I get the integration $$ \displaystyle\int \limits^{2\pi }_{0}\ln\left| 2\sin \left( x\right) +3\cos \left( x\right) \right| dx$$ I tried to solve it and find the close form but I couldn't. And I use W|A but it was useless.. my question is how I can get the close form for this integration?.
Integral $\int \limits^{2\pi }_{0}\ln\left| 2\sin \left( x\right) +3\cos \left( x\right) \right| dx$
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Why do you think there is "nice" closed form? – Galc127 Feb 26 '16 at 08:53
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What we have to do is essentially same as Computing the integral of $\log(\sin x)$, after a few steps of reducing your problem into this type by noticing $$2\sin x+3\cos x=\sqrt{13}\sin(x+\alpha)\quad where\quad \sin\alpha=3/\sqrt{13},\ \cos\alpha=2/\sqrt{13}$$ and $$\begin{array}{l}\int_0^{2\pi}\log|2\sin x+3\cos x|\mathrm{d}x=\\\int_0^{\pi-\alpha}\log(\sqrt{13}\sin(x+\alpha))\mathrm{d}x+\int_{\pi-\alpha}^{2\pi-\alpha}\log(-\sqrt{13}\sin(x+\alpha))\mathrm{d}x\\+\int_{2\pi-\alpha}^{2\pi}\log(\sqrt{13}\sin(x+\alpha))\mathrm{d}x.\end{array}$$ Note that the first term and third term of the RHS can be combined into $$\int_0^{\pi}\log(\sqrt{13}\sin x)\mathrm{d}x.$$ After all, we can obtain the answer by evaluating the above integral and multiply the result by 2.
Katie Imach
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2(+1) The final answer is so $\pi\log\left(\frac{13}{4}\right)$. – Jack D'Aurizio Feb 26 '16 at 09:21