知っていると便利な積分のテクニック

$$\tan {\displaystyle \frac{x}{2}=t}$$

$$\sin x={\displaystyle \frac{2t}{1+t^2}}$$

$$\cos x={\displaystyle \frac{1-t^2}{1+t^2}}$$

$$\tan x={\displaystyle \frac{2t}{1-t^2}}$$

$$dx={\displaystyle \frac{2}{1+t^2}dt}$$

$$\begin{array}{rl}\tan x& =\tan 2{\displaystyle \frac{x}{2}}\\ \\ & ={\displaystyle \frac{2\tan {\displaystyle \frac{x}{2}}}{1-{\tan }^2{\displaystyle \frac{x}{2}}}}\\ \\ & ={\displaystyle \frac{2t}{1-t^2}}\end{array}$$

$$\begin{array}{rl}\sin x& =\sin 2{\displaystyle \frac{x}{2}}\\ \\ & =2\sin {\displaystyle \frac{x}{2}\cos {\displaystyle \frac{x}{2}}}\\ \\ & =2\tan {\displaystyle \frac{x}{2}{\cos }^2{\displaystyle \frac{x}{2}}}\\ \\ & =2\tan {\displaystyle \frac{x}{2}{\displaystyle \frac{1}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}}\\ \\ & ={\displaystyle \frac{2t}{1+t^2}}\end{array}$$

$$\begin{array}{rl}\cos x& =\cos 2{\displaystyle \frac{x}{2}}\\ \\ & =2{\cos }^2{\displaystyle \frac{x}{2}-1}\\ \\ & ={\displaystyle \frac{2}{1+{\tan }^2{\displaystyle \frac{x}{2}}}-1}\\ \\ & ={\displaystyle \frac{2}{1+t^2}-1}\\ \\ & ={\displaystyle \frac{1-t^2}{1+t^2}}\end{array}$$

$$\begin{array}{rl}dx& =2{\cos }^2{\displaystyle \frac{x}{2}dt}\\ \\ & ={\displaystyle \frac{2}{1+{\tan }^2{\displaystyle \frac{x}{2}}}dt}\\ \\ & ={\displaystyle \frac{2}{1+t^2}dt}\end{array}$$

$${\displaystyle {\int }_0^{\frac{1}{2}\pi }{\displaystyle \frac{1}{3+\cos x}dx}}$$

$$\begin{array}{ccccc}x& :& 0& \to & {\displaystyle \frac{1}{2}\pi }\\ \\ t& :& 0& \to & 1\end{array}$$

$$\begin{array}{rl}{\displaystyle {\int }_0^{\frac{1}{2}\pi }{\displaystyle \frac{1}{3+\cos x}dx}}& ={\displaystyle {\int }_0^1{\displaystyle \frac{1}{3+{\displaystyle \frac{1-t^2}{1+t^2}}}{\displaystyle \frac{2}{1+t^2}dt}}}\\ \\ & ={\displaystyle {\int }_0^1{\displaystyle \frac{1}{2+t^2}dt}}\end{array}$$

$$\begin{array}{rl}t& =\sqrt{2}\tan \theta \\ \\ dt& ={\displaystyle \frac{\sqrt{2}}{{\cos }^2\theta }d\theta }\end{array}$$

$$\begin{array}{ccccc}t& :& 0& \to & 1\\ \\ \theta & :& 0& \to & \arctan {\displaystyle \frac{1}{\sqrt{2}}}\end{array}$$

$$\begin{array}{rl}{\displaystyle {\int }_0^1{\displaystyle \frac{1}{2+t^2}dt}}& ={\displaystyle {\int }_0^{\arctan \frac{1}{\sqrt{2}}}{\displaystyle \frac{1}{2+2{\tan }^2\theta }{\displaystyle \frac{\sqrt{2}}{{\cos }^2\theta }d\theta }}}\\ \\ & ={\displaystyle \frac{1}{\sqrt{2}}{\displaystyle {\int }_0^{\arctan \frac{1}{\sqrt{2}}}d\theta }}\\ \\ & ={\displaystyle \frac{1}{\sqrt{2}}{\left[\theta \right]}_0^{\arctan \frac{1}{\sqrt{2}}}}\\ \\ & ={\displaystyle \frac{1}{\sqrt{2}}\arctan {\displaystyle \frac{1}{\sqrt{2}}}}\end{array}$$

