Question

integration of sin^2 x / (1+sin x cos x) with upper limit as pi/2 and lower limit as 0

Answer 1

$$\begin{gathered} {\int }_0^{\mathrm{\pi }/2}\frac{{\sin }^{2}x}{1+\sin x\cos x}dx \\ LetI={\int }_0^{\mathrm{\pi }/2}\frac{{\sin }^{2}x}{1+\sin x\cos x}dx................\left(1\right) \\ thenbyformula \\ {\int }_0^{\mathrm{\pi }/2}\frac{{\sin }^{2}x}{1+\sin x\cos x}dx \\ ={\int }_0^{\mathrm{\pi }/2}\frac{{\sin }^2(\mathrm{\pi }/2-x)}{1+\sin (\mathrm{\pi }/2-x)\cos (\mathrm{\pi }/2-x)}dx \\ I={\int }_0^{\mathrm{\pi }/2}\frac{{\cos }^{2}x}{1+\sin x\cos x}dx...................\left(2\right) \\ Adding\left(1\right)and\left(2\right)weget \\ 2I={\int }_0^{\mathrm{\pi }/2}\frac{{\cos }^{2}x}{1+\sin x\cos x}dx+{\int }_0^{\mathrm{\pi }/2}\frac{{\sin }^{2}x}{1+\sin x\cos x}dx \\ 2I={\int }_0^{\mathrm{\pi }/2}\frac{{\sin }^{2}x+{\cos }^{2}x}{1+\sin x\cos x}dx \\ 2I={\int }_0^{\mathrm{\pi }/2}\frac{1}{1+\sin x\cos x}dx \\ 2I={\int }_0^{\mathrm{\pi }/2}\frac{1}{1+{\displaystyle \frac{1}{2}}2\sin x\cos x}dx \\ 2I={\int }_0^{\mathrm{\pi }/2}\frac{1}{1+{\displaystyle \frac{1}{2}}\sin 2x}dx \\ 2I={\int }_0^{\mathrm{\pi }/2}\frac{2}{2+\sin 2x}dx \\ I={\int }_0^{\mathrm{\pi }/2}\frac{1}{2+\sin 2x}dx \\ Put\sin 2x=\frac{2\tan x}{1+{\tan }^{2}x} \\ I={\int }_0^{\mathrm{\pi }/2}\frac{1}{2+\frac{2\tan x}{1+{\tan }^{2}x}}dx \\ I={\int }_0^{\mathrm{\pi }/2}\frac{1+{\tan }^{2}x}{2(1+{\tan }^{2}x)+2\tan x}dx \\ I=\frac{1}{2}{\int }_0^{\mathrm{\pi }/2}\frac{se{c}^{2}x}{1+{\tan }^{2}x+\tan x}dx \\ Let\tan x=tthenlimitbecomest=0tot=\infty \\ I=\frac{1}{2}{\int }_0^{\infty }\frac{1}{1+t^2+t}dx \\ I=\frac{1}{2}{\int }_0^{\infty }\frac{1}{1+t^2+t+{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{4}}}dx \\ I=\frac{1}{2}{\int }_0^{\infty }\frac{1}{(t+{\displaystyle \frac{1}{2}}{)}^2+{\left[{\displaystyle \frac{\sqrt{3}}{2}}\right]}^2}dx \\ I=\frac{1}{2}\left[\frac{2}{\sqrt{3}}{\tan }^{-1}\frac{t+{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{3}}{2}}}\right]t=0to\infty \\ I=\frac{1}{\sqrt{3}}[{\tan }^{-1}\infty -{\tan }^{-1}0] \\ I=\frac{1}{\sqrt{3}}\frac{\mathrm{\pi }}{2} \\ I=\frac{\mathrm{\pi }}{2\sqrt{3}} \end{gathered}$$