How to integrate 2x sec^2 x dx?

1 Answer
May 11, 2018

int2xsec^2xdx=2xtanx+2ln|cosx|+C

Explanation:

We have int2xsec^2xdx.

Apply Integration by Parts, making the following selections:

u=2x

du=2dx

dv=sec^2xdx

v=intsec^2xdx=tanx

uv-intvdu=2xtanx-2inttanxdx

inttanxdx=intsinx/cosxdx

We can solve this with a simple substitution:

u=cosx

du=-sinxdx

-int(du)/u=-ln|u|+C
=-ln|cosx|+C

Thus,

int2xsec^2xdx=2xtanx+2ln|cosx|+C