$${\displaystyle {\int }_a^{b}f(x)dx={\displaystyle {\int }_a^{b}f(a+b-x)dx}}$$

$${\displaystyle {\int }_a^{b}f(x)dx={\displaystyle \frac{1}{2}{\displaystyle {\int }_a^{b}f(x)+f(a+b-x)dx}}}$$

$${\displaystyle {\int }_a^{b}f(a+b-x)dx}$$

$$\begin{array}{rl}t& =a+b-x\\ \\ dt& =(-1)dx\end{array}$$

$$\begin{array}{ccccc}x& :& a& \to & b\\ \\ t& :& b& \to & a\end{array}$$

$$\begin{array}{rl}{\displaystyle {\int }_a^{b}f(a+b-x)dx}& ={\displaystyle {\int }_b^{a}f(t)(-1)dt}\\ \\ & ={\displaystyle {\int }_a^{b}f(t)dt}\end{array}$$

$${\displaystyle {\int }_a^{b}f(t)dt={\displaystyle {\int }_a^{b}f(x)dx}}$$

$${\displaystyle {\int }_a^{b}f(a+b-x)dx={\displaystyle {\int }_a^{b}f(x)dx}}$$

$$\begin{array}{rl}{\displaystyle {\int }_a^{b}f(x)dx}& ={\displaystyle \frac{1}{2}\left({\displaystyle {\int }_a^{b}f(x)dx+{\displaystyle {\int }_a^{b}f(x)dx}}\right)}\\ \\ & ={\displaystyle \frac{1}{2}\left({\displaystyle {\int }_a^{b}f(x)dx+{\displaystyle {\int }_a^{b}f(a+b-x)dx}}\right)}\\ \\ & ={\displaystyle \frac{1}{2}{\displaystyle {\int }_a^{b}f(x)+f(a+b-x)dx}}\end{array}$$

$${\displaystyle {\int }_0^{\pi }{\displaystyle \frac{x\sin x}{1+{\cos }^{2}x}dx}}$$

$$\begin{array}{rl}{\displaystyle {\int }_0^{\pi }{\displaystyle \frac{x\sin x}{1+{\cos }^{2}x}dx}}& ={\displaystyle \frac{1}{2}{\displaystyle {\int }_0^{\pi }{\displaystyle \frac{x\sin x}{1+{\cos }^{2}x}+{\displaystyle \frac{(\pi -x)\sin (\pi -x)}{1+{\cos }^2(\pi -x)}dx}}}}\\ \\ & ={\displaystyle \frac{1}{2}{\displaystyle {\int }_0^{\pi }{\displaystyle \frac{x\sin x}{1+{\cos }^{2}x}+{\displaystyle \frac{(\pi -x)\sin x}{1+{\cos }^{2}x}dx}}}}\\ \\ & ={\displaystyle \frac{1}{2}{\displaystyle {\int }_0^{\pi }{\displaystyle \frac{\pi \sin x}{1+{\cos }^{2}x}dx}}}\\ \\ & ={\displaystyle \frac{1}{2}\pi {\displaystyle {\int }_0^{\pi }{\displaystyle \frac{\sin x}{1+{\cos }^{2}x}dx}}}\end{array}$$

$$\begin{array}{rl}t& =\cos x\\ \\ dt& =-\sin xdx\end{array}$$

$$\begin{array}{ccccc}x& :& 0& \to & \pi \\ \\ t& :& 1& \to & -1\end{array}$$

$$\begin{array}{rl}{\displaystyle \frac{1}{2}\pi {\displaystyle {\int }_0^{\pi }{\displaystyle \frac{\sin x}{1+{\cos }^{2}x}dx}}}& ={\displaystyle \frac{1}{2}\pi {\displaystyle {\int }_1^{-1}{\displaystyle \frac{-1}{1+t^2}dt}}}\\ \\ & ={\displaystyle \frac{1}{2}\pi {\displaystyle {\int }_{-1}^1{\displaystyle \frac{1}{1+t^2}dt}}}\\ \\ & ={\displaystyle \frac{1}{2}\pi {[\arctan t]}_{-1}^1}\\ \\ & ={\displaystyle \frac{1}{4}{\pi }^2}\end{array}$$

$${\displaystyle \int fgdx=fG-{\displaystyle \int f^{\prime }Gdx}}$$

$$\begin{array}{ccccc}{\displaystyle \int }& f& g& dx& =\\ & \downarrow & \uparrow & & \\ & f^{\prime }& G& & \end{array}$$

