Category: Integration

Question Number 87121 by M±th+et£s last updated on 03/Apr/20 $${\int }_0^{\frac{\pi }{2}}\frac{1-x^4}{1+x^4}dx$$ Answered by redmiiuser last updated on 03/Apr/20 $$\left(\mathrm{1}+{x}^{\mathrm{4}} \right)^{−\mathrm{1}}…

Question Number 87103 by hamdhan last updated on 02/Apr/20 $$\int \frac{dx}{(1+x)\sqrt{x-x^2}}$$ Commented by abdomathmax last updated on 02/Apr/20 $$I=\int \frac{dx}{(x+1)\sqrt{-(x^2-x)}}=\int \frac{dx}{(x+1)\sqrt{-(x^2-x+\frac{1}{4}-\frac{1}{4})}}$$$$=\int\:\:\:\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}}…

Question Number 87052 by lémùst last updated on 02/Apr/20 $$f\left(x\right)={\int }_0^{\pi /2}\frac{{sin}^2\left(t\right)}{1+{xsin}^2\left(t\right)}dt$$ Commented by mathmax by abdo last updated on 02/Apr/20…

Question Number 152580 by aupo14 last updated on 29/Aug/21 Answered by Lordose last updated on 29/Aug/21 $$\mathrm{I}\:=\:\int\mathrm{e}^{\mathrm{sec}^{\mathrm{2}} \left(\mathrm{x}\right)} \mathrm{tan}\left(\mathrm{x}\right)\mathrm{dx}\:\overset{\mathrm{u}=\mathrm{sec}\left(\mathrm{x}\right)} {=}\int\frac{\:\mathrm{e}^{\mathrm{u}^{\mathrm{2}} } }{\mathrm{u}}\mathrm{du}\:\overset{\mathrm{y}=\mathrm{u}^{\mathrm{2}} } {=}\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\:\mathrm{e}^{\mathrm{y}} }{\mathrm{y}}\mathrm{dy}…

Question Number 152576 by Tawa11 last updated on 31/Aug/21 $$\int \frac{\mathrm{t}+13}{\sqrt[3]{{\mathrm{t}}^2+5\mathrm{t}+6}}\mathrm{dt}$$ Commented by MATHkingElsenK last updated on 31/Aug/21 $$hello5xno5t$$ Terms of…