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$$$1 - \sec^{2}{\left(x \right)}$$$の積分
入力内容
$$$\int \left(1 - \sec^{2}{\left(x \right)}\right)\, dx$$$ を求めよ。
解答
項別に積分せよ:
$${\color{red}{\int{\left(1 - \sec^{2}{\left(x \right)}\right)d x}}} = {\color{red}{\left(\int{1 d x} - \int{\sec^{2}{\left(x \right)} d x}\right)}}$$
$$$c=1$$$ に対して定数則 $$$\int c\, dx = c x$$$ を適用する:
$$- \int{\sec^{2}{\left(x \right)} d x} + {\color{red}{\int{1 d x}}} = - \int{\sec^{2}{\left(x \right)} d x} + {\color{red}{x}}$$
$$$\sec^{2}{\left(x \right)}$$$ の不定積分は $$$\int{\sec^{2}{\left(x \right)} d x} = \tan{\left(x \right)}$$$ です:
$$x - {\color{red}{\int{\sec^{2}{\left(x \right)} d x}}} = x - {\color{red}{\tan{\left(x \right)}}}$$
したがって、
$$\int{\left(1 - \sec^{2}{\left(x \right)}\right)d x} = x - \tan{\left(x \right)}$$
積分定数を加える:
$$\int{\left(1 - \sec^{2}{\left(x \right)}\right)d x} = x - \tan{\left(x \right)}+C$$
解答
$$$\int \left(1 - \sec^{2}{\left(x \right)}\right)\, dx = \left(x - \tan{\left(x \right)}\right) + C$$$A