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 77232 by naka3546 last updated on 04/Jan/20

$$\underset{0}{\int }\stackrel{1}{}\ln (\sqrt{x}+\sqrt{1-x})dx=?$$

Answered by MJS last updated on 04/Jan/20

$$\int \ln (\sqrt{x}+\sqrt{1-x})dx=$$$$[t=\arcsin \sqrt{x}\to dx=2\sqrt{x(1-x)}dt]$$$$=2\int \cos t\sin t\ln (\cos t+\sin t)dt=$$$$\mathrm{by}\mathrm{parts}$$$$u=\ln (\cos t+\sin t)\to u^{\prime }=\frac{\cos t-\sin t}{\cos t+\sin t}$$$$v^{\prime }=\cos t\sin t\to v=-\frac{1}{2}{\cos }^{2}t$$$$=-{\cos }^{2}t\ln (\cos t+\sin t)+\int \frac{\cos t-\sin t}{\cos t+\sin t}{\cos }^{2}tdt=$$$$$$$$\int \frac{\cos t-\sin t}{\cos t+\sin t}{\cos }^{2}tdt=$$$$[u=\tan t\to dt={\cos }^{2}tdu]$$$$=-\int \frac{u-1}{(u+1){(u^2+1)}^2}du=$$$$=-\int \frac{u}{{(u^2+1)}^2}du-\frac{1}{2}\int \frac{u-1}{u^2+1}du+\frac{1}{2}\int \frac{du}{u+1}=$$$$=\frac{1}{2(u^2+1)}-\frac{1}{4}\ln (u^2+1)+\frac{1}{2}\arctan u+\frac{1}{2}\ln (u+1)=$$$$=\frac{1}{2}{\cos }^{2}t+\frac{1}{2}\ln \cos x+\frac{1}{2}t+\frac{1}{2}\ln (1+\tan x)$$$$$$$$=\frac{1}{2}(t+{\cos }^{2}t+\ln (\cos t+\sin t))=$$$$=\frac{1}{2}(-x+\ln (\sqrt{x}+\sqrt{1-x})+\arcsin \sqrt{x})+C$$$$\Rightarrow \mathrm{answer}\mathrm{is}\frac{\pi }{4}-\frac{1}{2}$$

Commented by naka3546 last updated on 04/Jan/20

$$\frac{\pi }{4}+\frac{1}{2}or\frac{\pi }{4}-\frac{1}{2}?$$

Commented by MJS last updated on 04/Jan/20

$$-$$$$\mathrm{sorry}\mathrm{typo}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com