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Question Number 27184 by abdo imad last updated on 02/Jan/18

$$calculateintermsofxf\left(x\right)={\int }_0^{\frac{\pi }{2}}\frac{dt}{1+xsint}.$$

Commented by abdo imad last updated on 05/Jan/18

$$letdothechangementtan\left(\frac{t}{2}\right)=\alpha \iff t=2arctan\alpha$$$$f\left(x\right)={\int }_0^1\frac{\frac{2d\alpha }{1+{\alpha }^2}}{1+\frac{2x\alpha }{1+{\alpha }^2}}={\int }_0^1\frac{2d\alpha }{1+{\alpha }^2+2x\alpha }={\int }_0^1\frac{2d\alpha }{{\alpha }^2+2x\alpha +1}$$$$={\int }_0^1\frac{2d\alpha }{{(\alpha +x)}^2+1-x^2}$$$$case1/x/<1wedothech.\alpha +x=\sqrt{1-x^2}t$$$$f\left(x\right)={\int }_{\frac{x}{\sqrt{1-x^2}}}^{\frac{x+1}{\sqrt{1-x^2}}}\frac{2\sqrt{1-x^2}dt}{(1-x^2)(1+t^2)}$$$$f\left(x\right)=\frac{2}{\sqrt{1-x^2}}{\left[arctan\left(t\right)\right]}_{\frac{x}{\sqrt{1-x^2}}}^{\frac{\sqrt{1+x}}{\sqrt{1-x}}}$$$$f\left(x\right)=\frac{2}{\sqrt{1-x^2}}(arctan\left(\frac{\sqrt{1+x}}{\sqrt{1-x}}\right)-arctan\left(\frac{x}{\sqrt{1-x^2}}\right))$$$$case2if/x/>1$$$$f\left(x\right)=\frac{1}{x}{\int }_0^{\frac{\pi }{2}}\frac{dt}{x^{-1}+sinx}andwecanusethesamemthode$$$$dueto/x^{-1}/<1.....$$

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