integrate $\int_0^{\frac{\pi}{4}}\frac{dx}{2+\tan x}$ 
$$\int^{\frac{\pi}{4}}_{0}\frac{dx}{2+\tan x}$$

$v=\tan(\frac{x}{2})$
$\tan x=\frac{2v}{1-v^2}$
$dx=\frac{2\,dv}{1+v^2}$
$$\int^{\frac{\pi}{4}}_0 \frac{dx}{2+\tan x}=\int^{\frac{\pi}{8}}_0 \frac{\frac{2\,dv}{1+v^2}}{2+\frac{2v}{1-v^2}}=\int^{\frac{\pi}{8}}_0 \frac{1-v^2}{(1+v^2)(-v^2+v+1)} \, dv$$
Using partial fractions 
$$-\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+4}{v^2+1}+\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+1}{-v^2+v+1}=\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{2v}{v^2+1}-\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{4}{v^2+1}+\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+1}{-v^2+v+1}$$
$$=\frac{1}{5}ln|v^2+1|-\frac{4}{5}\arctan(v)+\frac{1}{5}ln|-v^2+v+1|$$ from $\frac{\pi}{8}$ to $0$
$0.02-0+0.299-0+0.04-0=0.359$
But it should come out 0.32
 A: Universal trigonometric substitution is done in order to get the tangents instead of sines and cosines. In this case it is no longer necessary, so:
$$t= \tan x,\quad x = \arctan t,\quad dx = \dfrac1{t^2+1},$$
$$J =\int\limits_0^{\pi/4}\frac{dx}{2+\tan x} = \int_0^1\dfrac{dt}{(t+2)(t^2+1)}.$$
Let
$$R(t) = \dfrac1{(t+2)(t^2+1)} = \dfrac A{t+2}+\dfrac{Bt+C}{t^2+1},$$
then
$$A = \lim_{t\to-2}(t+2)R(t) = \lim_{t\to-2}\dfrac1{t^2+1} = \dfrac15,$$
$$A+B = \lim_{t\to\infty}tR(t) = 0,\quad B=-\dfrac15,$$
$$\dfrac A2+C = R(0) = \dfrac12,\quad C = \dfrac25.$$
Thus,
$$J=\dfrac15\int_0^1\dfrac{dt}{t+2} - \dfrac15\int_0^1\dfrac{t\,dt}{t^2+1} + \dfrac25\int_0^1\dfrac{dt}{t^2+1},$$
$$J = \dfrac15\log(t+2)\biggr|_0^1 - \dfrac1{10}\log(t^2+1)\biggr|_0^1 +\dfrac25\arctan t\biggr|_0^1,$$
$$J = \dfrac15\log\dfrac32 - \dfrac1{10}\log2 +\dfrac25\dfrac\pi4,$$
$$\boxed{J = \dfrac1{10}\left(\pi +\log\dfrac98\right)}$$
Another way (hinted by Wolfram Alpha):
$$J = \int\limits_0^{\pi/4}\frac{dx}{2+\tan x} = \int\limits_0^{\pi/4}\frac{\cos x\,dx}{\sin x + 2\cos x},$$
$$\cos x = A(\sin x + 2\cos x) + B(\cos x - 2\sin x),$$
$$
\begin{cases}
2A + B = 1\\
A - 2B = 0,
\end{cases}\quad 
A = \dfrac25,\quad
B = \dfrac15,
$$
$$J = \dfrac25\int\limits_0^{\pi/4}\,dx + \dfrac15\int\limits_0^{\pi/4}\frac{d(\sin x + 2\cos x)}{\sin x + 2\cos x}\,dx,$$
$$J = \dfrac{2x}5\biggr|_0^{\pi/4} + \dfrac15\log(\sin x + 2\cos x)\biggr|_0^{\pi/4},$$
$$J = \dfrac{2}5\dfrac\pi4 + \dfrac15\log\dfrac3{\sqrt2} - \dfrac15\log2,$$
$$\boxed{J = \dfrac1{10}\left(\pi +\log\dfrac98\right)}$$
A: The integrand is 
$$f(x):=\frac{\cos(x)}{2\cos(x)+\sin(x)}.$$
We can form a linear combination to let the derivative of the denominator appear at the numerator:
$$af(x)+b=\frac{a\cos(x)+b(2\cos(x)+\sin(x))}{2\cos(x)+\sin(x)}=\frac{-2\sin(x)+\cos(x)}{2\cos(x)+\sin(x)},$$ is obtained with
$$b=-2,a=5.$$
Then by integrating,
$$5F(x)-2x=\ln(|2\cos(x)+\sin(x)|),$$
from $0$ to $\dfrac\pi4$,
$$5I-\frac\pi2=\ln\left(\frac{\frac3{\sqrt2}}{2}\right),$$
we get
$$I=\frac{\ln(9)-\ln(8)+\pi}{10}\approx0.3259375689\cdots$$ as claimed.
