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Question Number 27781 by abdo imad last updated on 14/Jan/18

$$findthevalueofF\left(x\right)={\int }_0^{\frac{\pi }{2}}\frac{ln(1+x{sin}^{2}t)}{{sin}^{2}t}dtknowingthat$$$$-1<x<1.$$

Commented by abdo imad last updated on 18/Jan/18

$$afterverifyingthatFisderivableon]-1,1[$$$$F^{{,}^{\prime }}\left(x\right)={\int }_0^{\frac{\pi }{2}}\frac{{sin}^{2}t}{{sin}^{2}t(1+x{sin}^{2}t)}dt$$$$={\int }_0^{\frac{\pi }{2}}\frac{dt}{1+x{sin}^{2}t}={\int }_0^{\frac{\pi }{2}}\frac{dt}{1+x\frac{1-cos\left(2t\right)}{2}}$$$$={\int }_0^{\frac{\pi }{2}}\frac{2dt}{2+x-xcos\left(2t\right)}letusethech.2t=\theta$$$$F^{}\left(x\right)={\int }_0^{\pi }\frac{d\theta }{2+x-xcos\theta }thenthech.tan\left(\frac{\theta }{2}\right)=tgive$$$$F^{}\left(x\right)={\int }_0^{+\propto }\frac{1}{2+x-x\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}$$$$={\int }_0^{+\propto }\frac{2dt}{(2+x)(1+t^2)-x+{xt}^2}$$$$={\int }_0^{+\propto }\frac{2dt}{2+x+(2+2x)t^2-x}={\int }_0^{+\propto }\frac{dt}{1+(1+x)t^2}ch.u=\sqrt{1+x}t$$$$={\int }_0^{+\propto }\frac{du}{\sqrt{1+x}(1+u^2)}=\frac{\pi }{2\sqrt{1+x}}$$$$\Rightarrow F\left(x\right)=\frac{\pi }{2}{\int }_0^x\frac{dt}{\sqrt{1+t}}+\lambda$$$$F\left(x\right)=\pi \sqrt{1+x}+\lambda and\lambda =F\left(0\right)-\pi =-\pi so$$$$F\left(x\right)=\pi (\sqrt{1+x}-1).$$

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