Question

${\int }_0^{\pi /2}\sin \varphi \cos \varphi \sqrt{(a^2{\sin }^2\varphi +b^2{\cos }^2\varphi )}d\varphi a\ne b(a>0,b>0)$

Answer 1

Given : ${\int }_0^{\frac{\pi }{2}}\sin \varphi \cos \varphi \sqrt{a^2{\sin }^2\varphi +b^2{\cos }^2\varphi }d\varphi$

$I={\int }_0^{\frac{\pi }{2}}\sin \varphi \cos \varphi \sqrt{a^2{\sin }^2\varphi +b^2{\cos }^2\varphi }d\varphi$

Let : ${\sin }^2\varphi =t\varphi \to 0t\to 0$

$2\sin \varphi \cos \varphi =dt\varphi \to \frac{\pi }{2}t\to 1I=\frac{1}{2}{\int }_0^1\sqrt{t^2(a^2-b^2)+b^2}dt=\frac{\sqrt{a^2-b^2}}{2a}{\int }_0^1\sqrt{t^2+\left(\frac{b^2}{a^2-b^2}\right)}I=\frac{1}{2}\sqrt{a^2-b^2}\left[\frac{1}{2}[t\sqrt{t^2+\left(\frac{b^2}{a^2-b^2}\right)}+\frac{b^2}{a^2-b^2}\log {|t^2+\left(\frac{b^2}{a^2-b^2}\right)|}_0^1]\right]I=\frac{1}{4}\sqrt{a^2-b^2}[\frac{a}{\sqrt{a^2-b^2}}+\frac{b^2}{a^2-b^2}\log |1+\frac{a}{\sqrt{a^2-b^2}}|-\frac{b^2}{a^2-b^2}\log \frac{b}{\sqrt{a^2-b^2}}]I=\frac{1}{4}\sqrt{a^2-b^2}[\frac{a}{\sqrt{a^2-b^2}}+\frac{b^2}{a^2-b^2}\log \left|\frac{\sqrt{a^2-b^2}+a}{b}\right|]$

Hence the correct answer is $\frac{1}{4}\sqrt{a^2-b^2}[\frac{a}{\sqrt{a^2-b^2}}+\frac{b^2}{a^2-b^2}\log \left|\frac{\sqrt{a^2-b^2}+a}{b}\right|]$