Question Number 153737 by ZiYangLee last updated on 09/Sep/21
$$\mathrm{Show}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\left(\mathrm{cos}\:\theta+\:\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }\:{d}\theta=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:} \\ $$
Answered by puissant last updated on 09/Sep/21
![=∫_0 ^(π/2) (1/([2((1/2)cosθ+((√3)/2)sinθ)]^2 ))dθ =(1/4)∫_0 ^(π/2) (1/(cos^2 ((π/3)−θ)))dθ u=(π/3)−θ → du=−dθ I=(1/4)∫_(π/3) ^(−(π/6)) (1/(cos^2 u))(−du) =(1/4)[tanu]_(−(π/6)) ^(π/3) = (1/4)(tan((π/3))−tan(−(π/6))) =(1/4)((√3)+(1/( (√3))))=(1/4)(((3+1)/( (√3))))=(1/4)((4/( (√3))))=(1/( (√3)..)) ⇒ ∫_0 ^(π/2) (1/((cosθ+(√3)sinθ)^2 ))dθ=(1/( (√3)))..](Q153739.png)
$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\left[\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}{cos}\theta+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sin}\theta\right)\right]^{\mathrm{2}} }{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{3}}−\theta\right)}{d}\theta \\ $$$${u}=\frac{\pi}{\mathrm{3}}−\theta\:\rightarrow\:{du}=−{d}\theta \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}\int_{\frac{\pi}{\mathrm{3}}} ^{−\frac{\pi}{\mathrm{6}}} \frac{\mathrm{1}}{{cos}^{\mathrm{2}} {u}}\left(−{du}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[{tanu}\right]_{−\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} =\:\frac{\mathrm{1}}{\mathrm{4}}\left({tan}\left(\frac{\pi}{\mathrm{3}}\right)−{tan}\left(−\frac{\pi}{\mathrm{6}}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{3}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}..} \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\left({cos}\theta+\sqrt{\mathrm{3}}{sin}\theta\right)^{\mathrm{2}} }{d}\theta=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}.. \\ $$
Answered by peter frank last updated on 09/Sep/21
