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Question Number 74995 by vishalbhardwaj last updated on 05/Dec/19

$$\mathrm{find}{\int }_0^{\frac{\pi }{2}}\mathrm{Log}\cos xdx$$

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

$$letI={\int }_0^{\frac{\pi }{2}}ln\left(cosx\right)dxandJ={\int }_0^{\frac{\pi }{2}}ln\left(sinx\right)dx$$$$wehsveJ{=}_{x=\frac{\pi }{2}-t}{\int }_{\frac{\pi }{2}}^{0}ln\left(cost\right)(-dt)={\int }_0^{\frac{\pi }{2}}ln\left(cost\right)dt=I\Rightarrow$$$$2I=I+J={\int }_0^{\frac{\pi }{2}}ln\left(cosxsinx\right)dx={\int }_0^{\frac{\pi }{2}}ln\left(\frac{1}{2}sin\left(2x\right)\right)dx$$$$=-\frac{\pi }{2}ln\left(2\right)+{\int }_0^{\frac{\pi }{2}}ln\left(sin\left(2x\right)\right)dxbut$$$${\int }_0^{\frac{\pi }{2}}ln\left(sin\left(2x\right)\right)dx{=}_{2x=t}\frac{1}{2}{\int }_0^{\pi }ln\left(sint\right)dt$$$$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}ln\left(sint\right)dt+\frac{1}{2}{\int }_{\frac{\pi }{2}}^{\pi }ln\left(sint\right)dt$$$${\int }_{\frac{\pi }{2}}^{\pi }ln\left(sint\right)dt{=}_{x=\frac{\pi }{2}+u}{\int }_0^{\frac{\pi }{2}}ln\left(cosu\right)du\Rightarrow$$$$2I=-\frac{\pi }{2}ln\left(2\right)+I\Rightarrow I=-\frac{\pi }{2}ln\left(2\right)wehaveprovedthat$$$${\int }_0^{\frac{\pi }{2}}ln\left(cosx\right)dx={\int }_0^{\frac{\pi }{2}}ln\left(sinx\right)dx=-\frac{\pi }{2}ln\left(2\right).$$

Answered by mind is power last updated on 05/Dec/19

$${\int }_0^{\frac{\pi }{2}}\log \left(\cos \left(\mathrm{x}\right)\right)\mathrm{dx}={\int }_0^{\frac{\pi }{2}}\log \left(\sin \left(\mathrm{x}\right)\right)\mathrm{dx},\therefore \mathrm{x}\to \frac{\pi }{2}-\mathrm{x}\therefore$$$${\int }_0^{\frac{\pi }{2}}\log \left(\sin \left(2\mathrm{x}\right)\right)\mathrm{dx}={\int }_0^{\pi }\log \left(2\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)\right)=\pi \log \left(2\right)+{\int }_0^{\frac{\pi }{2}}\log \left(\sin \left(\mathrm{x}\right)\right)\mathrm{dx}+{\int }_0^{\frac{\pi }{2}}\log \left(\cos \left(\mathrm{x}\right)\right)\mathrm{dx}$$$$=\pi \log \left(2\right)+2{\int }_0^{\frac{\pi }{2}}\log \left(\sin \left(\mathrm{x}\right)\right)\mathrm{dx}$$$$\mathrm{u}=2\mathrm{x}\Rightarrow ={\int }_0^{\pi }\frac{\log \left(\sin \left(\mathrm{u}\right)\right)\mathrm{du}}{2}$$$${\int }_0^{\pi }\log \left(\sin \left(\mathrm{u}\right)\right)\mathrm{du}={\int }_0^{\frac{\pi }{2}}\log \left(\sin \left(\mathrm{u}\right)\right)\mathrm{du}+{\int }_{\frac{\pi }{2}}^{\pi }\log \left(\sin \left(\mathrm{u}\right)\right)\mathrm{du}$$$${\int }_{\frac{\pi }{2}}^{\pi }\log \left(\sin \left(\mathrm{u}\right)\right)\mathrm{du}={\int }_0^{\frac{\pi }{2}}\log \left(\sin (\frac{\pi }{2}+\mathrm{u})\right)\mathrm{d}(\mathrm{u}+\frac{\pi }{2})={\int }_0^{\frac{\pi }{2}}\log \left(\cos \left(\mathrm{u}\right)\right)\mathrm{du}$$$$\Rightarrow {\int }_0^{\pi }\frac{\log \left(\sin \left(\mathrm{u}\right)\right)\mathrm{du}}{2}={\int }_0^{\frac{\pi }{2}}\log \left(\sin \left(\mathrm{u}\right)\right)\mathrm{du}$$$$\iff \pi \log \left(2\right)+2{\int }_0^{\frac{\pi }{2}}\log \left(\sin \left(\mathrm{u}\right)\right)\mathrm{du}={\int }_0^{\frac{\pi }{2}}\log \left(\sin \left(\mathrm{u}\right)\right)\mathrm{du}$$$$\Rightarrow {\int }_0^{\frac{\pi }{2}}\log \left(\sin \left(\mathrm{u}\right)\right)\mathrm{du}=-\pi \log \left(2\right)$$

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

$$errorofcalculussirmind..$$

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