Question

Integration (0 to pi/2) sin 2 x/sinx + cosx

Answer 1

We use substitution for tan x/2,  then resolve the function into partial fractions and then integrate each part.. a bit long.  perhaps there is a shorter method.

f(x) = sin 2x / [ sin x + cos x]
Divide the numerator and denom by  cos x

f(x) = 2 sin x /[1 + tan x]
let  tan x/2  = t         =>   2 dt = (1+t^2) dx
when x =0  , t = 0  and  x = π/2,   t = 1

$f(x)dx =2 \frac{2t}{1+t^2}*\frac{1-t^2}{1-t^2+2t}*\frac{2}{1+t^2}\ dt\\\\ f(t)=\frac{A+Bt}{1+t^2}+\frac{C+Dt}{(1+t^2)^2}+\frac{E+Ft}{1-t^2+2t}=\frac{8t-8t^3}{(1+t^2)^2\ (1-t^2+2t)}\\\\A+C+E=0,2A+B+2C+D+F=8, \\2B+2D-C+2E=0,2A-D+2F=-8,\\ 2B-A+E=0, F-B=0\\\\A=-2 ,B=F=0 ,C = D =4, E =-2\\\\f(t)=\frac{-2}{1+t^2}+\frac{4t}{(1+t^2)^2}+\frac{4}{(1+t^2)^2}-\frac{2}{1-t^2+2t}$

$For\ \frac{4}{(1+t^2)^2}dt:\ \ \ t=tanz,\ dt=sec^2z\ dz,\\\\ \implies Cos^2z\ dz \implies integrating : 2z+sin2z=2(tan^{-1}t+\frac{t}{1+t^2})\\\\For\ \frac{-2}{1-t^2+2t}dt=\frac{-2}{2-(1-t)^2}dt =\frac{-dt}{1-(\frac{1-t}{\sqrt2})^2}\\\\w=\frac{1-t}{\sqrt2},\ dt=-\sqrt2\ dw,\ \implies \frac{\sqrt2}{1-w^2}dw=\frac{1}{\sqrt2}[\frac{1}{1-w}+\frac{1}{1+w}]dw\\\\integrating: \frac{1}{\sqrt2}Ln | \frac{1+w}{1-w} |=\frac{1}{\sqrt2}Ln | \frac{\sqrt2+1-t}{\sqrt2-1+t} |$

After integration we get

$-2 tan^{-1}t-\frac{2}{1+t^2}+4*\frac{1}{2} [tan^{-1}t+\frac{t}{1+t^2}]+\frac{1}{\sqrt2}Ln | \frac{\sqrt2+1-t}{\sqrt2-1+t} |\\\\=-\frac{2(1-t)}{1+t^2}+\frac{1}{\sqrt2}Ln | \frac{\sqrt2+1-t}{\sqrt2-1+t} |\\\\taking\ limits:\\\\2-\sqrt2*Ln(1+\sqrt2)$

That is the correct answer.  That was a long one.....