I am trying to get result of this integral, but with no success. I know that I have to use integration by parts, but still I am lost. Thanks for your advice.

$$\int \sin (x) \ln (\tan (x))dx$$

Source (in czech language, it's the second integral):


Edit: when I use integration by parts I get

$$- \ln(\tan (x)) \cos (x) +\int \frac{1}{\tan (x)} \frac{1}{\cos (x)}dx $$

  • $\begingroup$ Which integration by parts did you try? Please be specific. $\endgroup$ – Did Feb 24 '14 at 12:59
  • $\begingroup$ In the formula $\int u \,dv=uv-\int v\,du$, use $dv=\sin x\,dx$, $u=\ln(\tan x)$. $\endgroup$ – David Mitra Feb 24 '14 at 12:59
  • $\begingroup$ You can also use the change of variable formula setting $u=\cos x$, then do something with the log and then integrate by parts. $\endgroup$ – Etienne Feb 24 '14 at 13:07
  • $\begingroup$ I edited my usage of integration, in this step I am lost $\endgroup$ – zdarsky.peter Feb 24 '14 at 13:11
  • 1
    $\begingroup$ You should have obtained $$-\ln(\tan x)\cos x +\int {1\over\tan x}{1\over \cos^2 x}\cdot\color{maroon}{\cos x}\,dx=-\ln(\tan x)\cos x +\int\csc x\,dx.$$ $\endgroup$ – David Mitra Feb 24 '14 at 13:33

Note that the derivative of $\log \tan x$ is $1/( \sin x\cos x) =2/\sin(2x)$ so your formula is wrong.

Using integration by parts, then

$$\int \sin(x) \log \tan(x) dx = -\cos x \log \tan x + \int \frac{1}{\sin x} dx$$

and that

$$\int \frac{1}{\sin x} dx = \int \frac{1}{2\sin (x/2) \cos(x/2)} dx = \log \tan(x/2)$$

  • 2
    $\begingroup$ definitely the simplest approach (+1) $\endgroup$ – robjohn Feb 24 '14 at 14:26

The easiest thing here is to separate the ln $$ \ln(\tan x) = \ln \left( \frac{\sin x}{\cos x}\right) = \ln(\sin x) - \ln(\cos x). $$

Then you get: $$ \int \sin x \ln(\sin x) \,dx - \int \sin x \ln(\cos x) \,dx. $$

The first one you can solve using integration by parts. A simple substitution will do for the second.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.