Question Number 137930 by mnjuly1970 last updated on 08/Apr/21
![.....nice ... ... calculus..... calculation of::: 𝛗=∫_2 ^( 6) ((1+((((x−2)(x−4)(x−6)))^(1/3) )cos^(2021) (πx))/(x^2 −8x+20))dx=? solution:: x−4=t ⇒{_( x=6 ⇒ t=2) ^(x=2 ⇒t=−2) 𝛗=∫_(−2) ^( 2) ((1+(((t+2)(t)(t−2)))^(1/3) cos^(2021) (πt))/(t^2 +4))dt =∫_(−2) ^( 2) (1/(t^2 +2^2 ))dt +∫_(−2) ^( 2) {((((t^3 −4t))^(1/3) )/(t^2 +4))cos^(2021) (πt)=odd function}dt =2[(1/2) Arctan((t/2))]_0 ^2 + 0=(π/4) ....... 𝛗=(π/4) ...... adaapted based on .. Q: 137891 ...mr bobhans.... solved by ...mr ..bemath..](https://www.tinkutara.com/question/Q137930.png)
$$\:\:\:\:\:\:\:\:\:\:\:\:\:…..{nice}\:\:…\:…\:{calculus}….. \\ $$$$\:\:\:\:\:{calculation}\:{of}::: \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{2}} ^{\:\mathrm{6}} \:\frac{\mathrm{1}+\left(\sqrt[{\mathrm{3}}]{\left({x}−\mathrm{2}\right)\left({x}−\mathrm{4}\right)\left({x}−\mathrm{6}\right)}\:\right){cos}^{\mathrm{2021}} \left(\pi{x}\right)}{{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{20}}{dx}=? \\ $$$$\:\:\:\:\:{solution}:: \\ $$$$\:\:\:\:\:{x}−\mathrm{4}={t}\:\Rightarrow\left\{_{\:{x}=\mathrm{6}\:\Rightarrow\:{t}=\mathrm{2}} ^{{x}=\mathrm{2}\:\Rightarrow{t}=−\mathrm{2}} \right. \\ $$$$\:\:\:\:\boldsymbol{\phi}=\int_{−\mathrm{2}} ^{\:\mathrm{2}} \frac{\mathrm{1}+\sqrt[{\mathrm{3}}]{\left({t}+\mathrm{2}\right)\left({t}\right)\left({t}−\mathrm{2}\right)}\:{cos}^{\mathrm{2021}} \left(\pi{t}\right)}{{t}^{\mathrm{2}} +\mathrm{4}}{dt} \\ $$$$\:\:\:\:\:\:\:\:=\int_{−\mathrm{2}} ^{\:\mathrm{2}} \frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }{dt}\:+\int_{−\mathrm{2}} ^{\:\mathrm{2}} \left\{\frac{\sqrt[{\mathrm{3}}]{{t}^{\mathrm{3}} −\mathrm{4}{t}}\:\:\:}{{t}^{\mathrm{2}} +\mathrm{4}}{cos}^{\mathrm{2021}} \left(\pi{t}\right)={odd}\:{function}\right\}{dt} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{2}\left[\frac{\mathrm{1}}{\mathrm{2}}\:{Arctan}\left(\frac{{t}}{\mathrm{2}}\right)\right]_{\mathrm{0}} ^{\mathrm{2}} +\:\mathrm{0}=\frac{\pi}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…….\:\:\boldsymbol{\phi}=\frac{\pi}{\mathrm{4}}\:…… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{adaapted}\:{based}\:{on}\:.. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Q}:\:\mathrm{137891}\:…{mr}\:\:{bobhans}…. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{solved}\:{by}\:…{mr}\:..{bemath}.. \\ $$
Commented by benjo_mathlover last updated on 08/Apr/21
$${nice}… \\ $$