Menu Close

Category: Integration

calculus-evaluate-I-0-pi-2-tan-2x-sin-4-x-4cos-2-x-cos-4-x-4sin-2-x-dx-m-n-july-1970-

Question Number 114472 by mnjuly1970 last updated on 19/Sep/20 $$\:\:\:\:\:\:\:\:\:…\:\:{calculus}… \\ $$$${evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{tan}\left(\mathrm{2}{x}\right)}{\:\sqrt{{sin}^{\mathrm{4}} \left({x}\right)+\mathrm{4}{cos}^{\mathrm{2}} \left({x}\right)}−\sqrt{{cos}^{\mathrm{4}} \left({x}\right)+\mathrm{4}{sin}^{\mathrm{2}} \left({x}\right)\:}}\:{dx}=\:??? \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:…{m}.{n}.{july}.\mathrm{1970}…. \\…

Question-179991

Question Number 179991 by mnjuly1970 last updated on 05/Nov/22 Answered by mindispower last updated on 05/Nov/22 $$=\underset{{n}\geqslant\mathrm{2}} {\sum}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{m}!}{x}^{{m}} {e}^{−{nx}} {dx}=\frac{\mathrm{1}}{{m}!}\int_{\mathrm{0}} ^{\infty} \left(\frac{{y}}{{n}}\right)^{{m}} {e}^{−{y}}…

Question-179862

Question Number 179862 by cherokeesay last updated on 03/Nov/22 Commented by CElcedricjunior last updated on 03/Nov/22 $$\begin{cases}{\boldsymbol{{y}}=\boldsymbol{{lnx}}}\\{\boldsymbol{{y}}=\mathrm{2}}\end{cases}=>\boldsymbol{{lnx}}=\mathrm{2}=>\boldsymbol{{x}}=\boldsymbol{{e}}^{\mathrm{2}} \\ $$$$\left.=\left.>\boldsymbol{{x}}\in\right]\mathrm{0};\boldsymbol{{e}}^{\mathrm{2}} \right]\:\boldsymbol{{et}}\:\boldsymbol{{y}}\in\left[\boldsymbol{{lnx}};\mathrm{2}\right] \\ $$$$\Leftrightarrow\boldsymbol{{A}}=\int_{\boldsymbol{{lnx}}} ^{\mathrm{2}} \int_{\mathrm{0}} ^{\boldsymbol{{e}}^{\mathrm{2}}…

xdx-x-1-x-2-

Question Number 179853 by AKSHAYTHAKUR last updated on 03/Nov/22 $$\int\frac{\boldsymbol{{xdx}}}{\left(\boldsymbol{{x}}−\mathrm{1}\right)\left(\boldsymbol{{x}}−\mathrm{2}\right)} \\ $$ Commented by CElcedricjunior last updated on 04/Nov/22 $$\int\:\frac{\boldsymbol{{xdx}}}{\left(\boldsymbol{{x}}−\mathrm{1}\right)\left(\boldsymbol{{x}}−\mathrm{2}\right)}=\int\left[\frac{\mathrm{1}}{\boldsymbol{{x}}−\mathrm{2}}+\frac{\mathrm{1}}{\left(\boldsymbol{{x}}−\mathrm{1}\right)\left(\boldsymbol{{x}}−\mathrm{2}\right)}\right]\boldsymbol{{dx}} \\ $$$$={ln}\mid\boldsymbol{{x}}−\mathrm{2}\mid+\int\left[−\frac{\mathrm{1}}{\boldsymbol{{x}}−\mathrm{1}}+\frac{\mathrm{1}}{\boldsymbol{{x}}−\mathrm{2}}\right]\boldsymbol{{dx}} \\ $$$$=\boldsymbol{{ln}}\mid\boldsymbol{{x}}−\mathrm{2}\mid−\boldsymbol{{ln}}\mid\boldsymbol{{x}}−\mathrm{1}\mid+\boldsymbol{{ln}}\mid\boldsymbol{{x}}−\mathrm{1}\mid \\…