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Question Number 31096 by abdo imad last updated on 02/Mar/18

$$find{\int }_0^{\pi }\frac{dx}{{(a+bcosx)}^2}witha>b>0thengivethe$$ $$valueof{\int }_0^{\pi }\frac{dx}{{(2+cosx)}^2}$$

Commented byabdo imad last updated on 04/Mar/18

$$letintroducethefunctionf\left(a\right)={\int }_0^{\pi }\frac{dx}{a+bcosx}wehave$$ $$f^{{}^{\prime }}\left(a\right)=-{\int }_0^{\pi }\frac{dx}{{(a+bcosx)}^2}\Rightarrow {\int }_0^{\pi }\frac{dx}{{(a+bcosx)}^2}=-f^{{}^{\prime }}\left(a\right)let$$ $$calculatef\left(a\right)ch.tan\left(\frac{x}{2}\right)=tgive$$ $$f\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{1}{a+b\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}={\int }_0^{\mathrm{\infty }}\frac{2dt}{a(1+t^2)+b(1-t^2)}$$ $$={\int }_0^{\mathrm{\infty }}\frac{2dt}{a+b+(a-b)t^2}={\int }_0^{\mathrm{\infty }}\frac{2dt}{(a+b)(1+\frac{a-b}{a+b}{t}^2)}then$$ $$weusethech.\sqrt{\frac{a-b}{a+b}}t=u\Rightarrow$$ $$f\left(a\right)=\frac{1}{a+b}{\int }_0^{\mathrm{\infty }}\frac{2}{(1+u^2)}\sqrt{\frac{a+b}{a-b}}du$$ $$=\frac{2}{\sqrt{a^2-b^2}}{\int }_0^{\mathrm{\infty }}\frac{du}{1+u^2}=\frac{\pi }{2}\frac{2}{\sqrt{a^2-b^2}}=\frac{\pi }{\sqrt{a^2-b^2}}$$ $$f\left(a\right)=\pi {(a^2-b^2)}^{-\frac{1}{2}}\Rightarrow f^{{}^{\prime }}\left(a\right)=\frac{-\pi }{2}\left(2a\right){(a^2-b^2)}^{-\frac{3}{2}}$$ $$=\frac{-\pi a}{(a^2-b^2)\sqrt{a^2-b^2}}\Rightarrow {\int }_0^{\pi }\frac{dx}{{(a+bcosx)}^2}=\frac{\pi a}{(a^2-b^2)\sqrt{a^2-b^2}}$$ $$2)lettakea=2andb=1\Rightarrow$$ $${\int }_0^{\pi }\frac{dx}{{(2+cosx)}^2}=\frac{2\pi }{(2^2-1^2)\sqrt{2^2-1^2}}=\frac{2\pi }{3\sqrt{3}}.$$

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