# Proving that $\int_0^{\pi/2} (\sin (\tan (x))+\cot (x) \cos (\tan (x))-\cot (x))\cot (x) \, dx=\frac{\pi(e-2)}{2e}$

I think one of the ways of doing it is by the use of the differentiation with parameter. Do you see
an easy way of calculating it by real methods?

$$\int_0^{\pi/2} (\sin (\tan (x))+\cot (x) \cos (\tan (x))-\cot (x))\cot (x) \, dx=\frac{\pi(e-2)}{2e}$$

What tools would you like to employ? Solutions are just optional.

EDIT_I: or in other words we might focus on the harder part, that is proving that

$$\int_0^{\pi/2} \cot ^2(x) (1-\cos (\tan (x))) \, dx=\frac{\pi}{2e}$$

EDIT_II: and after some work all gets reduced to calculating

$$\int_0^{\infty} \frac{\cos(x)}{1+x^2} \ dx$$ that is well-known, this being the last part.

EDIT_III: we're done. Thanks!

• Maybe you can write the solution as an answer. It may be useful for someone. – Marco Cantarini Aug 25 '15 at 12:53
• Be careful, don't forget about the $\cot^2 x$ factor in front. – John Joy Aug 25 '15 at 12:57
• @MarcoCantarini yeah, I might develop later my idea with more details. – user 1357113 Aug 25 '15 at 13:07
• @JohnJoy OK, I'll be very careful. ;) – user 1357113 Aug 25 '15 at 13:08
• Maybe you want to add that the only thing you really need is $x=\arctan(y), dx=\frac{1}{1+y^2}$ to bring your second integral to a standard form – tired Aug 25 '15 at 13:21

## 2 Answers

$$I = \int_0^{\frac{\pi }{2}} {\sin (\tan x)\cot xdx} - \int_0^{\frac{\pi }{2}} {{{\cot }^2}x[1 - \cos (\tan x)]dx}$$

For the first integral, making $\tan x = u$ gives $$\int_0^{\frac{\pi }{2}} {\sin (\tan x)\cot xdx} = \int_0^{ + \infty } {\frac{{\sin x}}{{x(1 + {x^2})}}dx} = \int_0^{ + \infty } {\frac{{\sin x}}{x}dx} - \int_0^{ + \infty } {\frac{{x\sin x}}{{1 + {x^2}}}dx}$$ Hence $$\int_0^{\frac{\pi }{2}} {\sin (\tan x)\cot xdx} =\frac{\pi }{2} - \frac{\pi }{{2e}}$$

For the second one, it's just the same to the first one, \eqalign{ & \int_0^{\frac{\pi }{2}} {{{\cot }^2}x[1 - \cos (\tan x)]dx} = \int_0^{ + \infty } {\frac{{1 - \cos x}}{{{x^2}(1 + {x^2})}}dx} \cr & = \int_0^{ + \infty } {\frac{{1 - \cos x}}{{{x^2}}}dx} - \int_0^{ + \infty } {\frac{{1 - \cos x}}{{1 + {x^2}}}dx} = \frac{\pi }{{2e}} \cr} Note that we have used some famous integrals here, all of them can be evaluated without complex number.

Finally, your original integral is $$I=\frac{\pi }{2} - \frac{\pi }{e}$$

Maybe it's interesting to note that we can also solve the integrals using complex analysis, noting that $$\int_{0}^{\infty}\frac{\sin\left(u\right)}{u\left(1+u^{2}\right)}du=\frac{1}{2}\textrm{Im}\left(\int_{-\infty}^{\infty}\frac{e^{iu}-1}{u\left(1+u^{2}\right)}du\right)$$ and $$\int_{0}^{\infty}\frac{\cos\left(u\right)-1}{u^{2}\left(1+u^{2}\right)}du=\frac{1}{2}\textrm{Re}\left(\int_{-\infty}^{\infty}\frac{e^{iu}-1}{u^{2}\left(1+u^{2}\right)}du\right)$$ or using the Feynman trick. Take $$I\left(a\right)=\int_{0}^{\infty}\frac{\sin\left(au\right)}{u\left(1+u^{2}\right)}du,\ a>0.$$ We can observe that $$-I''\left(a\right)+I\left(a\right)=\int_{0}^{\infty}\frac{\sin\left(au\right)}{u}du=\frac{\pi}{2}$$ hence $$I\left(a\right)=\frac{\pi}{2}+Ae^{a}+Be^{-a}$$ and evaluating $I\left(\infty\right)$ and $I'\left(0^{+}\right)$ we get $$A=0,\ B=-\frac{\pi}{2}$$ then $$I\left(a\right)=\frac{\pi}{2}\left(1-e^{-a}\right).$$ Now if we put $$J\left(a\right)=\int_{0}^{\infty}\frac{\cos\left(au\right)-1}{u^{2}\left(1+u^{2}\right)}du$$ we note that $$J'\left(a\right)=-\int_{0}^{\infty}\frac{\sin\left(au\right)}{u\left(1+u^{2}\right)}du=-I\left(a\right)$$ then $$J\left(a\right)=-\frac{\pi}{2}\left(a+e^{-a}\right)+C$$ and observing that $$0=J\left(0\right)=-\frac{\pi}{2}+C$$ we get $$C=\frac{\pi}{2}$$ so the integral is equal to$$\lim_{a\rightarrow1}\frac{\pi}{2}\left(1-e^{-a}\right)-\frac{\pi}{2}\left(a+e^{-a}\right)+\frac{\pi}{2}=\frac{\pi}{2}-\frac{\pi}{e}.$$

• Good to mention it. (+1) – user 1357113 Aug 25 '15 at 18:49
• @RaymondManzoni Fixed, thank you! – Marco Cantarini Aug 25 '15 at 18:49