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Question Number 165741 by metamorfose last updated on 07/Feb/22

$${\int }_0^{\pi }\frac{dt}{1-sina.cost}=???,a\in ]0,\frac{\pi }{2}[$$

Answered by MJS_new last updated on 08/Feb/22

$$a\in ]0;\frac{\pi }{2}[\Rightarrow 0<\sin a<1\Rightarrow \mathrm{let}\sin a=A;0<A<1$$$$\int \frac{dt}{1-A\cos t}=$$$$[u=\tan \frac{t}{2}\Rightarrow dt=\frac{2du}{u^2+1}]$$$$=2\int \frac{du}{(1+A)u^2+(1-A)}=$$$$[\mathrm{let}B=1+A\wedge C=1-A\Rightarrow B,C>0]$$$$=2\int \frac{du}{{Bu}^2+C}=$$$$[u=\frac{\sqrt{C}}{\sqrt{B}}v\to du=\frac{\sqrt{C}}{\sqrt{B}}dv]$$$$=\frac{2}{\sqrt{BC}}\int \frac{dv}{v^2+1}=\frac{2}{\sqrt{BC}}\arctan v=$$$$=\frac{2}{\sqrt{1-A^2}}\arctan \frac{\sqrt{1+\mathrm{A}}u}{\sqrt{1-A}}=$$$$=\frac{2}{\cos a}\arctan \frac{\sqrt{1+\sin a}\tan \frac{t}{2}}{\sqrt{1-\sin a}}+C$$$$\Rightarrow$$$$\mathrm{answer}\mathrm{is}\underset{t\to {\pi }^{-}}{\lim }\left(\frac{2}{\cos a}\arctan \frac{\sqrt{1+\sin a}\tan \frac{t}{2}}{\sqrt{1-\sin a}}\right)=\frac{\pi }{\cos a}$$

Commented by metamorfose last updated on 09/Feb/22

$$thnxsir$$

Answered by Mathspace last updated on 08/Feb/22

$$letsina=\alpha \Rightarrow$$$$I={\int }_0^{\pi }\frac{dt}{1-\alpha cost}$$$${=}_{tan\left(\frac{t}{2}\right)=y}{\int }_0^{\mathrm{\infty }}\frac{2dy}{(1+y^2)(1-\alpha \frac{1-y^2}{1+y^2})}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{dy}{1+y^2-\alpha +\alpha y^2}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{dy}{1-\alpha +(1+\alpha )y^2}$$$$=\frac{2}{1-\alpha }{\int }_0^{\mathrm{\infty }}\frac{dy}{1+\frac{1+\alpha }{1-\alpha }{y}^2}$$$${=}_{z=\sqrt{\frac{1+\alpha }{1-\alpha }}y}\frac{2}{1-\alpha }{\int }_0^{\mathrm{\infty }}\frac{dz}{\sqrt{\frac{1+\alpha }{1-\alpha }}(1+z^2)}$$$$=\frac{2}{1-\alpha }\frac{\sqrt{1-\alpha }}{\sqrt{1+\alpha }}\times \frac{\pi }{2}$$$$=\frac{\pi }{\sqrt{1-{\alpha }^2}}=\frac{\pi }{\sqrt{1-{sin}^2\alpha }}=\frac{\pi }{\mid cos\alpha \mid }$$$$or0<\alpha <\frac{\pi }{2}\Rightarrow I=\frac{\pi }{cos\alpha }$$

Commented by metamorfose last updated on 09/Feb/22

$$thnxsir$$

Commented by Mathspace last updated on 08/Feb/22

$$I=\frac{\pi }{cosa}$$

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