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प्रश्न
उत्तर
\[\int \sec^4 \text{x . tan x dx}\]
\[ = \int \sec^2 x . \sec^2 \text{x tan x dx}\]
\[ = \int \left( 1 + \tan^2 x \right) \sec^2 \text{x . tan x dx}\]
\[ = \int \left( \tan x + \tan^3 x \right) \sec^2 \text{x dx}\]
\[\text{Let tan x} = t\]
\[ \Rightarrow \sec^2 \text{x dx} = dt\]
\[Now, \int \left( \tan x + \tan^3 x \right) \sec^2\text{ x dx}\]
\[ = \int\left( t + t^3 \right) dt\]
\[ = \frac{t^2}{2} + \frac{t^4}{4} + C\]
\[ = \frac{\tan^2 x}{2} + \frac{\tan^4 x}{4} + C\]
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