Question Number 124004 by john_santu last updated on 30/Nov/20 $$\:\int\:\frac{{dx}}{\:\sqrt[{\mathrm{3}}]{\mathrm{tan}\:{x}}}\:?\: \\ $$ Answered by liberty last updated on 30/Nov/20 $${T}\:=\:\int\:\frac{{dx}}{\:\sqrt[{\mathrm{3}}]{\mathrm{tan}\:{x}}}\:;\:\left[\:{let}\:{u}^{\mathrm{3}} =\mathrm{tan}\:^{\mathrm{2}} {x}\:\wedge\:{dx}\:=\frac{\mathrm{3}{u}^{\mathrm{2}} }{\mathrm{2}{u}^{\mathrm{3}/\mathrm{2}} \:\left({u}^{\mathrm{3}} +\mathrm{1}\right)}{du}\:\right]…
Question Number 123998 by john_santu last updated on 30/Nov/20 $$\:\int_{\:\mathrm{0}} ^{\:\infty} \:\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx}\: \\ $$ Answered by liberty last updated on 30/Nov/20 $$\:{substituting}\:{x}\:=\:\mathrm{tan}\:{q}\:{with}\:{upper}\:{limit}\:\frac{\pi}{\mathrm{2}}…
Question Number 123977 by mnjuly1970 last updated on 29/Nov/20 Answered by mathmax by abdo last updated on 29/Nov/20 $$\mathrm{let}\:\mathrm{A}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{cosx}\right)\mathrm{dx}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{A}\:=−\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)\:\mathrm{also} \\ $$$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\frac{\mathrm{e}^{\mathrm{ix}}…
Question Number 123967 by Algoritm last updated on 29/Nov/20 Answered by TANMAY PANACEA last updated on 29/Nov/20 $${t}^{\mathrm{6}} ={x}+\mathrm{2} \\ $$$$\int\frac{{t}}{{t}^{\mathrm{2}} +{t}^{\mathrm{3}} }×\mathrm{6}{t}^{\mathrm{5}} {dt} \\…
Question Number 123964 by mindispower last updated on 29/Nov/20 $$\int\sqrt{{x}^{\mathrm{3}} +{ax}+{b}}{dx} \\ $$ Commented by MJS_new last updated on 29/Nov/20 $${x}_{\mathrm{1}} ={p} \\ $$$${x}_{\mathrm{2}} =−\frac{{p}}{\mathrm{2}}−\sqrt{{q}}…
Question Number 189496 by Tawa11 last updated on 17/Mar/23 $$\int\:\mathrm{xe}^{\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} \:\mathrm{dx} \\ $$ Answered by BaliramKumar last updated on 18/Mar/23 $${let}\:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:=\:{y} \\ $$$$\frac{\mathrm{2}{x}}{\mathrm{2}}\:=\:\frac{{dy}}{{dx}}\:\Rightarrow\:{xdx}\:=\:{dy}…
Question Number 189489 by mnjuly1970 last updated on 17/Mar/23 $$ \\ $$$$\:\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\infty} {e}^{\:−{x}} {cos}\left({x}\right){ln}\left({x}\right){dx}=? \\ $$$$\:\:\:\:\:−−− \\ $$$$\:\:\:\:\:\:{f}\:\left({a}\:\right)=\:\int_{\mathrm{0}} ^{\:\infty} {e}^{\:−{x}} {cos}\left({x}\right){x}^{\:{a}} \:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…
Question Number 123937 by mnjuly1970 last updated on 29/Nov/20 $$\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:\:\:\:\:\:{calculus}… \\ $$$$\:{prove}\:{that}:: \\ $$$$\:\int_{\frac{−\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \frac{\left(\pi−\mathrm{4}{x}\right){tan}\left({x}\right)}{\mathrm{1}−{tan}\left({x}\right)}{dx}\overset{???} {=}\pi{ln}\left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$$$ \\ $$ Answered by Dwaipayan…
Question Number 123921 by john_santu last updated on 29/Nov/20 $$\int\:\frac{\mathrm{1}}{\:\sqrt{{x}}\:\left({x}+\mathrm{1}\right)\left(\left(\mathrm{tan}^{−\mathrm{1}} \sqrt{{x}}\right)^{\mathrm{2}} +\mathrm{9}\right)}{dx} \\ $$ Answered by mindispower last updated on 29/Nov/20 $$=\int\frac{\mathrm{2}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({tan}^{−} \left({t}\right)^{\mathrm{2}} +\mathrm{9}\right)}\:…
Question Number 123920 by john_santu last updated on 29/Nov/20 $$\:\int\:\frac{\left({x}−\mathrm{1}\right)\sqrt{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)}\:{dx}\:? \\ $$ Answered by liberty last updated on 29/Nov/20 $$\left(\bullet\right)\:\frac{{x}−\mathrm{1}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)}\:=\:\frac{{x}^{\mathrm{2}}…