Evaluate $ \int_{\pi/4}^{\pi/2} \frac{2\sin x+\cos x}{\sin x+2\cos x}\,dx$ 
Evaluate 
  $$ \int_{\pi/4}^{\pi/2} \frac{2\sin x+\cos x}{\sin x+2\cos x}\,dx,$$

My attempt : $u=\tan\frac{x}{2} \rightarrow x=2\arctan(u) \rightarrow \frac{2}{1+u^2}du=dx$
$$\sin x= \frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}=\frac{2u}{1+u^2}$$
$$\cos x= \frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}= \frac{1-u^2}{1+u^2}$$
I was looking to see if this is the right way to go and also i am not sure how to evaluate the boundaries . Could you help me out ?


\begin{align} I&=\int_{\pi/4}^{\pi/2} \frac{2\sin x+\cos x}{\sin x+2\cos x}\,dx\tag{1}\label{1}.\end{align} 

\begin{align} 
I&=
\int_{\pi/4}^{\pi/2} 
\frac{2\cot x+1}{\cot x+2}\,dx
\overset{\color{blue}{t=\cot x}}
{=}
\int_0^1
 \frac{2+t}{(1+2t)(1+t^2)}
 \, dt
\tag{2}\label{2}
\\
&=
\int_0^1
\frac{2+t}{(1+2t)(1+t^2)}
\, dt
=
\int_0^1
\tfrac 65\,\frac 1{1+2t}
+\tfrac45\,\frac1{1+t^2}
-\tfrac35\,\frac {t}{1+t^2}
\, dt
\tag{3}\label{3}
\\
&=
\left(
\tfrac 65\cdot\tfrac12\,\ln(1+2t)
\right)_0^1
+
\left(
\tfrac45\cdot\arctan(t)
\right)_0^1
-
\left(
\tfrac35\cdot
\tfrac12\,\ln(1+t^2)
\right)_0^1
\\
&=
\tfrac35\,\ln(3)+\tfrac\pi5-\tfrac3{10}\,\ln(2)
\approx 1.07954175
.
\end{align}
