Category: Integration

Question Number 126323 by mnjuly1970 last updated on 19/Dec/20 $$\dots advancedcalculus\dots$$$$provethat:::$$$$\frac{\mathrm{\Gamma }\left(\frac{1-x}{2}\right)\mathrm{\Gamma }\left(x\right)}{\mathrm{\Gamma }\left(\frac{x}{2}\right)}\stackrel{???}{=}\frac{2^{x-1}\sqrt{\pi }}{cos\left(\frac{\pi x}{2}\right)}$$$$$$ Terms of Service Privacy Policy…

Question Number 126316 by benjo_mathlover last updated on 19/Dec/20 $$\underset{\sqrt{2}}{\stackrel{2e^{\sqrt{2}}}{\int }}\ln \left(\frac{e^x+e^{-x}}{9\sqrt{x}}\right)dx?$$ Commented by liberty last updated on 19/Dec/20 $$\int\:\left(\mathrm{ln}\left(\:{e}^{{x}}…

Question Number 126314 by benjo_mathlover last updated on 19/Dec/20 $$\int \frac{dx}{(1+x^2)\sqrt{1-x^2}}?$$ Answered by Dwaipayan Shikari last updated on 19/Dec/20 $$\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\:\:\:\:\:\:\:{x}={sin}\theta…

Question Number 126274 by bramlexs22 last updated on 19/Dec/20 Answered by liberty last updated on 19/Dec/20 $$lettingx=\sqrt{2}\sec \ell with\to \{\begin{array}{l}\ell =\frac{\pi }{2}\\ \ell =\frac{\pi }{4}\end{array}$$$${\int }_{\pi /4}^{\pi /2}\frac{\sqrt{2}\sec \ell \tan \ell }{\sqrt{2}\sec \ell \sqrt{2\tan {}^2\ell }}d\ell =$$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\:{d}\ell\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\left[\:\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\:\right]\:=\:\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}\: \…

Question Number 60739 by Forkum Michael Choungong last updated on 25/May/19 $$evaluate$$$$i.\int \left(\frac{x+1}{x-1}\right)dx$$$$ii.{\int }_0^{\pi }\left(2cosxsinx\right)dx$$$$iii.{\int }_{\frac{\pi }{3}}^{\pi }\left(\frac{sin2x}{cos2x}\right)dx$$ Commented…