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Question
\[\int \sin^5 x\ dx\]
Sum
Solution
\[\text{ Let I }= \int \sin^5 x \text{ dx }\]
\[ = \int \sin^4 x \cdot \text{ sin x dx}\]
\[ = \int \left( \sin^2 x \right)^2 \text{ sin x dx}\]
\[ = \int \left( 1 - \cos^2 x \right)^2 \text{ sin x dx}\]
\[ = \int\left( \cos^4 x - 2 \cos^2 x + 1 \right) \text{ sin x dx}\]
\[\text{ Putting cos x = t}\]
\[ \Rightarrow - \text{ sin x dx} = dt\]
\[ \Rightarrow \text{ sin x dx} = - dt\]
\[ \therefore I = - \int\left( t^4 - 2 t^2 + 1 \right) dt\]
\[ = - \int t^4 dt + 2\int t^2 dt - \int dt\]
\[ = \frac{- t^5}{5} + \frac{2 t^3}{3} - t + C\]
\[ = \frac{- \cos^5 x}{5} + \frac{2}{3} \text{ cos}^3 x - \cos x + C .......\left[ \because t = \cos x \right]\]
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