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Question Number 67525 by mathmax by abdo last updated on 28/Aug/19

$$leta>b>0calculate{\int }_0^{2\pi }\frac{dx}{{(a+bsinx)}^2}$$

Commented by ~ À ® @ 237 ~ last updated on 29/Aug/19

$$letconsiderf(a,b)={\int }_0^{2\pi }\frac{dx}{a+bsinx}=\frac{1}{b}{\int }_0^{2\pi }\frac{dx}{\frac{a}{b}+sinx}$$$${\int }_0^{2\pi }\frac{dx}{c+sinx}=2i\pi \sum _{\mid z_k\mid <1}Res(f\left(z\right),z_k)withf\left(z\right)=\frac{1}{iz}.\frac{1}{c+\frac{z-z^{-1}}{2i}}=\frac{2}{z^2+2icz-1}=\frac{2}{(z-z_0)(z-z_1)}with{z}_0=-i(c+\sqrt{c^2-1})and{z}_1=-i(c-\sqrt{c^2-1})$$$$wehave\mid z_0\mid >1and\mid z_1\mid <1$$$${\int }_0^{2\pi }\frac{dx}{c+sinx}=2i\pi \frac{2}{z_1-z_0}=\frac{2\pi }{\sqrt{c^2-1}}$$$$Nowf(a,b)=\frac{1}{b}.\frac{2\pi }{\sqrt{{\left(\frac{a}{b}\right)}^2-1}}=\frac{2\pi }{\sqrt{a^2-b^2}}$$$$\frac{\partial f(a,b)}{\partial a}={\int }_0^{2\pi }\frac{dx}{{(a+bsinx)}^2}=2\pi \frac{-\frac{2a}{2\sqrt{a^2-b^2}}}{(a^2-b^2)}=\frac{-4\pi a}{{(a^2-b^2)}^{\frac{3}{2}}}$$$$$$

Commented by mathmax by abdo last updated on 29/Aug/19

$$letf\left(a\right)={\int }_0^{2\pi }\frac{dx}{a+bsinx}wehave{f}^{{}^{\prime }}\left(a\right)=-{\int }_0^{2\pi }\frac{dx}{{(a+bsinx)}^2}\Rightarrow$$$${\int }_0^{2\pi }\frac{dx}{{(a+bsinx)}^2}=-f^{{}^{\prime }}\left(a\right)changement{e}^{ix}=zgive$$$$f\left(a\right)={\int }_{\mid z\mid =1}\frac{dz}{iz\{a+b\frac{z-z^{-1}}{2i}\}}={\int }_{\mid z\mid =1}\frac{2idz}{iz\{2ai+bz-{bz}^{-1}\}}$$$$={\int }_{\mid z\mid =1}\frac{2idz}{-2az+{biz}^2-bi}={\int }_{\mid z\mid =1}\frac{2idz}{2i^{2}az+{biz}^2-bi}$$$$={\int }_{\mid z\mid =1}\frac{2dz}{2iaz+{bz}^2-b}letW\left(z\right)=\frac{2}{{bz}^2+2iaz-b}polesofW?$$$${\mathrm{\Delta }}^{{}^{\prime }}=-a^2+b^2=-(a^2-b^2)={\left(i\sqrt{a^2-b^2}\right)}^2\Rightarrow$$$$z_1=\frac{-ia+i\sqrt{a^2-b^2}}{b}and{z}_2=\frac{-ia-i\sqrt{a^2-b^2}}{b}$$$$\mid z_1\mid -1=\frac{\mid -a+\sqrt{a^2-b^2}\mid }{b}-1=\frac{\sqrt{a^2-b^2}-a}{b}-1=\frac{\sqrt{a^2-b^2}-a-b}{b}<0$$$$\Rightarrow \mid z_1\mid <1wehave{z}_1.z_2=-1\Rightarrow \mid z_2\mid =\frac{1}{\mid z_1\mid }>1$$$${\int }_{\mid z\mid =1}W\left(z\right)dz=2i\pi Res(W,z_1)$$$$W\left(z\right)=\frac{2}{b(z-z_1)(z-z_2)}\Rightarrow Res(W,z_1)=\frac{2}{b(z_1-z_2)}=\frac{2}{b2i\frac{\sqrt{a^2-b^2}}{b}}$$$$=\frac{1}{i\sqrt{a^2-b^2}}\Rightarrow {\int }_{\mid z\mid =1}W\left(z\right)dz=2i\pi \times \frac{1}{i\sqrt{a^2-b^2}}=\frac{2\pi }{\sqrt{a^2-b^2}}=f\left(a\right)$$$$f\left(a\right)=2\pi {(a^2-b^2)}^{-\frac{1}{2}}\Rightarrow f^{{}^{\prime }}\left(a\right)=2\pi (-\frac{1}{2}){(a^2-b^2)}^{-\frac{3}{2}}$$$$=-\pi {(a^2-b^2)}^{-\frac{3}{2}}\Rightarrow {\int }_0^{2\pi }\frac{dx}{{(a+bsinx)}^2}=\frac{\pi }{{(a^2-b^2)}^{\frac{3}{2}}}$$$$$$

Commented by mathmax by abdo last updated on 29/Aug/19

$$erroratfinallinesf\left(a\right)=2\pi {(a^2-b^2)}^{-\frac{1}{2}}\Rightarrow$$$$f^{{}^{\prime }}\left(a\right)=2\pi (-\frac{1}{2})\left(2a\right){(a^2-b^2)}^{-\frac{3}{2}}=\frac{-2\pi a}{{(a^2-b^2)}^{\frac{3}{2}}}\Rightarrow$$$${\int }_0^{2\pi }\frac{dx}{{(a+bsinx)}^2}=\frac{2\pi a}{{(a^2-b^2)}^{\frac{3}{2}}}$$

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