7}=\int\frac{dx}{(x-2)^2+3}={1\over\sqrt{3}}{\rm Tan}^{-1}\frac{x-2}{\sqrt{3}}\)

\((7)\ \displaystyle\int\frac{dx}{x^2-4x+1}=\int\frac{dx}{(x-2)^2-3}={1\over2\sqrt{3}}\ln\left|\frac{x-2-\sqrt{3}}{x-2+\sqrt{3}}\right|\)

\((8)\ \displaystyle\int\frac{dx}{\sqrt{4x-1-x^2}}=\int\frac{dx}{\sqrt{3-(x-2)^2}}={\rm Sin}^{-1}{x-2\over\sqrt{3}}\)

\((9)\ \displaystyle\int\frac{dx}{\sqrt{x^2-4x+7}}=\int\frac{dx}{\sqrt{(x-2)^2+3}}=\ln(x-2+\sqrt{x^2-4x+7})\)

\((10)\ \displaystyle\int\frac{dx}{x^2-4x+3}=\int\frac{dx}{(x-3)(x-1)}={1\over2}\ln\left|\frac{x-3}{x-1}\right|\)

\((11)\ \displaystyle\int(3^x+\cos{x})dx=\frac{3^x}{\ln3}+\sin{x}\)

\((12)\ \displaystyle\int(10^x+1)^2dx=\int(10^{2x}+2\cdot10^x+1)dx=\frac{10^{2x}}{\ln100}+\frac{2\cdot10^x}{\ln10}+x\)

다음 공식들도 자주 이용된다.

\[\begin{split}&\int\frac{f'(x)}{f(x)}dx=\ln|f(x)|\\&\int\{f(x)\}^\alpha f'(x)dx={1\over\alpha+1}\{f(x)\}^{\alpha+1}\quad(\alpha\ne-1)\end{split}\]

<증명>

첫번째 공식의 증명 :

\[(\ln|f(x)|)'=\frac{f'(x)}{f(x)}\]

두번째 공식의 증명 :

\[\left({1\over\alpha+1}\{f(x)\}^{\alpha+1}\right)'={f(x)}^\alpha f'(x)\]

위의 첫번째 공식의 특별한 경우로서 다음 공식이 성립한다.

\[\begin{align}&\int\tan{x}dx=-\ln|\cos{x}|=\ln|\sec{x}|\\&\int\cot{x}dx=\ln|\sin{x}|=-\ln|\csc{x}|\end{align}\]

<증명>

첫번째 공식의 증명 :

\[\int\tan{x}dx=\int\frac{\sin{x}}{\cos{x}}dx=-\int\frac{(\cos{x})'}{\cos{x}}dx=-\ln|\cos{x}|\]

두번째 공식의 증명 :

\[\int\cot{x}dx=\int\frac{\cos{x}}{\sin{x}}dx=\int\frac{(\sin{x})'}{\sin{x}}dx=\ln|\sin{x}|\]

[예제 3] 다음 적분을 구하여라.

\[\int\frac{x+1}{x^2+x+1}dx\]

<풀이>

\[\begin{split}\int\frac{x+1}{x^2+x+1}&={1\over2}\int\left\{\frac{2x+1}{x^2+x+1}+\frac{1}{\left(x+{1\over2}\right)^2+\left({\sqrt{3}\over2}\right)^2}\right\}dx\\&={1\over2}\ln(x^2+x+1)+{1\over\sqrt{3}}{\rm Tan}^{-1}\frac{2x+1}{\sqrt{3}}\end{split}\]

[예제 4] 다음 적분을 구하여라.

\((1)\ \sin{px}\sin{qx}\ (p^2\ne q^2)\qquad(2)\ \sin^2px\ (p^2\ne0)\qquad(3)\ \sin^3px\ (p\ne0)\)

<풀이> 삼각함수 항등식 참조

\((1)\ \begin{split}\int\sin{px}\sin{qx}dx&={1\over2}\int\{\cos(p-q)x-\cos(p+q)\}dx\\&={1\over2(p-q)}\sin(p-q)x-{1\over2(p+q)}\sin(p+q)x\end{split}\)

\((2)\ \displaystyle\int\sin^2pxdx={1\over2}\int(1-\cos2px)dx={x\over2}-{1\over4p}\sin2px\)

\((3)\ \displaystyle\int\sin^3pxdx={1\over4}\int(3\sin{px-\sin3px})dx=-{3\over4p}\cos{px}+{1\over12p}\cos3px\)

《문        제》

1. 다음 함수를 적분하여라.

