Question Number 221830 by Nicholas666 last updated on 11/Jun/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\sqrt{{x}}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx} \\ $$$$ \\ $$ Terms of Service Privacy Policy…
Question Number 221831 by Nicholas666 last updated on 11/Jun/25 $$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\:\frac{\sqrt{{x}}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx} \\ $$$$ \\ $$ Answered by MathematicalUser2357 last updated on 25/Jun/25…
Question Number 221769 by MrGaster last updated on 10/Jun/25 $$\int_{\mathrm{0}} ^{\pi} \frac{{a}}{{a}−\mathrm{cos}^{\mathrm{2}{n}} {x}}{dx}=?,{a}>\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{2}}{\mathrm{2}−\mathrm{cos}^{\mathrm{4}} {x}}{dx}=? \\ $$$$\underset{{m}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{cos}^{\mathrm{2}{n}} \left(\mathrm{2}{mx}\right)}{{a}−\mathrm{cos}^{\mathrm{2}{n}} {x}}{dx}=?,{a}>\mathrm{1},{n}\in\mathbb{N}^{+}…
Question Number 221770 by MrGaster last updated on 10/Jun/25 $$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{arcsin}{x}}{\mathrm{1}+{x}^{\mathrm{4}} }{dx}=\frac{\sqrt{\mathrm{2}}\pi^{\mathrm{2}} }{\mathrm{16}}−\frac{\sqrt{\mathrm{2}}\pi}{\mathrm{8}}\mathrm{ln}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)+\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)}\mid{z}_{\mathrm{0}} \mid^{\mathrm{2}{n}+\mathrm{1}} \mathrm{sin}\left(\frac{\pi}{\mathrm{4}}−\left(\mathrm{2}{n}+\mathrm{1}\right)\beta\right) \\ $$ Terms of Service Privacy…
Question Number 221663 by MrGaster last updated on 09/Jun/25 $$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{k}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−{x}^{{k}} \right){dx}=\frac{\mathrm{4}\pi\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{23}}}\centerdot\frac{\mathrm{sinh}\frac{\sqrt{\mathrm{23}}\pi}{\mathrm{6}}}{\mathrm{2}\:\mathrm{cosh}\frac{\sqrt{\mathrm{23}}\pi}{\mathrm{3}}−\mathrm{1}} \\ $$ Commented by MrGaster last updated on 09/Jun/25 It is difficult for me to give an analytical solution to that integral.…
Question Number 221680 by ajfour last updated on 09/Jun/25 $${I}\:{suspect} \\ $$$$\pi={i}\underset{\boldsymbol{{i}}} {\overset{−\mathrm{1}} {\int}}\frac{\left({z}−\mathrm{1}\right){dz}}{\:{z}\sqrt{{z}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${someone}\:{please}\:{help}\:{confirm}\:{or}\:{reject}! \\ $$ Commented by fantastic last updated on…
Question Number 221588 by Nicholas666 last updated on 08/Jun/25 $$ \\ $$$$\:\:\:\:\int\:\frac{\mathrm{8}{t}\:−\:\mathrm{8}{t}^{\:\mathrm{3}} }{{t}^{\:\mathrm{6}} \:+\:\mathrm{6}{t}^{\mathrm{5}} \:+\:\mathrm{3}{t}^{\:\mathrm{4}} \:−\:\mathrm{20}{t}^{\mathrm{3}} \:+\:\mathrm{3}{t}^{\mathrm{2}} \:+\:\mathrm{6}{t}\:+\:\mathrm{1}}\:{dt}\:\:\:\: \\ $$$$ \\ $$ Answered by Frix…
Question Number 221586 by Nicholas666 last updated on 08/Jun/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{1}\:+\:\mathrm{3}{x}\:}\:{dx} \\ $$$$ \\ $$ Answered by MrGaster last updated on 08/Jun/25 Commented by…
Question Number 221587 by Tawa11 last updated on 08/Jun/25 $$\int_{\:\mathrm{2}} ^{\:\mathrm{3}} \:\frac{\mathrm{tan}^{−\:\mathrm{1}} \left(\mathrm{x}\right)}{\mathrm{1}\:\:−\:\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx} \\ $$ Answered by maths2 last updated on 09/Jun/25 $$=\int_{\mathrm{2}} ^{\mathrm{3}}…
Question Number 221582 by MrGaster last updated on 08/Jun/25 $$\left(\mathrm{1}\right):\int_{\mathrm{0}} ^{{u}} {x}^{\nu−\mathrm{1}} \left({u}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\varrho−\mathrm{1}} {e}^{\mu{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mu^{{n}} }{{n}!}\int_{\mathrm{0}} ^{{u}} {x}^{\nu+{n}−\mathrm{1}} \left({u}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\varrho−\mathrm{1}}…