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