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Question Number 93249 by 1549442205 last updated on 12/May/20

calculate∫_0 ^((Π/2) ) (dx/(2+cos2x))

$${calculate}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}\:} \frac{{dx}}{\mathrm{2}+{cos}\mathrm{2}{x}} \\ $$

Commented by 1549442205 last updated on 12/May/20

Calculate its value?

$${Calculate}\:{its}\:{value}? \\ $$

Commented by abdomathmax last updated on 12/May/20

I = ∫_0 ^(π/2)  (dx/(2+cos(2x))) ⇒I =_(2x=t)   (1/2)∫_0 ^π  (dt/(2+cost))  =_(tan((t/2))=u)    (1/2)∫_0 ^∞    ((2du)/((1+u^2 )(2+((1−u^2 )/(1+u^2 )))))  =∫_0 ^∞   (du/(2+2u^2  +1−u^2 )) =∫_0 ^∞  (du/(3+u^2 )) =_(u=(√3)z)   =∫_0 ^∞   (((√3)dz)/(3(1+z^2 ))) =(1/(√3))[arctanz]_0 ^(+∞)  =(1/(√3))×(π/2) ⇒  I =(π/(2(√3)))

$${I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dx}}{\mathrm{2}+{cos}\left(\mathrm{2}{x}\right)}\:\Rightarrow{I}\:=_{\mathrm{2}{x}={t}} \:\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \:\frac{{dt}}{\mathrm{2}+{cost}} \\ $$$$=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{2}+\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\right)} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\mathrm{2}+\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{1}−{u}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{du}}{\mathrm{3}+{u}^{\mathrm{2}} }\:=_{{u}=\sqrt{\mathrm{3}}{z}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\sqrt{\mathrm{3}}{dz}}{\mathrm{3}\left(\mathrm{1}+{z}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\left[{arctanz}\right]_{\mathrm{0}} ^{+\infty} \:=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}×\frac{\pi}{\mathrm{2}}\:\Rightarrow \\ $$$${I}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$ \\ $$

Commented by 1549442205 last updated on 12/May/20

Thank you very much!

$${Thank}\:{you}\:{very}\:{much}! \\ $$

Commented by mathmax by abdo last updated on 12/May/20

you are welcome.

$${you}\:{are}\:{welcome}. \\ $$

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