[미적분학1] 쌍곡선 함수로 치환하여 적분하기

a>0 일 때

\int \frac{1}{\sqrt{a^2+x^2}}dx=\int \frac{1}{\sqrt{a^2+a^2sinh^2\theta}}acosh\theta d\theta \\ { } \hspace{93} =\int\frac{acosh\theta}{achsh\theta} d\theta = \int d\theta = \theta + C \\ { } \hspace{93} = sinh^{-1}\frac{x}{a} +C \\ { } \hspace{93} =ln\left(\frac{x}{a}+\sqrt{\left(\frac{x}{a} \right )^2+1} \right) +C \hspace{10}

\int\frac{1}{\sqrt{x^2-a^2}}dx = \int \frac{1}{\sqrt{a^2cosh^2\theta-a^2}}asinh\theta d\theta \\ { } \hspace{93} = \int \frac{asinh\theta}{asinh\theta}d\theta = \int d\theta = \theta + C \\ { } \hspace{93} = cosh^{-1}\frac{x}{a} + C \\ { } \hspace{93} = ln \left( \frac{x}{a}+\sqrt{\left(\frac{x}{a} \right )^2-1} \right ) + C \hspace{10}, \hspace{5}x \geq a

\int\frac{1}{a^2-x^2}dx = \int \frac{1}{a^2-a^2tanh^2\theta}asech^2\theta d\theta \\ { } \hspace{93} = \int \frac{asech^2\theta}{asech^2\theta}d\theta = \int d\theta = \theta + C \\ { } \hspace{93} = tanh^{-1}\frac{x}{a} + C \\ { } \hspace{93} = \frac{1}{2}ln \left( \frac{1+\frac{x}{a}}{1-\frac{x}{a}} \right ) + C \hspace{10}, \hspace{5} -a< x< a