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I need help. I have to integrate $\cos^{3} \cdot \ln(\sin(x))$ and I don´t know how to solve it. In our book it is that we have to solve using the substitution method. If somebody knows it, you will help me..please

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substitute $u(x)=\sin(x)$ – Blah Feb 25 '12 at 9:50
How can you use Pythagorean trigonometric identity to procede with Blah's method? – savick01 Feb 25 '12 at 9:53
The expression is meaningless as written; is that first factor supposed to be $\cos^3 x$? – Brian M. Scott Feb 25 '12 at 9:55
I think you mean $\cos^3x\log(\sin x)$. Is that what you want to integrate? – Gerry Myerson Feb 25 '12 at 9:56
Oh, I sorry, I forgot argument of the function x, of course I thought cos^{3}x* ln(sin x). I've counted, thank you for your help. – Clare Feb 25 '12 at 10:33

Substitute :

$\sin x =t \Rightarrow \cos x dx =dt$ , hence :

$I=\int (1-t^2)\cdot \ln (t) \,dt$

This integral you can solve using integration by parts method .

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If you mean $\cos^3 x\ln(\sin x)$, let $u=\sin x$. Then $du=\cos x dx$, and $$\begin{align*} \cos^3 x\ln(\sin x)dx&=\cos^2 x\ln(\sin x)\Big(\cos x dx\Big)\\ &=\cos^2 x\ln u \,du\\ &=(1-\sin^2 x)\ln u\,du\\ &=(1-u^2)\ln u\,du\;, \end{align*}$$

which can be integrated by parts.

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