Evaluate ∫tan4⁡xdx

∫tan4⁡xdx=∫tan2⁡xtan2⁡xdx=∫tan2⁡xsec2⁡x−1dx∫tan2⁡xsec2⁡xdx−∫tan2⁡xdx=tan3⁡x3−∫sec2⁡x−1dx =tan3⁡x3−tan⁡x+x+C

Evaluate ∫tan4⁡xdx

∫tan4⁡xdx=∫tan2⁡xtan2⁡xdx=∫tan2⁡xsec2⁡x−1dx∫tan2⁡xsec2⁡xdx−∫tan2⁡xdx=tan3⁡x3−∫sec2⁡x−1dx =tan3⁡x3−tan⁡x+x+C

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$Evaluate \int \tan^{4} x d x$

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$\frac{\sec^{3} x}{3} - \tan x + C$

$\frac{\sec^{3} x}{3} - \tan x + x + C$

$\frac{\tan^{3} x}{3} + \tan x + x + C$

$\frac{\tan^{3} x}{3} - \tan x + x + C$

answer is D.

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Detailed Solution

$\begin{array}{l} \int \tan^{4} x d x = \int \tan^{2} x \tan^{2} x d x = \int \tan^{2} x (\sec^{2} x - 1) d x \\ \int \tan^{2} x \sec^{2} x d x - \int \tan^{2} x d x = \frac{\tan^{3} x}{3} - \int (\sec^{2} x - 1) d x \\ = \frac{\tan^{3} x}{3} - \tan x + x + C \end{array}$