\(\displaystyle(1)\ (2x+3)^n\ (n\ne-1)\qquad(2)\ \frac{1}{1-t}\qquad\qquad\quad\ \,(3)\ \frac{1}{(a-x)^2}\)

\(\displaystyle(4)\ \frac{1}{x^2-4x+4}\qquad\qquad\ \ \ (5)\ \sqrt{ax}\ (a>0)\qquad\ \ \,(6)\ \sqrt{3-x}\)

\(\displaystyle(7)\ \sqrt[3]{2x+5}\qquad\qquad\qquad\ \ (8)\ \sin(2x-5)\qquad\quad(9)\ \frac{x^3}{x-1}\)

\(\displaystyle(10)\ \frac{2x-1}{x+1}\qquad\qquad\qquad\ (11)\ \tan^2(2x+3)\qquad(12)\ (e^x-e^{-x})^3\)

<풀이>

\(\displaystyle(1)\ \int(2x+3)^ndx=\frac{(2x+3)^n}{2(n+1)}\)

\(\displaystyle(2)\ \int\frac{dt}{1-t}=-\ln|1-t|=\ln\frac{1}{|1-t|}\)

\(\displaystyle(3)\ \int\frac{dx}{(a-x)^2}=\frac{(a-x)^{-2+1}}{-(-2+1)}=\frac{1}{a-x}\)

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

\(\displaystyle(5)\ \int\sqrt{ax}dx=\frac{(ax)^{1\over2+1}}{a\left({1\over2}+1\right)}=\frac{2\sqrt{a}}{3}x\sqrt{x}\)

\(\displaystyle(6)\ \int\sqrt{3-x}dx=\frac{(3-x)^{1\over2+1}}{a\left({1\over2}+1\right)}=-{2\over3}(3-x)\sqrt{3-x}\)

\(\displaystyle(7)\ \int\sqrt[3]{2x+5}dx=\frac{(2x+5)^{{1\over3}+1}}{2\left({1\over3}+1\right)}={3\over8}(2x+5)\sqrt[3]{2x+5}\)

\(\displaystyle(8)\ \int\sin(2x-5)dx={1\over2}\cos(2x-5)\)

\(\displaystyle(9)\ \int\frac{x^3}{x-1}dx=\int\left(x^2+x+1+{1\over x-1}\right)dx={x^3\over3}+{x^2\over2}+\ln|x-1|\)

\(\displaystyle(10)\ \int\frac{2x-1}{x+1}dx=\int\left(2-\frac{3}{x+1}\right)dx=2x-3\ln|x+1|\)

\(\displaystyle(11)\ \int\tan^2(2x+3)dx=\int\{\sec^2(2x+3)-1\}dx=\frac{\tan(2x+3)}{2}-x\)

\(\displaystyle(12)\ \int(e^x-e^{-x})^3dx=\int(e^{3x}-3e^x+3e^{-x}-e^{-3x})dx={e^{3x}\over3}-3e^x-3^{-x}+{e^{-3x}\over3}\)

2. 다음 함수를 적분하여라.

\(\displaystyle(1)\ \frac{x^2-1}{x^2+5}\qquad\qquad\qquad(2)\ \frac{1}{a^2x^2+b^2}\ (ab\ne0)\qquad\qquad\ (3)\ \frac{1}{x^2-5}\)

\(\displaystyle(4)\ \frac{1}{x^2-4x+2}\qquad\quad\ \ \,(5)\ \frac{1}{x^2-5x+4}\qquad\qquad\qquad\quad(6)\ \frac{1}{\sqrt{x^2-4x+2}}\)

\(\displaystyle(7)\ \frac{1}{\sqrt{x^2-5x+4}}\qquad\ \ \,(8)\ \frac{1}{\sqrt{3-2x^2}}\qquad\qquad\qquad\qquad(9)\ (2^x+2^{-x})^2\)

