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$\int \sec x \tan ^3 x d x=$
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The correct answer is:
$\frac{1}{3} \sec ^3 x-\sec x+c$
$\begin{aligned} & \int \sec x \tan ^3 x d x=\int \sec x\left(\sec ^2 x-1\right) \tan x d x \\ & =\int \sec x \tan x \sec ^2 x d x-\int \sec x \tan x d x \\ & =\frac{\sec ^3 x}{3}-\sec x+c, \quad \text { (Putting } \sec x=t \text { in first part). }\end{aligned}$
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