$$\begin{array}{cccccc}{\displaystyle \int }& f& g& dx& =& fG\\ & \downarrow & \uparrow & & & \\ & f^{\prime }& G& & & \end{array}$$

$$\begin{array}{ccccccc}{\displaystyle \int }& f& g& dx& =& fG& {\displaystyle \int f^{\prime }Gdx}\\ & \downarrow & \uparrow & & & & \\ & f^{\prime }& G& & & & \end{array}$$

$$\begin{array}{cccccc}{\displaystyle \int }& f& g& dx& =& fG-{\displaystyle \int f^{\prime }Gdx}\\ & \downarrow & \uparrow & & & \\ & f^{\prime }& G& & & \end{array}$$

$${\displaystyle \int fgdx=fG-{\displaystyle \int f^{\prime }Gdx}}$$

$${\displaystyle \int x\cos xdx}$$

$$\begin{array}{ccccc}{\displaystyle \int }& x& \cos x& dx& =\\ & \downarrow & \uparrow & & \\ & 1& \sin x& & \end{array}$$

$$\begin{array}{cccccc}{\displaystyle \int }& x& \cos x& dx& =& x\sin x\\ & \downarrow & \uparrow & & & \\ & 1& \sin x& & & \end{array}$$

$$\begin{array}{ccccccc}{\displaystyle \int }& x& \cos x& dx& =& x\sin x& {\displaystyle \int 1\sin xdx}\\ & \downarrow & \uparrow & & & & \\ & 1& \sin x& & & & \end{array}$$

$$\begin{array}{cccccc}{\displaystyle \int }& x& \cos x& dx& =& x\sin x-{\displaystyle \int 1\sin xdx}\\ & \downarrow & \uparrow & & & \\ & 1& \sin x& & & \end{array}$$

$$\begin{array}{rl}{\displaystyle \int x\cos xdx}& =x\sin x-{\displaystyle \int 1\sin xdx}\\ \\ & =x\sin x+\cos x+C\end{array}$$

$${\displaystyle \int x^3\cos xdx}$$

$$\begin{array}{ccccc}{\displaystyle \int }& x^3& \cos x& dx& =\\ & \downarrow & \uparrow & & \\ & 3x^2& \sin x& & \\ & \downarrow & \uparrow & & \\ & 6x& -\cos x& & \\ & \downarrow & \uparrow & & \\ & 6& -\sin x& & \end{array}$$

$$\begin{array}{cccccccc}{\displaystyle \int }& x^3& \cos x& dx& =& x^3\sin x& -3x^2\cos x& -6x\sin x\\ & \downarrow & \uparrow & & & & & \\ & 3x^2& \sin x& & & & & \\ & \downarrow & \uparrow & & & & & \\ & 6x& -\cos x& & & & & \\ & \downarrow & \uparrow & & & & & \\ & 6& -\sin x& & & & & \end{array}$$

$$\begin{array}{ccccccccc}{\displaystyle \int }& x^3& \cos x& dx& =& x^3\sin x& -3x^2\cos x& -6x\sin x& {\displaystyle \int -6\sin xdx}\\ & \downarrow & \uparrow & & & & & & \\ & 3x^2& \sin x& & & & & & \\ & \downarrow & \uparrow & & & & & & \\ & 6x& -\cos x& & & & & & \\ & \downarrow & \uparrow & & & & & & \\ & 6& -\sin x& & & & & & \end{array}$$

$$\begin{array}{cccccc}{\displaystyle \int }& x^3& \cos x& dx& =& x^3\sin x-(-3x^2\cos x)+(-6x\sin x)-{\displaystyle \int -6\sin xdx}\\ & \downarrow & \uparrow & & & \\ & 3x^2& \sin x& & & \\ & \downarrow & \uparrow & & & \\ & 6x& -\cos x& & & \\ & \downarrow & \uparrow & & & \\ & 6& -\sin x& & & \end{array}$$

$$\begin{array}{rl}{\displaystyle \int x^3\cos xdx}& =x^3\sin x-(-3x^2\cos x)+(-6x\sin x)-{\displaystyle \int -6\sin xdx}\\ \\ & =x^3\sin x+3x^2\cos x-6x\sin x+{\displaystyle \int 6\sin xdx}\\ \\ & =x^3\sin x+3x^2\cos x-6x\sin x-6\cos x+C\end{array}$$