Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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

2 Answers 2

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 .

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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