Is there any other method to compute $\int_0^{\frac{\pi}{2}} \frac{x}{\sec x+\csc x} d x$?

Question

Rationalization sometimes makes our life easier

Letting $x\mapsto \frac{\pi}{2}-x$ transforms the integral to $\displaystyle I=\frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{1}{\sec x+\csc x} d x=\frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{\sin x \cos x}{\sin x+\cos x} d x\tag*{}\\ $

‘Rationalize’ the integrand gives

$\displaystyle \begin{aligned} \int_0^{\frac{\pi}{2}} \frac{\sin x \cos x}{\sin x+\cos x} d x = & \int_0^{\frac{\pi}{2}} \frac{\sin x \cos x(\sin x-\cos x)}{\sin ^2 x-\cos ^2 x} d x \\= & \int_0^{\frac{\pi}{2}} \frac{\sin ^2 x}{2 \sin ^2 x-1} d(\sin x)+\int_0^{\frac{\pi}{2}} \frac{\cos ^2 x}{2 \cos ^2 x-1} d(\cos x)\end{aligned}\tag*{} $

Putting $y=\sin x$ in the first integral and then $y=\cos x$ in the second yields the same definite integrals as:

$\displaystyle \begin{aligned}\int_0^{\frac{\pi}{2}} \frac{\sin x \cos x}{\sin x+\cos x} d x & =2 \int_0^1 \frac{y^2}{2 y^2-1} d y \\& =\int_0^1\left(1+\frac{1}{2 y^2-1}\right) d y \\& =\left[y+\frac{1}{2 \sqrt{2}} \ln \left|\frac{\sqrt{2 y}-1}{\sqrt{2 y}+1}\right|\right]_0^1 \\& =1+\frac{1}{\sqrt{2}} \ln (\sqrt{2}-1)\end{aligned}\tag*{} $ Now we can conclude that $\displaystyle \boxed{I=\frac{\pi}{4 \sqrt{2}}[\sqrt{2}+\ln (\sqrt{2}-1)]}\tag*{} $

Is there any other simpler method?

Your comments and alternative methods are highly appreciated.

Answer 1

I'm not sure it's any shorter, but applying the tangent half-angle substitution $$x = 2 \arctan t, \qquad dx = \frac{2 \,dt}{t^2 + 1}$$ is arguably more straightforward. That substitution transforms the integral $\int_0^{\frac{\pi}{2}} \frac{dx}{\sec x + \csc x}$ to $$4 \int_0^1 \frac{t^3 - t}{(t^2 + 1)^2 (t^2 - 2 t - 1)}\,dt .$$ Decomposing the integrand using the method of partial fractions gives $$\int_0^1\left[-\frac{1}{t^2 + 1} + 2\frac{t + 1}{(t^2 + 1)^2} + \frac{1}{t^2 - 2 t - 1}\right] dt = 1 - \frac{1}{\sqrt{2}} \operatorname{artanh} \frac{1}{\sqrt{2}} ,$$ hence $$\int_0^{\frac{\pi}{2}} \frac{x \,dx}{\sec x + \csc x} = \frac{\pi}{4} \left(1 - \frac{1}{\sqrt{2}} \operatorname{artanh} \frac{1}{\sqrt{2}}\right) .$$

Answer 2

Alternatively, the translation $$y = x - \frac{\pi}{4}$$ transforms the integrand to be symmetric about $y = 0$, yielding (after applying the angle sum identities for $\sin$ and $\cos$ and simplifying) $$\int_0^\frac{\pi}{2} \frac{dx}{\sec x + \csc x} = \frac{1}{2 \sqrt{2}} \int_{-\frac{\pi}{4}}^\frac{\pi}{4} \left(\cos y - \frac{\sin^2 y}{\cos y}\right) \,dy .$$ One can handle the second integrand by rewriting it as $$\frac{\sin^2 y \cos y}{\cos^2 y} = \frac{\sin^2 y \cos y}{1 - \sin^2 y} = \left(1 - \frac{1}{1 - \sin^2 y}\right) \cos y ,$$ which is then amendable to the substitution $u = \sin y$, $du = \cos y \,dy$.

Answer 3

Based on the relation you have obtained, one has that \begin{align*} \frac{\cos(x)\sin(x)}{\cos(x) + \sin(x)} & = \frac{\sin(2x)}{2(\cos(x) + \sin(x))}\\\\ & = \frac{(\cos(x) + \sin(x))^{2} - 1}{2(\cos(x) + \sin(x))}\\\\ & = \frac{\cos(x) + \sin(x)}{2} - \frac{1}{2(\cos(x) + \sin(x))} \end{align*}

where the first term is easy to integrate. Now it remains to integrate the second term above, which can be done through the substitution $x = 2\arctan(t)$.

Answer 4

Wherever I see an integral with rational function of sines and cosines, I think that using complex exponential is the simplest way to go. $$ \int \frac{1}{\sec x+\csc x} dx=\int \frac{1}{\frac{1}{\sin x}+\frac{1}{\cos x}} dx$$ using Euler's formula gives you $$\int \frac{1}{\sec x+\csc x} dx=\int \frac{1}{\frac{2i}{\exp(ix)-\exp(-ix)}+\frac{2}{\exp(ix)+\exp(-ix)}} dx$$ and with some algebra and the substitution $e^{ix}=u\Rightarrow e^{ix}dx=-idu$ we obtain an integral of an rational function $$\frac{1}{4}\int\frac{(1-u^4)(1-i+(i+1)u^2)}{u^6+u^2}du$$ which can be solved with Partial fractions.

The advantage is that the algebra is simple and the substitution is easy to remember and to apply.

Answer 5

Integrate by parts \begin{align} I=&\int_0^{\frac{\pi}{4}} \frac{2\sin x \cos x}{\sin x+\cos x} d x =\int_0^{\frac{\pi}{4}} \frac{\sin x -\cos x}{\sqrt{\cos 2x}} d (\sqrt{\cos 2x})\\ =&\ 1- \frac1{\sqrt2}\int_0^{\frac{\pi}{4}} \sec(x-\frac\pi4)dx =1-\frac{\ln(\sqrt2+1)}{\sqrt2} \end{align}