now we have an integral \(\int x\cos x\ln\lvert \sin x\rvert dx\). though this may look more complex, it is actually more simple. we can apply integration by parts again. The requirement for integration by parts is that one variable must be easy to differentiate, in this case it is \(x\), and one variable must be easy to integrate, in this case \(\cos x\ln\lvert\sin x\rvert dx\). We know this is easy to integrate as we can see there is a \(\cos x\) and a \(\sin x\), so we can use a simple u-substitution. To make things simple, i shall removed the modulus sigh, but u can factor it back in and u will get 2 answers for the integral

this last integral \(\int (\sin x\ln(\sin x)-\sin x)dx\)is essentially \(\int\sin x\ln(\sin x)dx - \int \sin x dx\)

\(\int\sin x\ln(\sin x) dx\) can be found by integration by parts. after finding that, sub everything back into the orginal equation for your final answer

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## Comments

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TopNewesthttp://www.wolframalpha.com/input/?i=integral+ln%7Csin%28x%29%7C I think you might've made a typo.

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what typo?

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Well the function you gave doesn't have an elementary antiderivative.

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If you can't see the link, Wolfram Alpha says this evaluates to something nasty that involves imaginaries.

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Integration by parts formula states that \(\int u_{}dv = uv-\int v_{}du\)

In this case we have \(\int \ln\lvert\sin x\rvert dx\)

comparing this to the integration by parts formula

\(u = \ln\lvert\sin x\rvert\Rightarrow du = \cos x\ln\lvert\sin x\rvert dx\)

\(dv = dx \Rightarrow v = x\)

Now we can apply the integration by parts formula

\(\int \ln\lvert\sin x\rvert dx = x\ln\lvert\sin x\rvert-\int x\cos x\ln\lvert\sin x\rvert dx\)

now we have an integral \(\int x\cos x\ln\lvert \sin x\rvert dx\). though this may look more complex, it is actually more simple. we can apply integration by parts again. The requirement for integration by parts is that one variable must be easy to differentiate, in this case it is \(x\), and one variable must be easy to integrate, in this case \(\cos x\ln\lvert\sin x\rvert dx\). We know this is easy to integrate as we can see there is a \(\cos x\) and a \(\sin x\), so we can use a simple u-substitution. To make things simple, i shall removed the modulus sigh, but u can factor it back in and u will get 2 answers for the integral

Applying integration by parts:

\(u = x\Rightarrow du = dx\)

\(dv = \cos x\ln(sin x) dx \Rightarrow v = \sin x\ln(\sin x)-\sin x\) <-- u-substitution

\(\int x\cos x\ln( \sin x) dx = x(\sin x\ln(\sin x) - \sin x)- \int (\sin x\ln(\sin x)-\sin x)dx\)

this last integral \(\int (\sin x\ln(\sin x)-\sin x)dx\)is essentially \(\int\sin x\ln(\sin x)dx - \int \sin x dx\)

\(\int\sin x\ln(\sin x) dx\) can be found by integration by parts. after finding that, sub everything back into the orginal equation for your final answer

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itu integralnya buat cari rumus apa? kecepatan, percepatan atau posisi? atau yang lain?

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By partial method you mean by the method of partial sums?

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i mean partial is integral of u dv=u.v-integral of v du. you can do it?

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