$$$\cos^{7}{\left(x \right)}$$$の積分
入力内容
$$$\int \cos^{7}{\left(x \right)}\, dx$$$ を求めよ。
解答
余弦を1つ取り出し、$$$\alpha=x$$$ を用いた公式 $$$\cos^2\left(\alpha \right)=-\sin^2\left(\alpha \right)+1$$$ により、残りはすべて正弦で表せ。:
$${\color{red}{\int{\cos^{7}{\left(x \right)} d x}}} = {\color{red}{\int{\left(1 - \sin^{2}{\left(x \right)}\right)^{3} \cos{\left(x \right)} d x}}}$$
$$$u=\sin{\left(x \right)}$$$ とする。
すると $$$du=\left(\sin{\left(x \right)}\right)^{\prime }dx = \cos{\left(x \right)} dx$$$(手順は»で確認できます)、$$$\cos{\left(x \right)} dx = du$$$ となります。
したがって、
$${\color{red}{\int{\left(1 - \sin^{2}{\left(x \right)}\right)^{3} \cos{\left(x \right)} d x}}} = {\color{red}{\int{\left(1 - u^{2}\right)^{3} d u}}}$$
Expand the expression:
$${\color{red}{\int{\left(1 - u^{2}\right)^{3} d u}}} = {\color{red}{\int{\left(- u^{6} + 3 u^{4} - 3 u^{2} + 1\right)d u}}}$$
項別に積分せよ:
$${\color{red}{\int{\left(- u^{6} + 3 u^{4} - 3 u^{2} + 1\right)d u}}} = {\color{red}{\left(\int{1 d u} - \int{3 u^{2} d u} + \int{3 u^{4} d u} - \int{u^{6} d u}\right)}}$$
$$$c=1$$$ に対して定数則 $$$\int c\, du = c u$$$ を適用する:
$$- \int{3 u^{2} d u} + \int{3 u^{4} d u} - \int{u^{6} d u} + {\color{red}{\int{1 d u}}} = - \int{3 u^{2} d u} + \int{3 u^{4} d u} - \int{u^{6} d u} + {\color{red}{u}}$$
$$$n=6$$$ 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:
$$u - \int{3 u^{2} d u} + \int{3 u^{4} d u} - {\color{red}{\int{u^{6} d u}}}=u - \int{3 u^{2} d u} + \int{3 u^{4} d u} - {\color{red}{\frac{u^{1 + 6}}{1 + 6}}}=u - \int{3 u^{2} d u} + \int{3 u^{4} d u} - {\color{red}{\left(\frac{u^{7}}{7}\right)}}$$
定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=3$$$ と $$$f{\left(u \right)} = u^{2}$$$ に対して適用する:
$$- \frac{u^{7}}{7} + u + \int{3 u^{4} d u} - {\color{red}{\int{3 u^{2} d u}}} = - \frac{u^{7}}{7} + u + \int{3 u^{4} d u} - {\color{red}{\left(3 \int{u^{2} d u}\right)}}$$
$$$n=2$$$ 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:
$$- \frac{u^{7}}{7} + u + \int{3 u^{4} d u} - 3 {\color{red}{\int{u^{2} d u}}}=- \frac{u^{7}}{7} + u + \int{3 u^{4} d u} - 3 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=- \frac{u^{7}}{7} + u + \int{3 u^{4} d u} - 3 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$
定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=3$$$ と $$$f{\left(u \right)} = u^{4}$$$ に対して適用する:
$$- \frac{u^{7}}{7} - u^{3} + u + {\color{red}{\int{3 u^{4} d u}}} = - \frac{u^{7}}{7} - u^{3} + u + {\color{red}{\left(3 \int{u^{4} d u}\right)}}$$
$$$n=4$$$ 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:
$$- \frac{u^{7}}{7} - u^{3} + u + 3 {\color{red}{\int{u^{4} d u}}}=- \frac{u^{7}}{7} - u^{3} + u + 3 {\color{red}{\frac{u^{1 + 4}}{1 + 4}}}=- \frac{u^{7}}{7} - u^{3} + u + 3 {\color{red}{\left(\frac{u^{5}}{5}\right)}}$$
次のことを思い出してください $$$u=\sin{\left(x \right)}$$$:
$${\color{red}{u}} - {\color{red}{u}}^{3} + \frac{3 {\color{red}{u}}^{5}}{5} - \frac{{\color{red}{u}}^{7}}{7} = {\color{red}{\sin{\left(x \right)}}} - {\color{red}{\sin{\left(x \right)}}}^{3} + \frac{3 {\color{red}{\sin{\left(x \right)}}}^{5}}{5} - \frac{{\color{red}{\sin{\left(x \right)}}}^{7}}{7}$$
したがって、
$$\int{\cos^{7}{\left(x \right)} d x} = - \frac{\sin^{7}{\left(x \right)}}{7} + \frac{3 \sin^{5}{\left(x \right)}}{5} - \sin^{3}{\left(x \right)} + \sin{\left(x \right)}$$
簡単化せよ:
$$\int{\cos^{7}{\left(x \right)} d x} = - \frac{\sin^{7}{\left(x \right)}}{7} + \frac{3 \sin^{5}{\left(x \right)}}{5} + \sin{\left(x \right)} \cos^{2}{\left(x \right)}$$
積分定数を加える:
$$\int{\cos^{7}{\left(x \right)} d x} = - \frac{\sin^{7}{\left(x \right)}}{7} + \frac{3 \sin^{5}{\left(x \right)}}{5} + \sin{\left(x \right)} \cos^{2}{\left(x \right)}+C$$
解答
$$$\int \cos^{7}{\left(x \right)}\, dx = \left(- \frac{\sin^{7}{\left(x \right)}}{7} + \frac{3 \sin^{5}{\left(x \right)}}{5} + \sin{\left(x \right)} \cos^{2}{\left(x \right)}\right) + C$$$A