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