How to integrate 2x sec^2 x dx?

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Answer 1 · 2018-05-11T17:10:31

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

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#