Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 33628 by naka3546 last updated on 20/Apr/18

Commented by MJS last updated on 20/Apr/18

$$\mathrm{this}\mathrm{seems}\mathrm{to}\mathrm{be}=0$$

Commented by naka3546 last updated on 21/Apr/18

$$howtoget0$$

Commented by prof Abdo imad last updated on 21/Apr/18

$$letputI={\int }_0^4\frac{ln\left(x\right)}{\sqrt{4x-x^2}}dxwehave$$$$-x^2+4x=-(x^2-4x)=-(x^2-4x+4-4)$$$$=-{(x-2)}^2+4=4-{(x-2)}^{2}ch.x-2=2sintgive$$$$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{ln(2+2sint)}{2\sqrt{1-{sin}^{2}t}}2costdt$$$$={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}ln(2+2sint)dt$$$$=\pi ln\left(2\right)+{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}ln(1+sint)dtletfindthevalue$$$$ofthisintegralletf\left(x\right)={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}ln(1+xsint)dt$$$$with\mid x\mid ⩽1andx\ne 0$$$$f^{{}^{\prime }}\left(x\right)={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{sint}{1+xsint}dt$$$$=\frac{1}{x}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{1+xsint-1}{1+xsint}dt=\frac{\pi }{x}-\frac{1}{x}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{dt}{1+xsint}$$$$chtan\left(\frac{t}{2}\right)=ugive$$$${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{dt}{1+xsint}={\int }_{-1}^1\frac{1}{1+x\frac{2u}{1+u^2}}\frac{2du}{1+u^2}$$$$=2{\int }_{-1}^1\frac{du}{1+u^2+2xu}=2{\int }_{-1}^1\frac{du}{u^2+2xu+1}$$$$=2{\int }_{-1}^1\frac{du}{{(u+x)}^2+1-x^2}(ch.u+x=\sqrt{1-x^2}\alpha )$$$$=2{\int }_{\frac{x-1}{\sqrt{1-x^2}}}^{\frac{x+1}{\sqrt{1-x^2}}}\frac{\sqrt{1-x^2}}{(1-x^2)(1+{\alpha }^2)}d\alpha$$$$=\frac{2}{\sqrt{1-x^2}}{\int }_{-\frac{1}{\sqrt{1+x}}}^{\frac{1}{\sqrt{1-x}}}\frac{d\alpha }{1+{\alpha }^2}=\frac{2}{\sqrt{1-x^2}}{\left[arctan\left(\alpha \right)\right]}_{-\frac{1}{\sqrt{1+x}}}^{\frac{1}{\sqrt{1-x}}}$$$$=\frac{2}{\sqrt{1-x^2}}(arctan\left(\frac{1}{\sqrt{1-x}}\right)+arctan\left(\frac{1}{\sqrt{1+x)}}\right))$$$$=\frac{2}{\sqrt{1-x^2}}(\pi -arctan(\sqrt{1-x)}-arctan\left(\sqrt{1+x}\right))$$$$=\frac{2\pi }{\sqrt{1-x^2}}-\frac{2}{\sqrt{1-x^2}}(arctan(\sqrt{1-x)}+arctan\left(\sqrt{1+x}\right))$$$$...becontinued....$$

Commented by prof Abdo imad last updated on 21/Apr/18

$$letputI={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}ln(1+sinx)dx$$$$J={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}ln(1-sinx)dx.ch.x=-t\Rightarrow$$$$J={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}ln(1+sint)dt=I\Rightarrow$$$$2I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}ln(1-{sin}^{2}x)dx={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}2ln\left(cosx\right)dx$$$$=4{\int }_0^{\frac{\pi }{2}}ln\left(cosx\right)dx=4(-\frac{\pi }{2}ln2)$$$$=-2\pi ln\left(2\right)\Rightarrow I=-\pi ln\left(2\right)so$$$${\int }_0^4\frac{ln\left(x\right)}{\sqrt{-x^2+4x}}dx=\pi ln\left(2\right)-\pi ln\left(2\right)=0.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com