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Category: Integration

Advanced-mathematics-i-prove-that-1-pi-0-1-x-2-x-1-2-x-dx-1-ii-evaluate-0-1-x-2-ln-x-ln-1-x-dx-

Question Number 114135 by mnjuly1970 last updated on 17/Sep/20 $$\:\:\:\:\:\:\:\:\:\:\:\:….\mathscr{A}{dvanced}\:\:{mathematics}\:… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{i}::\:{prove}\:\:{that}\::\:\:\:\:\Omega=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{{x}}}{dx}\:=\mathrm{1}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{ii}::{evaluate}\:::\:\:\Phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\mathrm{2}} \:{ln}\left({x}\right)\:{ln}\left(\mathrm{1}−{x}\right){dx}=??? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….{m}.{n}.{july}.\:\mathrm{1970}…. \\ $$$$\:…

Evaluate-6-arctan-8-w-dw-

Question Number 179655 by Acem last updated on 31/Oct/22 $${Evaluate}\:\int\mathrm{6}\:\mathrm{arctan}\:\frac{\mathrm{8}}{{w}}\:{dw} \\ $$ Answered by CElcedricjunior last updated on 05/Nov/22 $$\int\mathrm{6}\boldsymbol{{arctan}}\left(\frac{\mathrm{8}}{\boldsymbol{\omega}}\right)\boldsymbol{{d}\omega}=\boldsymbol{{k}} \\ $$$$\boldsymbol{{posons}}\:\begin{cases}{\boldsymbol{{u}}=\boldsymbol{{arctan}}\left(\frac{\mathrm{8}}{\boldsymbol{\omega}}\right)}\\{\boldsymbol{{v}}'=\mathrm{1}}\end{cases}=>\begin{cases}{\boldsymbol{{u}}'=−\frac{\mathrm{8}}{\boldsymbol{\omega}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{\mathrm{8}}{\boldsymbol{\omega}}\right)^{\mathrm{2}} }\:\right)}\\{\boldsymbol{{v}}=\boldsymbol{\omega}}\end{cases} \\…

dx-tan-x-sin-x-

Question Number 114102 by bemath last updated on 17/Sep/20 $$\int\:\frac{{dx}}{\mathrm{tan}\:{x}−\mathrm{sin}\:{x}} \\ $$ Answered by bobhans last updated on 17/Sep/20 $$\int\:\frac{{dx}}{\mathrm{tan}\:{x}−\mathrm{sin}\:{x}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cosec}\:^{\mathrm{2}} {x}\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cot}\:{x}\:\mathrm{cosec}\:{x}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{cot}\:{x}+\mathrm{cosec}\:{x}\mid\:+\:{c} \\ $$…

ln-x-sin-1-x-dx-

Question Number 114094 by Lordose last updated on 17/Sep/20 $$\int\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{sin}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{dx}} \\ $$ Answered by Olaf last updated on 17/Sep/20 $${u}'\:=\:\mathrm{ln}{x},\:{u}\:=\:{x}\mathrm{ln}{x}−{x} \\ $$$${v}\:=\:\mathrm{arcsin}{x},\:{v}'\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\…

1-sinx-cosx-dx-

Question Number 114072 by Lordose last updated on 17/Sep/20 $$\int\frac{\mathrm{1}}{\boldsymbol{\mathrm{sinx}}\:+\:\boldsymbol{\mathrm{cosx}}}\boldsymbol{\mathrm{dx}} \\ $$ Answered by Olaf last updated on 17/Sep/20 $${x}\:=\:\frac{\pi}{\mathrm{4}}−{u} \\ $$$$\mathrm{sin}{x}+\mathrm{cos}{x}\:=\: \\ $$$$\left(\mathrm{sin}\frac{\pi}{\mathrm{4}}\mathrm{cos}{u}−\mathrm{sin}{u}\mathrm{cos}\frac{\pi}{\mathrm{4}}\right)+\left(\mathrm{cos}\frac{\pi}{\mathrm{4}}\mathrm{cos}{u}+\mathrm{sin}\frac{\pi}{\mathrm{4}}\mathrm{sin}{u}\right) \\…

old-and-unanswered-Mr-Mathdave-x-2-ln-1-x-ln-1-x-dx-

Question Number 114044 by Her_Majesty last updated on 16/Sep/20 $${old}\:{and}\:{unanswered}…\:{Mr}\:{Mathdave}??? \\ $$$$\int{x}^{\mathrm{2}} {ln}\left(\mathrm{1}−{x}\right){ln}\left(\mathrm{1}+{x}\right){dx}=? \\ $$ Answered by mathdave last updated on 17/Sep/20 $${sokution} \\ $$$${put}\:{x}=\left(\mathrm{2}{y}−\mathrm{1}\right)\:\:\left({wat}\:{i}\:{did}\:{here}\:{is}\:{logical}\right)…

advanced-calculus-i-prove-that-0-1-ln-1-ln-1-x-ln-1-x-dx-n-1-n-1-n-2-ii-prove-that-0-1-ln-1-x-x-1

Question Number 114045 by mnjuly1970 last updated on 17/Sep/20 $$\:\:\:\:\:\:\:\:…\:\:{advanced}\:{calculus}… \\ $$$$ \\ $$$${i}\::\:\:{prove}\:\:{that}\::: \\ $$$$\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{ln}\left(\mathrm{1}−{x}\right)\right)}{{ln}\left(\mathrm{1}−{x}\right)}\:{dx}\:\overset{?} {=}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\Gamma\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$${ii}:\: \\…