$$$\frac{\sin^{4}{\left(x \right)}}{\cos^{6}{\left(x \right)}}$$$の積分

$$$\int \frac{\sin^{4}{\left(x \right)}}{\cos^{6}{\left(x \right)}}\, dx$$$ を求めよ。

分子と分母に$$$\cos^{4}{\left(x \right)}$$$を掛け、$$$\frac{\sin^{4}{\left(x \right)}}{\cos^{4}{\left(x \right)}}$$$を$$$\tan^{4}{\left(x \right)}$$$に変換します。:

$${\color{red}{\int{\frac{\sin^{4}{\left(x \right)}}{\cos^{6}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\tan^{4}{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x}}}$$

$$$\frac{1}{\cos^{2}{\left(x \right)}}$$$ を $$$\sec^{2}{\left(x \right)}$$$ に変換する:

$${\color{red}{\int{\frac{\tan^{4}{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\tan^{4}{\left(x \right)} \sec^{2}{\left(x \right)} d x}}}$$

$$$u=\tan{\left(x \right)}$$$ とする。

すると $$$du=\left(\tan{\left(x \right)}\right)^{\prime }dx = \sec^{2}{\left(x \right)} dx$$$（手順は»で確認できます）、$$$\sec^{2}{\left(x \right)} dx = du$$$ となります。

したがって、

$${\color{red}{\int{\tan^{4}{\left(x \right)} \sec^{2}{\left(x \right)} d x}}} = {\color{red}{\int{u^{4} d u}}}$$

$$$n=4$$$ を用いて、べき乗の法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ を適用します:

$${\color{red}{\int{u^{4} d u}}}={\color{red}{\frac{u^{1 + 4}}{1 + 4}}}={\color{red}{\left(\frac{u^{5}}{5}\right)}}$$

次のことを思い出してください $$$u=\tan{\left(x \right)}$$$:

$$\frac{{\color{red}{u}}^{5}}{5} = \frac{{\color{red}{\tan{\left(x \right)}}}^{5}}{5}$$

したがって、

$$\int{\frac{\sin^{4}{\left(x \right)}}{\cos^{6}{\left(x \right)}} d x} = \frac{\tan^{5}{\left(x \right)}}{5}$$

積分定数を加える:

$$\int{\frac{\sin^{4}{\left(x \right)}}{\cos^{6}{\left(x \right)}} d x} = \frac{\tan^{5}{\left(x \right)}}{5}+C$$

$$$\int \frac{\sin^{4}{\left(x \right)}}{\cos^{6}{\left(x \right)}}\, dx = \frac{\tan^{5}{\left(x \right)}}{5} + C$$$A