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प्रश्न
\[\int \sin^3 x \cos^4 x\ \text{ dx }\]
योग
उत्तर
\[\text{ Let I }= \int \sin^3 x \cdot \cos^4 x\ dx\]
\[ = \int \sin^2 x \cdot \sin x \cdot \cos^4 x\ dx\]
\[ = \int\left( 1 - \cos^2 x \right) \cdot \cos^4 x \cdot \sin x\ dx \]
\[ = \int\left( \cos^4 x - \cos^6 x \right) \cdot \sin x\ dx\]
\[\text{ Putting cos x = t}\]
\[ \Rightarrow - \sin x\ dx = dt\]
\[ \Rightarrow \sin x\ dx = - dt\]
\[ \therefore I = - \int\left( t^4 - t^6 \right)dt\]
\[ = \int\left( t^6 - t^4 \right)dt\]
\[ = \frac{t^7}{7} - \frac{t^5}{5} + C\]
\[ = \frac{\cos^7 x}{7} - \frac{\cos^5 x}{5} + C......... \left[ \because t = \cos x \right]\]
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