How do you find the antiderivative of $ln (cos x) tan x d x$ $ln(cosx)tanxdx$ ?

Question

2651 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-04T15:15:06

$- \frac{{(ln (cos x))}^{2}}{2} + C$ $-(ln(cosx))^2/2+C$

We wish to find:

$\int ln (cos x) tan x d x$ $intln(cosx)tanxdx$

Use substitution:

If $u = ln (cos x)$ $u=ln(cosx)$ , then $d u = \frac{- sin x}{cos x} d x = - tan x d x$ $du=(-sinx)/cosxdx=-tanxdx$ .

Thus,

$\int ln (cos x) tan x d x = - \int ln (cos x) (- tan x) d x = - \int u d u$ $intln(cosx)tanxdx=-intln(cosx)(-tanx)dx=-intudu$

This becomes

$- \frac{u^{2}}{2} + C = - \frac{{(ln (cos x))}^{2}}{2} + C$ $-u^2/2+C=-(ln(cosx))^2/2+C$

How do you find the antiderivative of ln(cosx)tanxdxln(cosx)tanxdxln(cosx)tanxdx?