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Question Number 70262 by mathmax by abdo last updated on 02/Oct/19

$$calculate{\int }_0^{\frac{\pi }{4}}ln\left(cosx\right)dxand{\int }_0^{\frac{\pi }{4}}ln\left(sinx\right)dx$$

Commented by mathmax by abdo last updated on 06/Oct/19

$$letI={\int }_0^{\frac{\pi }{4}}ln\left(cosx\right)dxandJ={\int }_0^{\frac{\pi }{4}}ln\left(sinx\right)dx$$$$I+J={\int }_0^{\frac{\pi }{4}}ln(cosx+sinx)dx={\int }_0^{\frac{\pi }{4}}ln\left(\sqrt{2}cos(x-\frac{\pi }{4})\right)dx$$$$=\frac{\pi }{8}ln\left(2\right)+{\int }_0^{\frac{\pi }{4}}ln\left(cos(\frac{\pi }{4}-x)\right)dxbut$$$${\int }_0^{\frac{\pi }{4}}ln\left(cos(\frac{\pi }{4}-x)\right)dx{=}_{\frac{\pi }{4}-x=t}{\int }_0^{\frac{\pi }{4}}ln\left(cost\right)dt=I\Rightarrow$$$$I+J=\frac{\pi }{8}ln\left(2\right)+I\Rightarrow J=\frac{\pi }{8}ln\left(2\right)$$$$I={\int }_0^{\frac{\pi }{4}}ln\left(cosx\right)dx{=}_{x=\frac{t}{2}}\frac{1}{2}{\int }_0^{\frac{\pi }{2}}ln\left(cos\left(\frac{t}{2}\right)\right)dt\Rightarrow$$$$2I={\int }_0^{\frac{\pi }{2}}ln\left(sin(\frac{\pi }{2}-\frac{t}{2})\right)dt{=}_{\frac{\pi -t}{2}=u}{\int }_{\frac{\pi }{2}}^{\frac{\pi }{4}}ln\left(sinu\right)(-2)du$$$$=2{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}ln\left(sinu\right)du=2\{{\int }_{\frac{\pi }{4}}^{0}ln\left(sinu\right)du+{\int }_0^{\frac{\pi }{2}}ln\left(sinu\right)du\}$$$$=2\{-J-\frac{\pi }{2}ln\left(2\right)\}=-2J-\pi ln\left(2\right)=-\frac{\pi }{4}ln\left(2\right)-\pi ln\left(2\right)$$$$=-\frac{5\pi }{4}ln\left(2\right)\Rightarrow I=-\frac{5\pi }{8}ln\left(2\right)$$$$$$

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