# Integration by parts

## Deriving integration by parts formula

$$\frac{d}{dx}(f(x)g(x))=f'(x)g(x)+f(x)g'(x)$$

$$f(x)g(x)$$は右辺を積分したものになりますから $$f(x)g(x)=\int f'(x)g(x)dx + \int f(x)g'(x)dx$$ 両辺から$$\int f'(x)g(x)dx$$を引きます $$f(x)g(x)-\int f'(x)g(x)dx = \int f(x)g'(x)dx$$ 両辺を入れ替えます

$$\int f(x)g'(x)dx = f(x)g(x)-\int f'(x)g(x)dx$$

この規則を使って積分を行います

$$\int x \cos x dx=?$$ $$f(x)=x \quad f'(x)=1 \quad g(x)= \sin x \quad g'(x)= \cos x$$ これから $$\int x \cos x dx=x \cdot \sin x - \int (1 \cdot \sin x)dx$$ $$\quad = x \cdot \sin x - (-\cos x) = x \cdot \sin x + \cos x + C$$

$$\int (\ln x) dx=?$$ これを書き換えます $$\int (\ln x) dx=\int (\ln x \cdot 1) dx$$ $$f(x)=\ln x \quad f'(x)=\frac{1}{x} \quad g(x)= x \quad g'(x)=1$$ これから $$\int (\ln x \cdot 1) dx =x \cdot \ln x - \int (\frac{1}{x} \cdot x)dx =x \cdot \ln x - x + C$$

$$\int (x^2e^x) dx=?$$ $$f(x)=x^2 \quad f'(x)=2x \quad g(x)= e^x \quad g'(x)=e^x$$ これから $$\int (x^2e^x) dx=x^2e^x-\int (2xe^x) dx =x^2e^x-2\int (xe^x) dx$$ $$\int (xe^x) dx$$について規則から $$f(x)=x \quad f'(x)=1 \quad g(x)= e^x \quad g'(x)=e^x$$ したがって $$\int (x^2e^x) dx=x^2e^x-\int (2xe^x) dx =\int (x^2e^x) dx=x^2e^x-2\int (xe^x) dx =x^2e^x - (2xe^x-2\int (1 \cdot e^x)dx) =x^2e^x - 2xe^x + 2e^x+C$$

$$\int (e^x \cos x) dx=?$$ $$f(x)=e^x \quad f'(x)=e^x \quad g(x)= \sin x \quad g'(x)=\cos x$$ よって $$\int (e^x \cos x) dx=e^x \sin x - \int (e^x \sin x)dx$$ ここで $$\int (e^x \sin x)dx=?$$ $$f(x)=e^x \quad f'(x)=e^x \quad g(x)= -\cos x \quad g'(x)=\sin x$$ $$\int (e^x \sin x)dx=-e^x\cos x +\int (e^x \cos x)$$ $$\int (e^x \cos x) dx=e^x \sin x - \int (e^x \sin x)dx =e^x \sin x + e^x\cos x - \int (e^x \cos x)$$ $$2\int (e^x \cos x) dx=e^x \sin x + e^x\cos x$$ $$\int (e^x \cos x) dx=\frac{e^x \sin x + e^x\cos x}{2}+C$$

# U-substitution

$$u=x^3+x^2$$ $$\frac{du}{dx}=3x^2+2x \to du = (3x^2+2x)dx$$ とします

$$\int \frac{4x^3}{x^4 + 7}dx$$

$$u=x^4 + 7$$ $$\frac{du}{dx}=4x^3 \to du=4x^3dx$$ 問題を書き換えます $$\int \frac{4x^3}{x^4 + 7}dx=\int \frac{4x^3dx}{x^4 + 7} =\int \frac{du}{u}=\frac{1}{u}du$$ $$\quad = ln|u| + C = ln|x^4 + 7| + C$$

$$\int \sqrt{7x+9}dx$$
$$u=7x+9$$ $$\frac{du}{dx}=7 \to du=7dx$$

ここで７という数字を使うために少し工夫します $$\int \frac{1}{7}\cdot 7\sqrt{7x+9}dx =\frac{1}{7} \int 7dx\sqrt{7x+9}$$ $$\quad = \frac{1}{7} \int \sqrt{u}du = \frac{1}{7} \int u^{\frac{1}{2}}du =\frac{1}{7}(\frac{2}{3}u^{\frac{3}{2}}+C) =\frac{2}{21}u^{\frac{3}{2}} + C$$ $$\quad = \frac{2}{21}(7x+9)^{\frac{3}{2}} + C$$

