Question Number 69167 by rajesh4661kumar@gmail.com last updated on 21/Sep/19 Answered by petrochengula last updated on 21/Sep/19 $${from}\:{tan}^{−\mathrm{1}} \left(\frac{{x}+{y}}{\mathrm{1}−{xy}}\right)={tan}^{−\mathrm{1}} {x}+{tan}^{−\mathrm{1}} {y} \\ $$$${tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{1}+{x}−{x}^{\mathrm{2}} }\right)={tan}^{−\mathrm{1}} \left(\frac{{x}+{x}−\mathrm{1}}{\mathrm{1}−{x}\left({x}−\mathrm{1}\right)}\right)={tan}^{−\mathrm{1}}…
Question Number 134693 by Engr_Jidda last updated on 06/Mar/21 $$\int{sec}^{\mathrm{4}} {xtanxdx}?? \\ $$ Answered by john_santu last updated on 06/Mar/21 $$\mathcal{X}=\int\left(\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{1}\right)\mathrm{tan}\:{x}\:\mathrm{sec}\:^{\mathrm{2}} {x}\:{dx} \\ $$$${let}\:\mathrm{tan}\:{x}\:=\:\rho\:…
Question Number 134672 by mnjuly1970 last updated on 06/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….{nice}\:\:\:\:{calculus}… \\ $$$$\:\:\:\:\:{prove}\:\:{that}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sqrt{\mathrm{5}\pi}\leqslant\:\:\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{3}}} \frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{2}\pi}\sqrt{\mathrm{8}{sin}\left({x}\right)−{sin}\left(\mathrm{2}{x}\right)}\:{dx}\leqslant\sqrt{\mathrm{6}\pi} \\ $$$$\:\:\:\:\:\:\:\:…{m}.{n}…. \\ $$ Terms of Service Privacy Policy…
Question Number 69133 by Henri Boucatchou last updated on 20/Sep/19 $$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int\left(\boldsymbol{\mathrm{x}}\:−\:\boldsymbol{\mathrm{tanx}}\right)^{\mathrm{2}} \:\boldsymbol{\mathrm{dx}}\:\:=\:? \\ $$$$ \\ $$ Commented by mathmax by abdo last updated…
Question Number 134660 by benjo_mathlover last updated on 06/Mar/21 $$\mathscr{M}\:=\:\int\:\frac{{dx}}{\mathrm{2cos}\:{x}+\mathrm{3sin}\:{x}}\: \\ $$ Answered by EDWIN88 last updated on 06/Mar/21 $$\mathrm{Let}\:\begin{cases}{\mathrm{2}\:=\:\mathrm{r}\:\mathrm{sin}\:\alpha}\\{\mathrm{3}\:=\:\mathrm{r}\:\mathrm{cos}\:\alpha}\end{cases}\:\Rightarrow\:\mathrm{r}^{\mathrm{2}} =\:\mathrm{13};\:\mathrm{r}\:=\sqrt{\mathrm{13}} \\ $$$$\:\mathrm{and}\:\mathrm{tan}\:\alpha\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:,\therefore\:\alpha\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right) \\…
Question Number 134667 by benjo_mathlover last updated on 06/Mar/21 $$\mathcal{G}\:=\:\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \:\mathrm{arccos}\:\left(\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{1}+\mathrm{2cos}\:\mathrm{x}}\right)\:\mathrm{dx}\: \\ $$ Commented by LUFFY last updated on 06/Mar/21 $${search}\:\mathrm{C}{oxeter}'{s}\:{integral} \\ $$ Terms…
Question Number 69110 by Fawole last updated on 20/Sep/19 $${Evaluate}\:\int_{−\infty} ^{\infty} \frac{{cos}\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$ Commented by mathmax by abdo last updated on 20/Sep/19 $$\:\:{let}\:{I}\:=\int_{−\infty}…
Question Number 134641 by benjo_mathlover last updated on 06/Mar/21 $$\mathscr{B}\:=\:\int\:\frac{\mathrm{1}−\mathrm{sin}\:\mathrm{6}{x}}{\mathrm{1}+\mathrm{sin}\:\mathrm{6}{x}}\:{dx}\: \\ $$ Answered by Dwaipayan Shikari last updated on 06/Mar/21 $$\int\frac{\mathrm{2}}{\mathrm{1}+{sin}\mathrm{6}{x}}−\mathrm{1}{dx}\:\:\:\:\:\:\:\mathrm{6}{x}={u} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{1}}{\mathrm{1}+{sinu}}{du}−{x}\:\: \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\int\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}}…
Question Number 134636 by bobhans last updated on 06/Mar/21 $$\mathcal{INTEGRAL} \\ $$$$\left(\mathrm{1}\right)\int_{\mathrm{0}} ^{\:\mathrm{ln}\:\mathrm{2}} \:{x}^{−\mathrm{2}} .{e}^{−\frac{\mathrm{1}}{{x}}} \:{dx}\:=? \\ $$$$\left(\mathrm{2}\right)\:\int\:\frac{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}}\:\mathrm{dx}\:=? \\ $$ Answered by benjo_mathlover…
Question Number 134619 by payamnazerian last updated on 05/Mar/21 $$\:\:\:\: \\ $$$$\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}−\mathrm{1}} \boldsymbol{{H}}_{{n}} }{\boldsymbol{{n}}^{\mathrm{2}} }=??? \\ $$ Terms of Service Privacy Policy Contact:…