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Question Number 40140 by maxmathsup by imad last updated on 16/Jul/18

$$calculate{\int }_0^{\frac{\pi }{2}}\frac{sin\left(x\right)dx}{{cos}^{2}x+a^2{sin}^{2}x}dx$$

Commented by math khazana by abdo last updated on 20/Jul/18

$$letI={\int }_0^{\frac{\pi }{2}}\frac{sinx}{{cos}^{2}x+a^2{sin}^{2}x}dxchangement$$$$cosx=tgive$$$$I=-{\int }_0^1\frac{-dt}{t^2+a^2(1-t^2)}$$$$={\int }_0^1\frac{dt}{a^2+(1-a^2)t^2}=\frac{1}{a^2}{\int }_0^1\frac{dt}{1+\frac{1-a^2}{a^2}{t}^2}$$$$(a\ne o)$$$$case11-a^2>0\iff \mid a\mid <1wedothechangement$$$$\sqrt{\frac{1-a^2}{a^2}}t=u\Rightarrow$$$$I=\frac{1}{a^2}{\int }_0^{\sqrt{\frac{1-a^2}{a^2}}}\frac{1}{1+u^2}\sqrt{\frac{a^2}{1-a^2}}du$$$$=\frac{\mid a\mid }{a^2\sqrt{1-a^2}}{\int }_0^{\frac{\sqrt{1-a^2}}{\mid a\mid }}\frac{du}{1+u^2}$$$$=\frac{\xi \left(a\right)}{a\sqrt{1-a^2}}arctan\left(\frac{\sqrt{1-a^2}}{\mid a\mid }\right)$$$$\xi \left(a\right)=1ifa>0and\xi \left(a\right)=-1ifa<0$$$$case21-a^2<0\Rightarrow \mid a\mid >1\Rightarrow$$$$I=\frac{1}{a^2}{\int }_0^1\frac{dt}{1-\frac{a^2-1}{a^2}{t}^2}wedothechangement$$$$\sqrt{\frac{a^2-1}{a^2}}t=u\Rightarrow$$$$I=\frac{1}{a^2}{\int }_0^{\frac{\sqrt{a^2-1}}{\mid a\mid }}\frac{1}{1-u^2}\frac{\mid a\mid }{\sqrt{a^2-1}}du$$$$=\frac{\xi \left(a\right)}{\sqrt{a^2-1}}\frac{1}{2}{\int }_0^{\frac{\sqrt{a^2-1}}{\mid a\mid }}(\frac{1}{1+u}+\frac{1}{1-u})du$$$$=\frac{\xi \left(a\right)}{2\sqrt{a^2-1}}{[ln\mid \frac{1+u}{1-u}\mid ]}_0^{\frac{\sqrt{a^2-1}}{\mid a\mid }}$$$$=\frac{\xi \left(a\right)}{2\sqrt{a^2-1}}ln\left(\frac{1+\frac{\sqrt{a^2-1}}{\mid a\mid }}{1-\frac{\sqrt{a^2-1}}{\mid a\mid }}\right)$$$$=\frac{\xi \left(a\right)}{2\sqrt{a^2-1}}ln\left(\frac{\mid a\mid +\sqrt{a^2-1}}{\mid a\mid -\sqrt{a^2-1}}\right).$$

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jul/18

$$t=cosxdt=-sinxdx$$$${\int }_1^0\frac{-dt}{t^2+a^2(1-t^2)}$$$${\int }_0^1\frac{dt}{a^2+t^2(1-a^2)}$$$$=\frac{1}{1-a^2}{\int }_0^1\frac{dt}{\frac{a^2}{1-a^2}+t^2}$$$$=\frac{1}{1-a^2}\times \frac{\sqrt{1-a^2}}{a}\mid {tan}^{-1}\left(\frac{t}{\sqrt{\frac{a^2}{1-a^2}}}\right){\mid }_0^1$$$$=\frac{\sqrt{1-a^2}}{(a-a^3)}\mid {tan}^{-1}\left(\frac{t\sqrt{1-a^2}}{a}\right){\mid }_0^1$$$$=\frac{\sqrt{1-a^2}}{a-a^3}{tan}^{-1}\left(\frac{\sqrt{1-a^2}}{a}\right)$$$$$$

Commented by math khazana by abdo last updated on 20/Jul/18

$$sirTanmayyoumuststudythecases\mid a\mid >1$$$$and\mid a\mid <1....$$

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