\(\displaystyle(10)\ \frac{1}{\sqrt{2ax-x^2}}\qquad\quad\ (11)\ \frac{1}{(x-a)(x-b)}\ (a<b)\qquad(12)\ \frac{1}{\sqrt{(x-a)(x-b)}}\ (a\ne b)\)

<풀이>

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

\(\displaystyle(2)\ \int\frac{dx}{a^2x^2+b^2}={1\over a^2}\int\frac{dx}{x^2+\left({b\over a}\right)^2}={1\over ab}{\rm Tan}^{-1}{ax\over b}\)

\(\displaystyle(3)\ \int\frac{dx}{x^2-5}={1\over\sqrt{5}}\ln\left|\frac{x-\sqrt{5}}{x+\sqrt{5}}\right|\)

\(\displaystyle(4)\ \int\frac{dx}{x^2-4x+2}=\int\frac{dx}{(x-2)^2-2}={1\over2\sqrt{2}}\ln\left|\frac{x-2-\sqrt{2}}{x-2+\sqrt{2}}\right|\)

\(\displaystyle(5)\ \int\frac{dx}{x^2-5x+4}=\int\frac{dx}{(x-4)(x-1)}={1\over3}\ln\left|\frac{x-4}{x-1}\right|\)

\(\displaystyle(6)\ \int\frac{dx}{x^2-4x+2}=\int\frac{dx}{\sqrt{(x-2)^2-2}}=\ln|x-2+\sqrt{x^2-4x+2}|\)

\(\displaystyle(7)\ \int\frac{dx}{\sqrt{x^2-5x+4}}=\int\frac{dx}{\left(x-{5\over2}\right)^2-{9\over4}}=\ln\left|x-{5\over2}+\sqrt{x^2-5x+4}\right|\)

\(\displaystyle(8)\ \int\frac{dx}{\sqrt{3-2x^2}}={1\over{2}}{\rm Sin}^{-1}\sqrt{2\over3}x\)

\(\displaystyle(9)\ \int(2^x+2^{-x})^2dx=\int(2^{2x}+2+2^{-2x})dx={1\over\ln4}(2^{2x}-2^{-2x})+2x\)

\(\displaystyle(10)\ \int\frac{dx}{\sqrt{2ax-x^2}}=\int\frac{dx}{\sqrt{a^2-(x-a)^2}}={\rm Sin}^{-1}\frac{x-a}{a}\)

\(\displaystyle(11)\ \int\frac{dx}{(x-a)(b-x)}=-\int\frac{dx}{(x-a)(x-b)}={1\over b-a}\ln\left|\frac{x-a}{x-b}\right|\)

\(\displaystyle(12)\ \int\frac{dx}{\sqrt{(x-a)(x-b)}}=\int\frac{dx}{\sqrt{\left(x-{a+b\over2}\right)^2-\left({a-b\over2}\right)^2}}\)

\(\displaystyle=\ln\left|x-{a+b\over2}+\sqrt{(x-a)(x-b)}\right|\)

3. 다음 함수를 적분하여라.

\(\displaystyle(1)\ \frac{2x-1}{x^2+1}\qquad\qquad\ (2)\ \frac{e^x-e^{-x}}{e^x+e^{-x}}\qquad\quad\ \ \ (3)\ \frac{3x+1}{x^2+1}\)

\(\displaystyle(4)\ \frac{x}{a^2-x^2}\qquad\qquad(5)\ \frac{\ln{x}}{x}\qquad\qquad\quad\ \ \ (6)\ \sin^3x\cos{x}\)

\(\displaystyle(7)\ \frac{1}{x\ln{x}}\qquad\qquad\ \ \,(8)\ \frac{\cos^3x}{\sin{x}}\qquad\qquad\quad(9)\ \frac{e^x}{e^x+1}\)

\(\displaystyle(10)\ \frac{x-1}{1+x-x^2}\qquad(11)\ \frac{2x-3}{x^2-4x+7}\qquad(12)\ \frac{x^3-8x+7}{x^2-4x+5}\)

<풀이>

\(\displaystyle(1)\ \int\frac{2x-1}{x^2+1}dx=\int\left(\frac{2x}{x^2+1}-\frac{1}{x^2+1}\right)dx=\ln(x^2+1)-{\rm Tan}^{-1}x\)