$$\int \frac{\pi}{x\ln x}dx$$
$$u=\ln x$$ $$\frac{du}{dx}=\frac{1}{x} \to du={1}{x}dx$$ 問題を書き換えます $$\int \frac{\pi}{x\ln x}dx =\pi \int \frac{1}{\ln x}\frac{1}{x}dx =\pi \int \frac{1}{u}du$$ $$\quad =\pi ln|u| + C =\pi ln|\ln x| + C$$

$$\int_{0}^{1}x^2 2^{x^{3}} dx$$
$$\frac{d}{dx}e^x=e^x$$ $$\int e^x dx = e^x + C$$ $$2=e^{\ln2} \to 2^{x^3}=(e^{\ln2})^{x^3}=e^{x^{3}\ln2}$$ 問題を書き換えます $$\int_{0}^{1}x^2 e^{x^{3}\ln2} dx$$ u-subsutitution $$u=x^{3}\ln2$$ $$\frac{du}{dx}=3x^2 \ln2=x^2 3\ln2 =x^2 \ln2^3 = \ln8 x^2 \to du=\ln8 x^2dx$$ $$\intx^2 e^{x^{3}\ln2} dx =\frac{1}{\ln8} \int \ln8 x^2 e^{x^{3}\ln2} dx =\frac{1}{\ln8} \int e^{u}du =\frac{1}{\ln8}e^{u}+C=\frac{1}{\ln8}e^{x^{3}\ln2}+C$$ $$\int_{0}^{1}x^2 2^{x^{3}} dx =\frac{1}{\ln8}e^{1^{3}\ln2}-\frac{1}{\ln8}e^{0^{3}\ln2} =\frac{1}{\ln8}*2 - \frac{1}{\ln8}*1 =\frac{1}{\ln8}$$

$$\int \frac{2^{\ln x}}{x} dx = \int \frac{1}{x}2^{\ln x}dx$$
$$u=\ln x$$ $$du = \frac{1}{x}dx$$

$$\int \frac{1}{x}2^{\ln x}dx =\int 2^{u}du =\int (e^{\ln2})^{u}du =\int (e^{(\ln2)u}du$$ $$\qquad \int e^{au}du=\frac{1}{a}e^{au}+C$$ $$=\frac{1}{\ln2}e^{u\ln2}+C$$ $$\qquad a\ln b = \ln b^a \to u\ln2=\ln 2^{u}$$ $$=\frac{1}{\ln2}2^{u}+C$$ $$=\frac{1}{\ln2}2^{\ln x}+C$$

$$\int (x+3)(x-1)^5 dx$$
$$u=x-1 \to x= u+1$$ $$du = dx$$

$$\int (x+3)(x-1)^5 dx =\int (u+1+3)u^5 du =\int (u+4)u^5 du =\int (u^6+4u^5) du =\frac{u^7}{7}+\frac{4u^6}{6} + C$$ $$=\frac{(x-1)^7}{7}+\frac{2}{3}(x-1)^6 + C$$

$$\int_{0}^{\sqrt{\pi}}x\sin(x^2)dx$$
$$u=x^2$$ $$du = 2xdx$$ 置き換えます $$\frac{1}{2}\int_{0}^{\sqrt{\pi}}2x\sin(x^2)dx =\frac{1}{2}\int_{u=0}^{u=\pi}\sin(u)du =\frac{1}{2}[-\cos u ]_{0}^{\pi}$$ $$=\frac{1}{2}[-\cos \pi + \cos 0] =\frac{1}{2}[1 + 1]=1$$

$$\int \frac{\cos(5x)}{e^{\sin(5x)}}dx$$
$$u=\sin(5x)$$ $$\frac{d}{dx}=5\cos(5x) \to du=5\cos(5x)dx$$

