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Question Number 124432 by mnjuly1970 last updated on 03/Dec/20

$$...nicecalculus...$$$$find::$$$$\varphi ={\int }_0^4\frac{ln\left(x\right)}{{(4x-x^2)}^{\frac{1}{2}}}dx=?$$

Answered by mindispower last updated on 03/Dec/20

$$x=4t$$$$q={\int }_0^1\frac{ln\left(4\right)+ln\left(t\right)}{{(t-t^2)}^{\frac{1}{2}}}dt$$$$w=ln\left(4\right){\int }_0^1\frac{1}{t^{\frac{1}{2}}{(1-t)}^{\frac{1}{2}}}+{\int }_0^1\frac{ln\left(t\right)}{\sqrt{t}.\sqrt{1-t}}dt$$$$=ln\left(4\right).{\int }_0^1{t}^{\frac{1}{2}-}{(1-t)}^{\frac{1}{2}-1}dt+{\partial }_c{\int }_0^1{t}^{\frac{1}{2}+c-1}{(1-t)}^{\frac{1}{2}-1}dt{\mid }_{c=0}$$$$=ln\left(4\right)\beta (\frac{1}{2},\frac{1}{2})+{\partial }_C\beta (\frac{1}{2}+c,\frac{1}{2}){\mid }_{c=0}$$$$=\frac{ln\left(4\right)\pi }{sin\left(\frac{\pi }{2}\right)}+\beta (\frac{1}{2}+c,\frac{1}{2})(\mathrm{\Psi }(\frac{1}{2}+c)-\mathrm{\Psi }(1+c)){\mid }_{c=0}$$$$=\pi ln\left(4\right)+\pi (\mathrm{\Psi }\left(\frac{1}{2}\right)-\mathrm{\Psi }\left(1\right))$$$$$$$$=\pi (ln\left(4\right)+\pi (-2ln\left(2\right)-\gamma +\gamma )$$$$=0$$

Commented by mnjuly1970 last updated on 03/Dec/20

$$merceymrmindspower..$$$$thanksalot...$$

Commented by mindispower last updated on 05/Dec/20

$$withepleasursir$$

Answered by mnjuly1970 last updated on 03/Dec/20

$$anotherway$$$$weknow:$$$${\int }_0^{\frac{\pi }{2}}ln\left(sin\left(x\right)\right)dx=\frac{-\pi }{2}ln\left(2\right)✓$$$$x=4y$$$$\varphi ={\int }_0^1\frac{ln\left(4\right)+ln\left(y\right)}{\sqrt{y}\sqrt{1-y}}dy$$$$\stackrel{y={sin}^2\left(t\right)}{=}{\int }_0^{\frac{\pi }{2}}\left\{\frac{2ln\left(2\right)+2ln(sin\left(t\right)}{sin\left(t\right)cos\left(t\right)}\right\}\left(2sin\left(t\right)cos\left(t\right)\right)dt$$$$=2\pi ln\left(2\right)+4{\int }_0^{\frac{\pi }{2}}ln\left(sin\left(t\right)\right)dt$$$$=2\pi ln\left(2\right)-2\pi ln\left(2\right)=0✓$$

Answered by mathmax by abdo last updated on 03/Dec/20

$$\mathrm{A}={\int }_0^4\frac{\mathrm{lnx}}{\sqrt{4\mathrm{x}-{\mathrm{x}}^2}}\mathrm{dx}\Rightarrow \mathrm{A}={\int }_0^4\frac{\mathrm{lnx}}{\sqrt{-({\mathrm{x}}^2-4\mathrm{x}+4-4)}}\mathrm{dx}$$$$={\int }_0^4\frac{\mathrm{lnx}}{\sqrt{4-{(\mathrm{x}-2)}^2}}\mathrm{dx}{=}_{\mathrm{x}-2=2\mathrm{cost}}{\int }_{\pi }^0\frac{\ln (2+2\mathrm{cost})}{2\mathrm{sint}}(-2\mathrm{sint})\mathrm{dt}$$$$={\int }_0^{\pi }\ln \left(2(1+\mathrm{cost})\right)\mathrm{dt}={\int }_0^{\pi }\ln \left({4\cos }^2\left(\frac{\mathrm{t}}{2}\right)\right)\mathrm{dt}$$$$=2\pi \ln \left(2\right)+2{\int }_0^{\pi }\ln \left(\cos \left(\frac{\mathrm{t}}{2}\right)\right)\mathrm{dt}(\to \frac{\mathrm{t}}{2}=\mathrm{x})$$$$=2\pi \ln \left(2\right)+2{\int }_0^{\frac{\pi }{2}}\ln \left(\mathrm{cosx}\right)\left(2\mathrm{dx}\right)=2\pi \ln \left(2\right)+4(-\frac{\pi }{2}\ln 2)$$$$=2\pi \ln 2-2\pi \ln \left(2\right)=0\Rightarrow \mathrm{A}=0$$

Commented by mnjuly1970 last updated on 03/Dec/20

$$thanksalotsirmax$$$$excellent$$

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