A: You have made a mistake in simplifying the integrand. You should have
$$\int \frac{\frac{2dv}{1+v^2}}{2+\frac{2v}{1-v^2}} = \int \frac{1-v^2}{(1+v^2)(1+v-v^2)}dv.$$
Now do partial practions. You get
$$ \frac{1-v^2}{(1+v^2)(1+v-v^2)} = \frac{2v-1}{5(v^2-v-1)} -\frac{2v-2}{5(v^2+1)} .$$
The first integral is a U-substitution. For the second, split the numerator up, then U-substitution and arctan respectively.
A: Before doing any substitutions are algebra, you can see that the value of this integral will be a finite positive number because you're integrating a positive function over a bounded interval.
Multiplying the numerator and denominator by $(1-v^2)(1+v^2)$, we get:
$$
\frac{\frac 2 {1+v^2}}{2+\frac{2v}{1-v^2}} = \frac{2(1-v^2)}{2(1+v^2)(1-v^2)+2v(1+v^2)} = \frac{1-v^2}{(1 - v^4) + v+v^3}.
$$
Since the denominator is not $0$ when $v$ is $1$ or $-1$, nothing cancels.  You may need numerical methods to factor this; I'm not sure.
Notice that as $x$ goes from $0$ to $\pi/4$, then $v$ goes from $0$ to $\tan\dfrac\pi8$, so your bounds of integration need to take that into account.
A: You have made a few mistakes in your calculation.
First of all, since $v=\tan\frac{x}{2}$ with $\tan\frac{\frac{\pi}{4}}{2}=\tan\frac{\pi}{8}=\sqrt 2-1$, you should have
$$\int_{0}^{\frac{\pi}{4}}\frac{dx}{2+\tan x}=\int_{0}^{\color{red}{\sqrt 2-1}}\frac{\frac{2dv}{1+v^2}}{2+\frac{2v}{1-v^2}}$$
Also, you should have
$$\begin{align}&\frac{1-v^2}{(1+v^2)(-v^2+v+1)}\\&=\color{red}{+}\frac 15\cdot\frac{-2v+4}{v^2+1}+\frac{1}{5}\cdot\frac{-2v+1}{-v^2+v+1}\\&=-\frac 15\cdot\frac{2v}{v^2+1}+\frac 45\cdot\frac{1}{v^2+1}+\frac{1}{5}\cdot\frac{-2v+1}{-v^2+v+1}\end{align}$$
Now these give
$$\begin{align}&\int_{0}^{\frac{\pi}{4}}\frac{dx}{2+\tan x}\\&=\left[-\frac 15\ln(v^2+1)+\frac 45\arctan v+\frac 15\ln(-v^2+v+1)\right]_{0}^{\sqrt 2-1}\\&=-\frac 15\ln(4-2\sqrt 2)+\frac 45\arctan (\sqrt 2-1)+\frac 15\ln(3\sqrt 2-3)\\&=\frac 15\ln\left(\frac{3\sqrt 2-3}{4-2\sqrt 2}\right)+\frac 45\cdot\frac{\pi}{8}\\&=\frac 15\ln\left(\frac{3\sqrt 2}{4}\right)+\frac{\pi}{10}\approx 0.32594\end{align}$$
A: I have done the sum u just look at it.

Here after step 2 I have used the method where u have to do like this
Af(x) +Bf'(x) =numerator. and here A and B are constant.
Here f(x)=denominator =2cosx +sinx
Hence f'(x) =-2sinx +cosx
So we get A = 2/5 B= 1/5
Thanks for asking such a nice question.
