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

Question

Aurora Alvarado · Accepted Answer

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#.

Ezekiel Alexander · Answer

undefined To integrate $ \int x \tan^2(x) \, dx $ using integration by parts, we choose $ u = x $ and $ dv = \tan^2(x) \, dx $. Then, $ du = dx $ and $ v = \tan(x) - x $.

Applying the integration by parts formula:
$$ \int u \, dv = uv - \int v \, du $$

Substitute the values of $ u $, $ dv $, $ du $, and $ v $:
$$ \int x \tan^2(x) \, dx = x \cdot (\tan(x) - x) - \int (\tan(x) - x) \, dx $$

Simplify the integral:
$$ \int x \tan^2(x) \, dx = x \tan(x) - x^2 - \int \tan(x) \, dx + \int x \, dx $$

$$ \int x \tan^2(x) \, dx = x \tan(x) - x^2 - \ln|\sec(x)| + \frac{1}{2}x^2 + C $$

So, the antiderivative of $ \int x \tan^2(x) \, dx $ using integration by parts is $ x \tan(x) - x^2 - \ln|\sec(x)| + \frac{1}{2}x^2 + C $, where $ C $ is the constant of integration.