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