# Reverse chain rule

### Chain Rule

$$\frac{d}{dx}g(f(x))=g'(f(x)) \cdot f'(x)$$

$$\qquad \downarrow$$

### Reverse Chain Rule

$$\int g'(f(x)) \cdot f'(x) dx = g(f(x)) + C$$

$$\int (\sin x)^2 \cos x dx = ?$$

$$u=\sin x$$ $$\frac{du}{dx}=\cos x \to du=\cos x dx$$

$$\int (\sin x)^2 \cos x dx =\int (u)^2 du = \frac{u^3}{3} + C =\frac{(\sin x)^3}{3} + C$$

$$\int \frac{x}{2}\sin(2x^2+2) dx = ?$$

$$f(x)=2x^2 + 2$$ $$f'(x)=4x$$

$$\int \frac{x}{2}\sin(2x^2+2) dx =\frac{1}{2}\frac{1}{4} \int 4x \sin(2x^2+2) dx =\frac{1}{8}\int f'(x) \sin(f(x)) dx =\frac{1}{8}(-cos(f(x)))+C =\frac{1}{8}(-cos(2x^2 + 2))+C$$

$$\int \tan x dx = ?$$

$$\int \tan x dx =\int \frac{\sin x}{\cos x}dx =\int \sin x \frac{1}{\cos x}dx$$

$$\qquad \int \frac{1}{x}= \ln|x| + c$$ $$\qquad \int f'(x)\frac{1}{f(x)}=\ln|f(x)|+C$$ $$\qquad f(x)= \cos x$$ $$\qquad f'(x)= -\sin x$$

$$=-\ln|\cos x| + C$$

# Integration using trigonometric identities

$$\int \cos^3 x dx = ?$$

$$\int \cos^3 x dx = \int (\cos x)(\cos^2 x)dx = \int (\cos x)(1 - \sin^2 x)dx$$

$$\qquad u=\sin x$$ $$\qquad du = \cos x dx$$

$$=\int (1-u^2)du = u - \frac{1}{3}u^3 + C = \sin x - \frac{1}{3}sin^3 x + C$$

$$\int \sin^2 x \cos^3 x dx = ?$$

$$\int \sin^2 x \cos^3 x dx = \int \sin^2 x \cos^2 x \cos x dx =\int \sin^2 x (1-\sin^2 x) \cos x dx =\int (sin^2 x - \sin^4 x) \cos x dx$$

$$\qquad u=\sin x$$ $$\qquad du = \cos x dx$$

$$=\int (u^2 - u^4) du =\frac{u^3}{3}-\frac{u^5}{5} + C =\frac{sin^3 x}{3}-\frac{sin^5 x}{5} + C$$

$$\int \sin^4 x dx = ?$$

$$\qquad sin^2 x = \frac{1}{2}(1 - \cos 2x)$$

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

$$\qquad \cos^2 2x = \frac{1}{2}(1 + cos 4x)$$ $$=\frac{1}{4} \int (1-2\cos 2x + \frac{1}{2} + \frac{1}{2}\cos 4x)dx =\frac{1}{4} \int (\frac{3}{2}-2\cos 2x+\frac{1}{8}4\cos 4x) dx =\frac{1}{4} (\frac{3}{2}x - \sin 2x + \frac{1}{8} \sin 4x) + C$$

# Trigonometric substitution

$$\int \frac{1}{\sqrt{1-x^2}} dx = ?$$

$$\qquad -1 \lt x \lt 1$$ $$\qquad x = \sin \theta$$ $$\qquad dx = \cos \theta d\theta$$ $$\qquad \theta = \arcsin x$$

$$\int \frac{1}{\sqrt{1-x^2}} dx =\int \frac{\cos \theta}{\sqrt{(1-sin^2 \theta)}}d\theta =\int \frac{\cos \theta}{\sqrt{cos^2 \theta}}d\theta =\int \frac{\cos \theta}{|cos \theta|}d\theta$$

$$\qquad \cos \theta \ge 0$$ $$=\int d\theta =\theta + C =\arcsin x + C$$

$$\int \frac{\pi}{\sqrt{8-2x^2}} dx = ?$$

$$\qquad a^2 - x^2 \to x=a\sin \theta$$ $$\qquad a^2 - a^2\sin^2 \theta) = a(1-sin^2 \theta)$$ $$\qquad x = 2\sin \theta$$ $$\qquad dx = 2\cos \theta d\theta$$ $$\qquad \theta = \arcsin(\frac{x}{2})$$ $$\qquad 8 - 2x^2 = 2(4-x^2)= 2(2^2-2^2sin^2 \theta) = 2\cdot 2^2(1-\sin^2 \theta)=8\cos^2 \theta$$

