How do you integrate $∫sec^2 (x) tan^3 (x)dx$ ?

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Answer 1 · 2015-03-06T02:43:57

Derivative of $tanx$ is $sec^2x$ , so use substitution.

$intsec^2xtan^3xdx=int(tanx)^3sec^2xdx$ .

You can probably figure it out from there, but just in case:

Let $u=tanx$ , then $du=sec^2xdx$ , so the integral becomes $intu^3du$ , which, of course, evaluates to $1/4tan^4x+C$ .

How do you integrate ∫sec^2 (x) tan^3 (x)dx∫sec^2 (x) tan^3 (x)dx?