1/(sin x)^3の積分


\int \frac{1}{\sin^3 x}dxを求めよ.

やってみることにする.
解)
\int \frac{1}{\sin^3 x}dx
=\int \frac{\sin x}{\sin^4 x}dx
=\int \frac{\sin x}{(1-\cos^2 x)^2}dx
=\int  \frac{\sin x}{4}(\frac{1}{1+\cos x}+\frac{1}{1-\cos x})^{2} dx(部分分数分解)
=\int  \frac{\sin x}{4}(\frac{1}{(1+\cos x)^2}+\frac{2}{(1+\cos x)(1-\cos x)}+\frac{1}{(1-\cos x)^2}) dx
=\int   \frac{\sin x}{4(1+\cos x)^2} dx+ \int \frac{\sin x}{2(1+\cos x)(1-\cos x)}dx+ \int \frac{\sin x}{4(1-\cos x)^2}dx
=\int   \frac{\sin x}{4(1+\cos x)^2} dx+ \int ( \frac{\sin x}{4(1+\cos x)}+\frac{\sin x}{8(1-\cos x)})dx+ \int \frac{\sin x}{4(1-\cos x)^2}dx(部分分数分解)
=\frac{-1}{4(1+\cos x)}-\frac{1}{4} \log |1+\cos x|+\frac{1}{4} \log |1-\cos x|+\frac{-1}{4(1-\cos x)}+C.(Cは積分定数

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