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


$\tan x=\frac{2v}{1-v^2}$


$$\int^{\frac{\pi}{4}}_0 \frac{dx}{2+\tan x}=\int^{\frac{\pi}{8}}_0 \frac{\frac{2\,dv}{1+v^2}}{2+\frac{2v}{1-v^2}}=\int^{\frac{\pi}{8}}_0 \frac{1-v^2}{(1+v^2)(-v^2+v+1)} \, dv$$

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 $0$


But it should come out 0.32

  • 2
    $\begingroup$ Please: Write $\tan x$, $\arctan x$, $\ln x$, not $tan x$, $arctan x$, $ln x$. See my edits. $\endgroup$ – Michael Hardy Feb 19 '16 at 0:24

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 $$\begin{align}&\int_{0}^{\frac{\pi}{4}}\frac{dx}{2+\tan x}\\&=\left[-\frac 15\ln(v^2+1)+\frac 45\arctan v+\frac 15\ln(-v^2+v+1)\right]_{0}^{\sqrt 2-1}\\&=-\frac 15\ln(4-2\sqrt 2)+\frac 45\arctan (\sqrt 2-1)+\frac 15\ln(3\sqrt 2-3)\\&=\frac 15\ln\left(\frac{3\sqrt 2-3}{4-2\sqrt 2}\right)+\frac 45\cdot\frac{\pi}{8}\\&=\frac 15\ln\left(\frac{3\sqrt 2}{4}\right)+\frac{\pi}{10}\approx 0.32594\end{align}$$


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): $$J = \int\limits_0^{\pi/4}\frac{dx}{2+\tan x} = \int\limits_0^{\pi/4}\frac{\cos x\,dx}{\sin x + 2\cos x},$$ $$\cos x = A(\sin x + 2\cos x) + B(\cos x - 2\sin x),$$ $$ \begin{cases} 2A + B = 1\\ A - 2B = 0, \end{cases}\quad A = \dfrac25,\quad B = \dfrac15, $$ $$J = \dfrac25\int\limits_0^{\pi/4}\,dx + \dfrac15\int\limits_0^{\pi/4}\frac{d(\sin x + 2\cos x)}{\sin x + 2\cos x}\,dx,$$ $$J = \dfrac{2x}5\biggr|_0^{\pi/4} + \dfrac15\log(\sin x + 2\cos x)\biggr|_0^{\pi/4},$$ $$J = \dfrac{2}5\dfrac\pi4 + \dfrac15\log\dfrac3{\sqrt2} - \dfrac15\log2,$$ $$\boxed{J = \dfrac1{10}\left(\pi +\log\dfrac98\right)}$$

  • $\begingroup$ nice...........+1 $\endgroup$ – Bhaskara-III Feb 23 '16 at 12:53

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.

  • $\begingroup$ how dis you get to $(1+v^2)(1+v-v^2)$? $\endgroup$ – gbox Feb 19 '16 at 0:47
  • 1
    $\begingroup$ @gbox This is just fraction manipulation really (common denominator, factoring, etc...) $\endgroup$ – GaussTheBauss Feb 19 '16 at 0:49
  • $\begingroup$ The second term should be wrong. $\endgroup$ – mathlove Feb 23 '16 at 6:09

The integrand is


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


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.


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.

  • $\begingroup$ I think you mean something like "a positive, bounded function over a bounded interval" in your first sentence, right? $\endgroup$ – mickep Feb 23 '16 at 6:57

I have done the sum u just look at it.

enter image description here

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.