# Integrate problem

• Nov 30th 2010, 09:41 AM
reino17
Integrate problem
im sorry but i do not know how to use many of the math features to set up symbols and what not. so bare with me a little.

integrate: tanX ln(cosX) dx

so far i have:

sinX
----- ln(cosX)dx
cosX

let u= cosX

sinX
----- ln(u)dx
u

then:
du = sinX ln(u) dx
• Nov 30th 2010, 09:43 AM
Liverpool
What about substituting $\displaystyle u=ln(cos x)$ ?!!
• Nov 30th 2010, 10:16 AM
harish21
Quote:

Originally Posted by reino17
im sorry but i do not know how to use many of the math features to set up symbols and what not. so bare with me a little.

integrate: tanX ln(cosX) dx

so far i have:

sinX
----- ln(cosX)dx
cosX

let u= cosX

sinX
----- ln(u)dx
u

then:
du = sinX ln(u) dx

You dont really need to break that tan into sin and cos..

$\displaystyle \displaystyle \int \tan(x)\log(\cos(x)) \;dx$

Follow what Liverpool has said in the above post..

let $\displaystyle u=\log(\cos(x)) \rightarrow du\;=\;\dfrac{1}{\cos(x)} ({-\sin(x)}) dx = -\tan(x)\;dx \rightarrow dx = \dfrac{du}{\tan(x)}$

$\displaystyle \displaystyle \int \tan(x)\log(\cos(x)) \;dx = \int\tan(x)\;u\;du\; \dfrac{-1}{\tan(x)} =-\int u\;du$