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