Category: Integration

Question Number 196983 by universe last updated on 05/Sep/23 Commented by Frix last updated on 06/Sep/23 $$\mathrm{I}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{solve}\mathrm{the}\mathrm{first}2\mathrm{but}\mathrm{wolframalpha}\mathrm{can}$$$$\left(\mathrm{C}\mathrm{is}\mathrm{the}\mathrm{Catalan}\mathrm{constant}\right)$$$$\mathrm{1}.\:{I}_{\mathrm{1}} =\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{x}^{\mathrm{3}} }{…}{dx}=\frac{\pi\left(−\mathrm{1440C}+\mathrm{1144}+\mathrm{15}\pi\left(−\mathrm{41}+\mathrm{20}\pi+\mathrm{24ln}\:\mathrm{2}\right)\right)}{\mathrm{3780}}…

Question Number 196817 by universe last updated on 01/Sep/23 Answered by witcher3 last updated on 04/Sep/23 $$\mathrm{are}\mathrm{You}\mathrm{sur}\mathrm{of}\mathrm{the}\mathrm{expression}?$$ Commented by universe last updated on…

Question Number 196688 by cortano12 last updated on 29/Aug/23 Answered by Frix last updated on 29/Aug/23 $$\int \sqrt{x-\sqrt{x^2-4}}dx\stackrel{t=\sqrt{x-\sqrt{x^2-4}}}{=}$$$$=\int\frac{{t}^{\mathrm{4}} −\mathrm{4}}{{t}^{\mathrm{2}} }{dt}=\frac{{t}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{4}}{{t}}=…

Question Number 196496 by RoseAli last updated on 26/Aug/23 Answered by witcher3 last updated on 26/Aug/23 $$\int \frac{1}{{\mathrm{tg}}^2\left(\mathrm{x}\right).{\cos }^6\left(\mathrm{x}\right)}$$$$=\int\frac{\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{tg}^{\mathrm{2}} \left(\mathrm{x}\right)}.\left(\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \left(\mathrm{x}\right)\right)^{\mathrm{2}} \mathrm{dx}…

Question Number 196487 by RoseAli last updated on 25/Aug/23 $$\int (\sec {}^{4}x-\cot {}^{4}x)dx$$ Answered by MM42 last updated on 26/Aug/23 $${I}_{\mathrm{1}} =\int{sec}^{\mathrm{4}} {xdx}=\int\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)\left(\mathrm{1}+{tan}^{\mathrm{2}}…