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Question Number 87052 by lémùst last updated on 02/Apr/20

$$f\left(x\right)={\int }_0^{\pi /2}\frac{{sin}^2\left(t\right)}{1+{xsin}^2\left(t\right)}dt$$

Commented by mathmax by abdo last updated on 02/Apr/20

$$f\left(x\right)={\int }_0^{\frac{\pi }{2}}\frac{{sin}^{2}t}{1+x{sin}^{2}t}dt=\frac{1}{x}{\int }_0^{\frac{\pi }{2}}\frac{{xsin}^{2}t+1-1}{1+{xsin}^{2}t}dt$$$$=\frac{\pi }{2x}-\frac{1}{x}{\int }_0^{\frac{\pi }{2}}\frac{dt}{1+{xsin}^{2}t}[wehave$$$${\int }_0^{\frac{\pi }{2}}\frac{dt}{1+{xsin}^{2}t}={\int }_0^{\frac{\pi }{2}}\frac{dt}{1+x\times \frac{1-cos\left(2t\right)}{2}}$$$${=}_{2x=u}{\int }_0^{\pi }\frac{du}{2(1+x\times \frac{1-cos\left(2u\right)}{2})}={\int }_0^{\pi }\frac{du}{2+x-xcosu}$$$${=}_{tan\left(\frac{u}{2}\right)=z}{\int }_0^{\mathrm{\infty }}\frac{2dz}{(1+z^2)(2+x-x\frac{1-z^2}{1+z^2})}={\int }_0^{\mathrm{\infty }}\frac{2dz}{2+x+(2+x)z^2-x+{xz}^2}$$$$={\int }_0^{\mathrm{\infty }}\frac{2dz}{2+(2+2x)z^2}={\int }_0^{\mathrm{\infty }}\frac{dz}{1+(1+x)z^2}$$$$case11+x>0wedochangementu=\sqrt{1+x}z\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{dz}{1+(1+x)z^2}={\int }_0^{\mathrm{\infty }}\frac{du}{\sqrt{1+x}(1+u^2)}=\frac{1}{\sqrt{1+x}}\times \frac{\pi }{2}\Rightarrow$$$$f\left(x\right)=\frac{\pi }{2x}-\frac{\pi }{2x\sqrt{1+x}}(x>-1andx\ne 0)$$$$case21+x<0\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{dz}{1+(1+x)z^2}={\int }_0^{\mathrm{\infty }}\frac{dz}{1-(-(1+x))z^2}$$$$={\int }_0^{\mathrm{\infty }}\frac{dz}{(1-\sqrt{-1-x}z)(1+\sqrt{-1-x}z)}$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}(\frac{1}{1-\sqrt{-1-x}z}+\frac{1}{1+\sqrt{-1-x}z})$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}(\frac{1}{-\alpha z+1}+\frac{1}{\alpha z+1})dz(\alpha =\sqrt{-1-x})$$$$=\frac{1}{2}{[\frac{1}{\alpha }ln\mid \alpha z+1\mid -\frac{1}{\alpha }ln\mid \alpha z-1\mid ]}_0^{+\mathrm{\infty }}$$$$\frac{1}{2\alpha }{[ln\mid \frac{\alpha z+1}{\alpha z-1}\mid ]}_0^{+\mathrm{\infty }}=0so$$$$f\left(x\right)=\frac{\pi }{2x}-\frac{\pi }{2x\sqrt{1+x}}ifx>-1andx\ne -1$$$$f\left(x\right)=0ifx<-1$$

Commented by lémùst last updated on 02/Apr/20

$$mercibeaucoup$$$$maisjepensequef\left(x\right)=\frac{\pi }{2x}six<-1$$

Commented by mathmax by abdo last updated on 02/Apr/20

$$ouituaraisonjaioublie\frac{\pi }{2x}merci.$$

Commented by Ar Brandon last updated on 03/Apr/20

$$genial!!$$

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