Question Number 32716 by caravan msup abdo. last updated on 31/Mar/18
$${find}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{cos}^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{3}{sin}^{\mathrm{2}} {x}}{dx}\:. \\ $$
Answered by Joel578 last updated on 02/Apr/18
![I = ∫ (((1 + cos 2x)/2)/(1 + ((3(1 − cos 2x))/2))) dx = ∫ ((1 + cos 2x)/(5 − 3cos 2x)) dx = ∫ ((1 + ((1 − t^2 )/(1 + t^2 )))/(5 − ((3(1 − t^2 ))/(1 + t^2 )))) ((dt/(1 + t^2 ))) → t = tan x = ∫ (2/(2 + 8t^2 ))((dt/(1 + t^2 ))) = ∫ (dt/((1 + 4t^2 )(1 + t^2 ))) = ∫ (4/3)((1/(1 + 4t^2 ))) − (1/3)((1/(1 + t^2 )) )dt = (2/3)tan^(−1) (2t) − (1/3)tan^(−1) (t) = (2/3)tan^(−1) (2tan x) − (x/3) I = 4 [(2/3)tan^(−1) (2tan x) − (x/3)]_0 ^(π/2) = 4[((2/3)tan^(−1) (2tan (π/2)) − (π/6)) − 0] = 4((π/3) − (π/6)) = ((2π)/3)](https://www.tinkutara.com/question/Q32808.png)
$${I}\:=\:\int\:\frac{\frac{\mathrm{1}\:+\:\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}}}{\mathrm{1}\:+\:\frac{\mathrm{3}\left(\mathrm{1}\:−\:\mathrm{cos}\:\mathrm{2}{x}\right)}{\mathrm{2}}}\:{dx}\:=\:\int\:\frac{\mathrm{1}\:+\:\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{5}\:−\:\mathrm{3cos}\:\mathrm{2}{x}}\:{dx} \\ $$$$\:\:\:=\:\int\:\frac{\mathrm{1}\:+\:\frac{\mathrm{1}\:−\:{t}^{\mathrm{2}} }{\mathrm{1}\:+\:{t}^{\mathrm{2}} }}{\mathrm{5}\:−\:\frac{\mathrm{3}\left(\mathrm{1}\:−\:{t}^{\mathrm{2}} \right)}{\mathrm{1}\:+\:{t}^{\mathrm{2}} }}\:\left(\frac{{dt}}{\mathrm{1}\:+\:{t}^{\mathrm{2}} }\right)\:\:\:\:\:\:\rightarrow\:{t}\:=\:\mathrm{tan}\:{x} \\ $$$$\:\:\:=\:\int\:\frac{\mathrm{2}}{\mathrm{2}\:+\:\mathrm{8}{t}^{\mathrm{2}} }\left(\frac{{dt}}{\mathrm{1}\:+\:{t}^{\mathrm{2}} }\right)\:=\:\int\:\frac{{dt}}{\left(\mathrm{1}\:+\:\mathrm{4}{t}^{\mathrm{2}} \right)\left(\mathrm{1}\:+\:{t}^{\mathrm{2}} \right)}\: \\ $$$$\:\:\:=\:\int\:\frac{\mathrm{4}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{4}{t}^{\mathrm{2}} }\right)\:−\:\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{1}\:+\:{t}^{\mathrm{2}} }\:\right){dt} \\ $$$$\:\:\:=\:\frac{\mathrm{2}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{t}\right)\:−\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \left({t}\right) \\ $$$$\:\:\:=\:\frac{\mathrm{2}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2tan}\:{x}\right)\:−\:\frac{{x}}{\mathrm{3}} \\ $$$$ \\ $$$${I}\:=\:\mathrm{4}\:\left[\frac{\mathrm{2}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2tan}\:{x}\right)\:−\:\frac{{x}}{\mathrm{3}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\:\:\:\:=\:\mathrm{4}\left[\left(\frac{\mathrm{2}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2tan}\:\frac{\pi}{\mathrm{2}}\right)\:−\:\frac{\pi}{\mathrm{6}}\right)\:−\:\mathrm{0}\right] \\ $$$$\:\:\:\:=\:\mathrm{4}\left(\frac{\pi}{\mathrm{3}}\:−\:\frac{\pi}{\mathrm{6}}\right)\:=\:\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$
Commented by Joel578 last updated on 02/Apr/18
$$\mathrm{My}\:\mathrm{first}\:\mathrm{try}. \\ $$$$\mathrm{Please}\:\mathrm{recheck}\:\mathrm{my}\:\mathrm{solution} \\ $$