How do you integrate $int xtan^2x$ using integration by parts?

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Answer 1 · 2016-08-26T17:53:50

Use $tan^2 x = sec^2 x -1$ first.

$x tan^2 x = xsec^2 x - x$

$-int x dx = -x^2/2$

$int x sec^2 x dx$

Let $u = x$ and $dv = sec^2 x dx$ , so that

$du = dx$ and $v = tan x$ to get

$int x sec^2 x dx = x tan x - int tan x dx$ .

Now integrate $tan x = sinx/cosx$ using substitution $u = cos x$ .

$int x sec^2 x dx = x tan x - (- ln abs(cosx)) = x tan x + ln abs cosx$

Finish by putting it all together.

$int x tan^2 x = -x^2/2 + x tan x + ln abs cosx +C$ .

How do you integrate int xtan^2xint xtan^2x using integration by parts?