\(\displaystyle(2)\ \int\frac{e^x-e^{-x}}{e^x+e^{-x}}dx=\int\frac{((e^x+e^{-x})'}{e^x+e^{-x}}dx=\ln(e^x+e^{-x})\)

\(\displaystyle(3)\ \int\frac{3x+1}{x^2+1}dx=\int\left({3\over2}\cdot\frac{2x}{x^2+1}+\frac{1}{x^2+1}\right)dx={3\over2}\ln(x^2+1)+{\rm Tan}^{-1}x\)

\(\displaystyle(4)\ \int\frac{x}{a^2-x^2}dx={1\over2}\int\frac{2x}{a^2-x^2}dx={1\over2}\ln{1\over|x^2-a^2|}\)

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

\(\displaystyle(6)\ \int\sin^3x\cos{x}dx=\int\sin^3x(\sin{x})'dx={\sin^4x\over4}\)

\(\displaystyle(7)\ \int\frac{dx}{x\ln{x}}=\int\frac{(\ln{x})'}{\ln{x}}dx=\ln|\ln{x}|\)

\(\displaystyle(8)\ \int\frac{\cos^3x}{\sin{x}}dx=\int(\cot{x}-\sin{x}\cos{x})dx=\ln|\sin{x}|-{\sin^2{x}\over2}\)

\(\displaystyle(9)\ \int\frac{e^x}{e^x+1}dx=\int\frac{(e^x+1)'}{e^x+1}dx=\ln(e^x+1)\)

\(\displaystyle(10)\ \int\frac{x-1}{1+x-x^2}dx=-{1\over2}\int\left\{\frac{2x-1}{x^2-x-1}-\frac{1}{\left(x-{1\over2}\right)^2-{5\over4}}\right\}dx\)

\(\displaystyle=-{1\over2}\ln|x^2-x-1|+{1\over2\sqrt{5}}\ln\left|\frac{2x-1-\sqrt{5}}{2x-1+\sqrt{5}}\right|\)

\(\displaystyle(11)\ \int\frac{2x-3}{x^2-4x+7}dx=\int\left\{\frac{2x-4}{x^2-4x+7}+\frac{1}{(x-2)^2+3}\right\}dx\)

\(\displaystyle=\ln(x^2-4x+7)+{1\over\sqrt{3}}{\rm Tan}^{-1}{x-2\over\sqrt{3}}\)

\(\displaystyle(12)\ \int\frac{x^3-8x+7}{x^2-4x+5}dx=\int\left\{x+4+{3\over2}\cdot\frac{2x-4}{x^2-4x+5}-\frac{7}{(x-2)^2+1}\right\}\)

\(\displaystyle={x^2\over2}+4x+{3\over2}\ln(x^2-4x+5)-7{\rm Tan}^{-1}(x-2)\)

4. 다음 함수를 적분하여라.

\((1)\ \sin{x}\cos{x}\qquad\qquad(2)\ \sin{px}\cos{px}\ (p^2\ne q^2)\qquad(3)\ \cos{px}\cos{qx}\ (p^2\ne q^2)\)

\((4)\ \cos^2px\ (p\ne0)\qquad(5)\ \cos^3x\)

<풀이>

\(\displaystyle(1)\ \int\sin{x}\cos{x}dx={1\over2}\int\sin2xdx=-\frac{\cos2x}{4}\)

\(\displaystyle(2)\ \int\sin{px}\cos{qx}dx={1\over2}\int\{\sin(p+q)+\sin(p-q)x\}dx=-\frac{\cos(p+q)x}{2(p+q)}-\frac{\cos(p-q)x}{2(p-q)}\)

\(\displaystyle(3)\ \int\cos{px}\cos{qx}dx={1\over2}\int\{\cos(p-q)x+\cos(p+q)x\}dx=\frac{\sin(p+q)x}{2(p+q)}+\frac{\sin(p-q)x}{2(p-q)}\)

\(\displaystyle(4)\ \int\cos^2pxdx={1\over2}\int(1+\cos2px)dx={x\over2}+{\sin2px\over4p}\)

\(\displaystyle(5)\ \int\cos^3xdx={1\over4}\int(\cos3x+3\cos{x})dx={\sin3x\over12}+{3\over4}\sin{x}\)