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Question
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Solution
\[\text{We have}, \]
\[I = \int \left( \sin^{- 1} x \right)^3 dx\]
\[\text{ Let,} \sin^{- 1} x = t\]
\[ \Rightarrow \sin t = x \Rightarrow \cos t = \sqrt{1 - x^2}\]
\[\text{Differentiating both sides we get}\]
\[\text{ cos t dt = dx}\]
\[\text{Now, integral becomes}\]
\[I = \int \left( \sin^{- 1} x \right)^3 dx\]
\[ = \int t^3 \text{ cos t dt }\]
\[ = t^3 \sin t - \int3 t^2 \text{ sin t dt}.......... \left( \text{ Using by parts} \right)\]
\[ = t^3 \sin t - 3\int t^2 \text{ sin t dt } \]
\[ = t^3 \sin t - 3\left[ - t^2 \cos t - \int - 2t \text{ cos t dt} \right]\]
\[ = t^3 \sin t + 3 t^2 \cos t - 6\int t \text{ cos t dt }\]
\[ = t^3 \sin t + 3 t^2 \cos t - 6\left[ t\sin t - \int\text{ sin t dt} \right]\]
\[ = t^3 \sin t + 3 t^2 \cos t - 6\left[ t \sin t + \cos t \right] + C\]
\[ = \left( \sin^{- 1} x \right)^3 x + 3 \left( \sin^{- 1} x \right)^2 \sqrt{1 - x^2} - 6\left( \sin^{- 1} x \right) x - 6\sqrt{1 - x^2} + C\]
\[ = x\left( \sin^{- 1} x \right)\left[ \left( \sin^{- 1} x \right)^2 - 6 \right] + 3\left[ \left( \sin^{- 1} x \right)^2 - 2 \right]\sqrt{1 - x^2} + C\]
