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