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Category: Integration

prove-that-0-ln-1-x-x-4-17x-2-16-dx-pi-60-ln-2-

Question Number 162535 by mnjuly1970 last updated on 30/Dec/21 $$ \\ $$$$\:\:\:\:\:\mathrm{prove}\:\mathrm{that} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:\:\mathrm{ln}\:\left(\frac{\mathrm{1}}{{x}}\:\right)}{\:{x}^{\:\mathrm{4}} \:+\:\mathrm{17}{x}^{\:\mathrm{2}} \:+\:\mathrm{16}}\:{dx}\overset{?} {=}\:\frac{\pi}{\mathrm{60}}\:\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$ Commented by…

let-give-the-function-f-x-0-pi-ln-1-xcos-d-with-x-lt-1-1-find-a-simple-form-of-f-x-2-calculate-0-pi-ln-1-cos-d-3-calculate-0-pi-ln-1-cos-d-

Question Number 31463 by abdo imad last updated on 08/Mar/18 $${let}\:{give}\:{the}\:{function}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{1}+{xcos}\theta\right){d}\theta\:{with}\:\mid{x}\mid<\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\pi} \:{ln}\left(\mathrm{1}−{cos}\theta\right){d}\theta \\ $$$$\left.\mathrm{3}\right){calculate}\:\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{1}+{cos}\theta\right){d}\theta. \\ $$…

find-in-terms-of-n-the-value-of-A-n-0-1-k-1-n-1-x-2-2xcos-kpi-n-1-dx-with-n-from-N-

Question Number 31460 by abdo imad last updated on 08/Mar/18 $${find}\:{in}\:{terms}\:{of}\:\:{n}\:{the}\:{value}\:{of} \\ $$$${A}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left({x}^{\mathrm{2}} \:−\mathrm{2}{xcos}\left(\frac{{k}\pi}{{n}}\right)\:+\mathrm{1}\right){dx}\:\:\:{with}\:{n}\:{from}\:{N}^{\bigstar} . \\ $$ Commented by abdo…

solve-cosec-2-x-2-dx-

Question Number 162512 by vishal1234 last updated on 30/Dec/21 $${solve}\:\int\sqrt{{cosec}^{\mathrm{2}} {x}−\mathrm{2}}\:{dx} \\ $$ Commented by MJS_new last updated on 30/Dec/21 $$\mathrm{2}\:\mathrm{steps}\:\mathrm{are}\:\mathrm{necessary} \\ $$$$\left(\mathrm{1}\right)\:{t}=\mathrm{cos}\:{x} \\ $$$$\left(\mathrm{2}\right)\:{u}=\sqrt{\mathrm{2}}{t}+\sqrt{\mathrm{2}{t}^{\mathrm{2}}…