$$\int \frac{\pi}{\sqrt{8-2x^2}} dx =\pi \int \frac{2\cos \theta}{\sqrt{8\cos^2 \theta}} d\theta =\pi \int \frac{2\cos \theta}{2\sqrt{2}\cos\theta} d\theta =\frac{\pi}{\sqrt{2}}\int d\theta =\frac{\pi}{\sqrt{2}}\theta + C =\frac{\pi}{\sqrt{2}}\arcsin(\frac{x}{2}) + C$$

$$\int \frac{1}{\sqrt{3-2x^2}} dx = ?$$

$$\qquad \frac{2}{3}x^2=sin^2 \theta$$ $$\qquad \frac{\sqrt{2}}{\sqrt{3}}x^2=sin \theta$$ $$\qquad \theta = \arcsin(\frac{\sqrt{2}}{\sqrt{3}}x)$$ $$\qquad x=\frac{\sqrt{3}}{\sqrt{2}}\sin \theta$$ $$\qquad dx=\frac{\sqrt{3}}{\sqrt{2}}\cos \theta d\theta$$

$$=\int \frac{\frac{\sqrt{3}}{\sqrt{2}}\cos \theta d\theta} {\sqrt{3(1-sin^2 x)}} =\int \frac{\frac{\sqrt{3}}{\sqrt{2}}\cos \theta d\theta} {\sqrt{3(cos^2 x)}} =\int \frac{\frac{\sqrt{3}}{\sqrt{2}}\cos \theta d\theta} {\sqrt{3}cos x)} =\int \frac{1}{\sqrt{2}}d\theta =\frac{1}{\sqrt{2}} \int d \theta$$ $$=\frac{1}{\sqrt{2}}\theta + C =\frac{1}{\sqrt{2}}\arcsin(\frac{\sqrt{2}}{\sqrt{3}}x) + C$$

$$\int x^3 \sqrt{9-x^2} dx = ?$$

$$\qquad 9-x^2 = 3^2 - x^2$$ $$\qquad x=a\sin \theta = 3\sin \theta \to \sin \theta = \frac{x}{3}$$ $$\qquad dx = 3\cos \theta d\theta$$ $$\qquad 3^2 - 3^2 \sin^2 \theta = 9(1-sin^2 \theta)$$ $$\qquad \sqrt{9-x^2}=\sqrt{9\cos^2 \theta}=3\cos \theta$$

$$\int x^3 \sqrt{9-x^2} dx =\int 27\sin^3 \theta 3\cos \theta \cdot 3\cos \theta d\theta$$ $$=243 \int sin^3 \theta \cos^2 \theta d\theta$$ $$=243 \int sin^2 \theta \cos^2 \theta \sin \theta d\theta$$ $$=243 \int (1-cos^2 \theta) \cos^2 \theta \sin \theta d\theta$$

$$\qquad u= \cos \theta$$ $$\qquad du = -\sin \theta d\theta$$

$$=-243 \int (1-cos^2 \theta) \cos^2 \theta -\sin \theta d\theta$$ $$=-243 \int (1-u^2) \u^2 du =-243 \int(u^2 - u^4) du =-243(\frac{u^3}{3}-\frac{u^5}{5}) + C$$

$$\qquad \sin \theta = \frac{x}{3}$$ $$\qquad \cos \theta = \frac{\sqrt{9-x^2}}{3} =\sqrt{\frac{1}{9}(9-x^2)} = (1-\frac{x^2}{9})^{\frac{1}{2}}=u$$ $$=243(\frac{u^5}{5}-\frac{u^3}{3}) + C$$ $$=243(\frac{(1-\frac{x^2}{9})^{\frac{5}{2}}}{5} -\frac{(1-\frac{x^2}{9})^{\frac{3}{2}}}{3}) + C$$

$$\int \frac{1}{9 + x^2} dx = ?$$

$$\qquad a^2 - x^2 \to x = a\sin \theta としてきましたが$$ $$\qquad a^2 + x^2 \to x = a\tan \theta とします$$ $$\qquad = a^2 + a^2\tan^2 \theta =a^2(1 + tan^2 \theta) =a^2(\frac{\cos^2 \theta}{\cos^2 \theta} + \frac{\sin^2 \theta}{\cos^2 \theta}) =a^2(\frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}) =a^2(\frac{1}{\cos^2 \theta}) =a^2\sec^2 \theta$$ $$\qquad 9+x^2 = 3^2 + x^2=3^2\sec^2 \theta$$ $$\qquad a = 3$$ $$\qquad x = 3\tan \theta$$ $$\qquad dx = 3\sec^2 \theta d\theta$$ $$\qquad \frac{x}{3}=\tan \theta$$ $$\qquad \theta = \arctan(\frac{x}{3})$$

