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Category: Integration

0-pi-4-cos-x-sin-x-16sin-2x-9-dx-

Question Number 122922 by bemath last updated on 21/Nov/20 $$\:\:\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\:\frac{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}{\mathrm{16sin}\:\mathrm{2}{x}+\mathrm{9}}\:{dx}\: \\ $$ Answered by liberty last updated on 21/Nov/20 $$\varsigma\:=\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\frac{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}{\mathrm{16sin}\:\mathrm{2}{x}+\mathrm{9}}\:{dx}\: \\…

Question-122919

Question Number 122919 by bemath last updated on 20/Nov/20 Answered by Dwaipayan Shikari last updated on 20/Nov/20 $$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{xe}^{\mathrm{2}{x}} }{\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}}…

nice-calculus-prove-that-0-1-x-1-ln-x-2-dx-5-ln-m-n-

Question Number 122882 by mnjuly1970 last updated on 20/Nov/20 $$\:\:\:\:\:\:\:\:\:…\:\:{nice}\:\:{calculus}… \\ $$$$\:\:\:\:{prove}\:{that}\:::: \\ $$$$\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{{x}^{\varphi} −\mathrm{1}}{{ln}\left({x}\right)}\right)^{\mathrm{2}} {dx}=\sqrt{\mathrm{5}}\:{ln}\left(\varphi\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.{m}.{n}. \\ $$ Answered by TANMAY…

x-2-1-dx-x-4-x-2-1-

Question Number 122875 by pipin last updated on 20/Nov/20 $$\:\int\frac{\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{1}\right)\boldsymbol{\mathrm{dx}}}{\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{1}}\:=\:… \\ $$$$\: \\ $$ Answered by som(math1967) last updated on 20/Nov/20 $$\int\frac{\frac{\mathrm{x}^{\mathrm{2}}…

sin-1-x-2-dx-

Question Number 122877 by bemath last updated on 20/Nov/20 $$\:\:\int\:\left(\mathrm{sin}^{−\mathrm{1}} \left({x}\right)\right)^{\mathrm{2}} \:{dx}\:? \\ $$ Commented by liberty last updated on 20/Nov/20 $$\:{let}\:{u}\:=\:\mathrm{sin}^{−\mathrm{1}} \left({x}\right)\:\Rightarrow{x}\:=\:\mathrm{sin}\:{u}\: \\ $$$$\Rightarrow\:{dx}\:=\:\mathrm{cos}\:{u}\:{du}\:…

Question-122867

Question Number 122867 by bemath last updated on 20/Nov/20 Answered by som(math1967) last updated on 20/Nov/20 $$\mathrm{I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\sqrt{\mathrm{cosx}}\mathrm{dx}}{\:\sqrt{\mathrm{cosx}}+\sqrt{\mathrm{sinx}}} \\ $$$$\mathrm{again}\:\mathrm{I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\sqrt{\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)}\mathrm{dx}}{\:\sqrt{\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)}+\sqrt{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)}} \\ $$$$=\underset{\mathrm{0}}…