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Category: Integration

sin-1-cos-x-dx-

Question Number 221957 by fantastic last updated on 13/Jun/25 $$\int\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{cos}\:{x}\right){dx} \\ $$ Answered by Ghisom last updated on 13/Jun/25 $$\int\mathrm{arcsin}\:\mathrm{cos}\:{x}\:{dx}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}:\:\begin{cases}{{u}'=\mathrm{1}\:\rightarrow\:{u}={x}}\\{{v}=\mathrm{arcsin}\:\mathrm{cos}\:{x}\:\rightarrow\:{v}'=−\sigma}\end{cases}\right] \\ $$$$\:\:\:\:\:\mathrm{where}\:\sigma=\mathrm{sign}\:\left(\mathrm{sin}\:{x}\right)…

Prove-0-pi-2-sin-x-K-2-sin-x-dx-pi-4-16-7-F-6-1-2-1-2-1-2-1-2-1-2-1-2-5-4-1-1-1-1-1-1-4-1-

Question Number 221882 by MrGaster last updated on 12/Jun/25 $$\mathrm{Prove}:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}\:{x}\:{K}^{\mathrm{2}} \:\mathrm{sin}\:{x}\:{dx}=\frac{\pi^{\mathrm{4}} }{\mathrm{16}}\:_{\mathrm{7}} {F}_{\mathrm{6}} \left(\frac{\mathrm{1}}{\:\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{5}}{\mathrm{4}};\mathrm{1},\mathrm{1},\mathrm{1},\mathrm{1},\mathrm{1},\frac{\mathrm{1}}{\mathrm{4}};\mathrm{1}\right) \\ $$ Answered by wewji12 last updated on 12/Jun/25…

Question-221902

Question Number 221902 by alvan545 last updated on 12/Jun/25 Answered by mahdipoor last updated on 12/Jun/25 $${x}−\mathrm{5}={u}\:\:\:\:\:\:\:{dx}={du} \\ $$$${x}=\mathrm{10}\equiv{u}=\mathrm{5}\:\:\:\:\:\:{x}=\mathrm{0}\equiv{u}−\mathrm{5} \\ $$$$\left.\int_{−\mathrm{5}} ^{\:\mathrm{5}} \left({u}+{u}^{\mathrm{2}} +{u}^{\mathrm{3}} \right){du}=\frac{{u}^{\mathrm{2}}…

let-a-0-1-find-all-continuous-function-f-0-1-0-such-that-0-1-f-x-dx-1-0-1-xf-x-dx-a-and-0-1-x-2-f-x-dx-a-2-how-many-such-function-are-there-

Question Number 221899 by universe last updated on 12/Jun/25 $$\:\:\:{let}\:{a}\in\left[\mathrm{0},\mathrm{1}\right]\:{find}\:{all}\:{continuous}\:{function} \\ $$$$\:\:\:\:\:{f}:\left[\mathrm{0},\mathrm{1}\right]\rightarrow\left[\mathrm{0},\infty\right)\:{such}\:{that}\:\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\:=\:\mathrm{1}\:\:, \\ $$$$\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {xf}\left({x}\right){dx}\:=\:{a}\:{and}\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} {f}\left({x}\right){dx}\:=\:{a}^{\mathrm{2}} \\ $$$$\:\:\:\:{how}\:{many}\:{such}\:{function}\:{are}\:{there}\:? \\ $$$$…

0-arctan-x-x-2-dx-

Question Number 221838 by Ghisom last updated on 11/Jun/25 $$\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\mathrm{arctan}\:{x}}{{x}}\right)^{\mathrm{2}} {dx}=? \\ $$ Answered by wewji12 last updated on 11/Jun/25 $$\mathrm{tan}^{−\mathrm{1}} \left({t}\right)=\rho\:\rightarrow\:\mathrm{d}{t}=\mathrm{sec}^{\mathrm{2}} \left(\rho\right)\mathrm{d}\rho…

f-x-lim-h-0-f-x-h-f-x-h-lim-h-0-c-c-h-lim-h-0-0-0-lim-h-0-x-h-2-x-2-h-lim-h-0-2xh-h-2-h-lim-h-0-2x-h-2x-lim-h-0-x-h-n-x-n-h-lim-h-0-k-0-n-n

Question Number 221832 by MrGaster last updated on 11/Jun/25 $${f}'\left({x}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{f}\left({x}+{h}\right)−{f}\left({x}\right)}{{h}} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{c}−{c}}{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}0}=\mathrm{0} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}+{h}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}{xh}+{h}^{\mathrm{2}} }{{h}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{2}{x}+{h}\right)=\mathrm{2}{x} \\ $$$$\underset{{h}\rightarrow\mathrm{0}}…

0-pi-a-a-cos-2n-x-dx-a-gt-1-0-pi-2-2-cos-4-x-dx-lim-m-0-pi-cos-2n-2mx-a-cos-2n-x-dx-a-gt-1-n-N-

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}^{+}…