$\int \sin ^{-3} x d x=\frac{-1}{-3} \sin ^{-4} x \cos x+\frac{-3-1}{-3} \int \sin ^{-5} x d x$

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The Answer of $\int \sin ^{-3} x d x=\frac{-1}{-3} \sin ^{-4} x \cos x+\frac{-3-1}{-3} \int \sin ^{-5} x d x$

∫sin⁡−3xdx=−1−3sin⁡−4xcos⁡x+−3−1−3∫sin⁡−5xdx\int \sin ^{-3} x d x=\frac{-1}{-3} \sin ^{-4} x \cos x+\frac{-3-1}{-3} \int \sin ^{-5} x d x∫sin−3xdx=−3−1​sin−4xcosx+−3−3−1​∫sin−5xdx