Question Number 207949 by mnjuly1970 last updated on 31/May/24
$$ \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:{Find}\:{the}\:{value}\:{of}\:: \\ $$$$ \\ $$$$\:\:\:\:\boldsymbol{\Omega}\:=\:\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\:\boldsymbol{{dx}}}{\boldsymbol{{sin}}^{\mathrm{6}} \boldsymbol{{x}}\:+\:\boldsymbol{{cos}}^{\mathrm{6}} \boldsymbol{{x}}}\:=\:?\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:βββββββββ \\ $$
Answered by Frix last updated on 31/May/24
![β«_0 ^(Ο/2) (dx/(cos^6 x +sin^6 x))=2β«_0 ^(Ο/4) (dx/(cos^6 x +sin^6 x))= =16β«_0 ^(Ο/4) (dx/(5+3cos 4x)) =^(t=tan 2x) 4β«_0 ^β (dt/(t^2 +4))= =2[tan^(β1) (t/2)]_0 ^β =Ο](https://www.tinkutara.com/question/Q207952.png)
$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{dx}}{\mathrm{cos}^{\mathrm{6}} \:{x}\:+\mathrm{sin}^{\mathrm{6}} \:{x}}=\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\frac{{dx}}{\mathrm{cos}^{\mathrm{6}} \:{x}\:+\mathrm{sin}^{\mathrm{6}} \:{x}}= \\ $$$$=\mathrm{16}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\frac{{dx}}{\mathrm{5}+\mathrm{3cos}\:\mathrm{4}{x}}\:\overset{{t}=\mathrm{tan}\:\mathrm{2}{x}} {=}\:\mathrm{4}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{4}}= \\ $$$$=\mathrm{2}\left[\mathrm{tan}^{β\mathrm{1}} \:\frac{{t}}{\mathrm{2}}\right]_{\mathrm{0}} ^{\infty} =\pi \\ $$
Commented by mnjuly1970 last updated on 01/Jun/24
$${thanks}\:{alot}\:{ali}… \\ $$
Answered by Berbere last updated on 01/Jun/24
$$\mathrm{1}=\left({sin}^{\mathrm{2}} \left({x}\right)+{cos}^{\mathrm{2}} \left({x}\right)\right)^{\mathrm{3}} ={sin}^{\mathrm{6}} \left({x}\right)+{cos}^{\mathrm{6}} \left({x}\right)+\mathrm{3}{sin}^{\mathrm{2}} \left({x}\right){cos}^{\mathrm{2}} \left({x}\right)\left({sin}^{\mathrm{2}} \left({x}\right)+{cos}^{\mathrm{2}} \left({x}\right)\right) \\ $$$$\Rightarrow{sin}^{\mathrm{6}} \left({x}\right)+{cos}^{\mathrm{6}} \left({x}\right)=\mathrm{1}β\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{2}{sin}\left({x}\right){cos}\left({x}\right)\right)^{\mathrm{2}} \\ $$$$=\mathrm{1}β\frac{\mathrm{3}{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}{\mathrm{4}}=\mathrm{1}β\frac{\mathrm{3}}{\mathrm{8}}\left(\mathrm{2}{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)=\mathrm{1}β\frac{\mathrm{3}}{\mathrm{8}}\left(\mathrm{1}β{cos}\left(\mathrm{4}{x}\right)\right)\right. \\ $$$${u}=\mathrm{4}{x} \\ $$$$\Omega=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{du}}{\mathrm{5}+\mathrm{3}{cos}\left({u}\right)} \\ $$$${cos}\left({u}\right)=\frac{{e}^{{iu}} +{e}^{β{iu}} }{\mathrm{2}},{z}={e}^{{iu}} \Rightarrow{dz}={izdu}\Leftrightarrow{du}=\frac{{dz}}{{iz}} \\ $$$$\Rightarrow\mathrm{2}\int_{{C}} \frac{{dz}}{{iz}\left(\mathrm{5}+\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{{z}^{\mathrm{2}} +\mathrm{1}}{{z}}\right)\right)}=\frac{\mathrm{4}}{{i}}\int_{{C}} \frac{{dz}}{\left(\mathrm{3}{z}^{\mathrm{2}} +\mathrm{10}{z}+\mathrm{3}\right)}=\Omega \\ $$$$\mathrm{3}{z}^{\mathrm{2}} +\mathrm{10}{z}+\mathrm{3}=\mathrm{0}\Rightarrow{z}\in\left\{β\mathrm{3},β\frac{\mathrm{1}}{\mathrm{3}}\right\} \\ $$$$\Omega=\mathrm{2}{i}\pi.\frac{\mathrm{4}}{{i}}{Res}\left(\frac{\mathrm{1}}{\left(\mathrm{3}{z}^{\mathrm{2}} +\mathrm{10}{z}+\mathrm{3}\right)},{z}=β\frac{\mathrm{1}}{\mathrm{3}}\right\}=\mathrm{8}\pi.\frac{\mathrm{1}}{\mathrm{6}{z}+\mathrm{10}}=\frac{\mathrm{8}\pi}{\mathrm{8}}=\pi \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 01/Jun/24
$${grateful}\:{sir}… \\ $$
Commented by Berbere last updated on 01/Jun/24
$${Withe}\:{pleasur} \\ $$