$$\int \frac{1}{9 + x^2} dx =\int \frac{3\sec^2 \theta d\theta}{9\sec^2 \theta} =\frac{1}{3}\int d\theta =\frac{1}{3} \theta + C$$ $$=\frac{1}{3} \arctan(\frac{x}{3}) + C$$

$$\int \frac{1}{36 + x^2} dx = ?$$

まず変形します $$=\int \frac{1}{36(1+\frac{x^2}{36})}$$

$$\qquad \frac{x^2}{36}= \tan^2 \theta \to \frac{x}{6} \to x=6\tan \theta$$ $$\qquad 1 + \tan^2 \theta = 1 + \frac{\sin^2 \theta}{\cos^2 \theta} =\frac{\cos^2 \theta}{\cos^2 \theta} + \frac{\sin^2 \theta}{\cos^2 \theta} =\frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta} =\frac{1}{\cos^2 \theta}=sec^2 \theta$$ $$\qquad dx=6\sec^2 \theta d\theta$$ $$\qquad \arctan \frac{x}{6} = \theta$$

$$=\int \frac{6\sec^2 \theta d\theta}{36(1+\tan^2 \theta)} =\int \frac{6\sec^2 \theta d\theta}{36\sec^2 \theta}$$ $$=\frac{1}{6}\int d\theta$$ $$=\frac{1}{6}\theta + C =\frac{1}{6}\arctan(\frac{x}{6}) + C$$

$$\int \sqrt{6x-x^2 -5} dx = ?$$

これを変形します $$=\int \sqrt{-5 +9 -(x^2-6x +9)} dx =\int \sqrt{4 - (x-3)^2} dx =\int \sqrt{4(1 - \frac{(x-3)^2}{4}} dx =\int 2 \sqrt{1-(\frac{x-3}{2})^2} dx$$

$$\qquad \cos^2 \theta = 1 - \sin^2 \theta$$ $$\qquad sin^2 \theta = (\frac{x-3}{2})^2 \to \sin \theta = \frac{x-3}{2}$$ $$\qquad \theta = \arcsin(\frac{x-3}{2})$$ $$\qquad x = 2\sin \theta + 3$$ $$\qquad dx = 2\cos \theta d\theta$$

$$=\int2 \sqrt{1-\sin^2 \theta} 2\cos \theta d\theta =\int 4\cos^2 \theta d\theta$$

$$\qquad \cos^2 \theta = \frac{1}{2}(1 - \cos2\theta)$$

$$=\int 4 \cdot \frac{1}{2}(1 - \cos2\theta)d\theta =\int 2(1 - \cos2\theta)d \theta =\int(2-2\cos2\theta) d \theta$$ $$=2\theta-\sin 2\theta + C$$

$$\qquad \sin 2\theta = \sin(\theta + \theta) = \sin \theta \cos \theta + \sin \theta + \cos \theta = =2\sin \theta \cos \theta$$

$$=2\theta + 2\sin \theta \cos \theta + C =2\theta + 2\sin \theta \sqrt{1-sin^2 \theta} + C$$ $$=2\theta + 2\sin \theta \sqrt{1-sin^2 \theta} +C$$ $$=2\arcsin(\frac{x-3}{2}) + 2\frac{x-3}{2} \sqrt{1-(\frac{x-3}{2})^2}+C$$ $$=2\arcsin(\frac{x-3}{2})+\frac{x-3}{2}\sqrt{4} \sqrt{1-\frac{(x-3)^2}{4}}+C$$ $$=2\arcsin(\frac{x-3}{2})+\frac{x-3}{2}\sqrt{4-(x-3)^2}+C$$ $$=2\arcsin(\frac{x-3}{2})+\frac{x-3}{2}\sqrt{4-(x^2-6x+9)}+C$$ $$=2\arcsin(\frac{x-3}{2})+\frac{x-3}{2}\sqrt{6x-x^2-5)}+C$$