Question Number 219597 by SdC355 last updated on 29/Apr/25
$$\mathrm{Evaluate}\:\mathrm{integral}\:\mathrm{by}\:\mathrm{Complex}\:\mathrm{integral}\:\mathrm{method} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \:\:\frac{\mathrm{1}}{{a}+{b}\centerdot\mathrm{cos}\left({n}\theta\right)}\:\mathrm{d}\theta \\ $$
Answered by Nicholas666 last updated on 29/Apr/25
$$\frac{\mathrm{2}\pi}{{n}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }} \\ $$
Commented by Nicholas666 last updated on 29/Apr/25
https://www.quora.com/profile/Bekicot-5/Solution-We-want-to-evalute-the-integral-math-I-int_0-2-pi-frac-1-a-b-cos-n-theta-d-theta-math-with?ch=10&oid=221082798&share=c38cf370&srid=5Xg5SU&target_type=post
Commented by SdC355 last updated on 29/Apr/25
$$\mathrm{Gooood}\:!!!! \\ $$
Commented by Frix last updated on 29/Apr/25
![Let a=5∧b=3∧n=4 ((2π)/(n(√(a^2 −b^2 ))))=(π/8) but ∫_0 ^(2π) (dx/(5+3cos 4x))=8∫_0 ^(π/4) (dx/(5+3cos 4x))= =[tan^(−1) ((tan 2x)/2)]_0 ^(π/4) =(π/2) ⇒ Your result is wrong](https://www.tinkutara.com/question/Q219607.png)
$$\mathrm{Let}\:{a}=\mathrm{5}\wedge{b}=\mathrm{3}\wedge{n}=\mathrm{4} \\ $$$$\frac{\mathrm{2}\pi}{{n}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}=\frac{\pi}{\mathrm{8}} \\ $$$$\mathrm{but} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\:\frac{{dx}}{\mathrm{5}+\mathrm{3cos}\:\mathrm{4}{x}}=\mathrm{8}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\frac{{dx}}{\mathrm{5}+\mathrm{3cos}\:\mathrm{4}{x}}= \\ $$$$=\left[\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{tan}\:\mathrm{2}{x}}{\mathrm{2}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} =\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{Your}\:\mathrm{result}\:\mathrm{is}\:\mathrm{wrong} \\ $$