How do you find the indefinite integral of ∫ x/ cos^2 x dx ?

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Answer 1 · 2018-04-14T22:15:10

#intx/cos^2xdx=xtanx+ln|cosx|+C#

So, we want

#intx/cos^2xdx#

We may rewrite as

#intxsec^2xdx# and then apply Integration by Parts.

#u=x#

#du=dx#

#dv=sec^2xdx#

#v=tanx#

Note why we made our selections this way, as #x# has a simple differential and #sec^2x# has a simple integral.

Thus,

#uv-intvdu=xtanx-inttanxdx#

Now,

#inttanxdx=intsinx/cosxdx#

A simple substitution will work:

#w=cosx#

#dw=-sinxdx#

#-dw=sinxdx#

#-int(dw)/w=-ln|w|=-ln|cosx|#

So,

#intx/cos^2xdx=xtanx+ln|cosx|+C#