How do you evaluate the integral of $int tan(x)ln(cosx) dx$ ?

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Answer 1 · 2016-05-03T23:02:07

$-ln^2(cosx)/2+C$

We can solve this integral using u-substitution, but it is fairly unapparent from the outset.

If we let $u=ln(cosx)$ , differentiating this reveals that

$du=(d/dx(cosx))/cosxdx=(-sinx)/cosxdx=-tanxdx$

Multiply the interior and exterior of the integral by $-1$ .

$inttanxln(cosx)dx=-int-tanxln(cosx)dx=-intudu$

Integrating this gives $-u^2/2+C$ , and since $u=ln(cosx)$ , this becomes $-ln^2(cosx)/2+C$ .

How do you evaluate the integral of int tan(x)ln(cosx) dxint tan(x)ln(cosx) dx?