Question

Help, quiz tomorrow andIdon't know how to do the Integral of tanx*ln(cosx) dx

Answer 1

\[\int\limits_{0}^{\pi/3}tanx*\ln(cosx)dx\]

Answer 2

\[\int_0^{\pi/3}\tan x \;\ln(\cos x) \;dx\\
\int_0^{\pi/3}\frac{\sin x}{\cos x} \;\ln(\cos x) \;dx\]
Have you tried a substitution?

Answer 3

I did try a u sub with u=cos x

Answer 4

that should work... ^^^

Answer 5

I got du= -sinx dx
cancelled the sin x , but then couldntdo the 1/u ln(u) integral :(

Answer 6

help!!!!!!!

Answer 7

try u =ln(cosx)

Answer 8

So, using the substitution
\[u=\cos x,\\
du=-\sin x \;dx\\
-du=\sin x\; dx,\]
the integral becomes
\[-\int_1^{\frac{1}{2}}\frac{\ln u}{u}du\\
\int_{\frac{1}{2}}^{1}\frac{\ln u}{u}du\]
Try another substitution. @joemath314159's sub is quicker, but I've a preference for step-by-step subs.

Answer 9

I am gonna try u=ln(cosx)

Answer 10

I got it, thank you joemath and giggles. I always feel like you cant do an f(g(x)) u substittution