# Evaluating $\int_0^{\pi/2} \frac{\sqrt{\tan x}}{\sin x(\cos x+\sin x)}\mathrm{d}x$

I was solving the following integral: $$I=\int_0^{\pi/2} \frac{\sqrt{\tan x}}{\sin x(\cos x+\sin x)}\mathrm{d}x$$ Using the identity $$\sin x=\frac{\tan x}{\sec x}$$, I simplified the integral in the following way: $$I=\int_0^{\pi/2}\frac{\sqrt{\tan x}}{\frac{\tan x}{\sec x}\left(\cos x+\frac{\tan x}{\sec x}\right)}\mathrm{d}x=\int_0^{\pi/2}\frac{\sec^2 x}{\sqrt{\tan x}(1+\tan x)}\mathrm{d}x=2\tan^{-1}(\sqrt{\tan x})\Bigg|^{\pi/2}_0=\lim_{x\to \pi/2} 2\tan^{-1}(\sqrt{\tan x})=\pi$$ However, I'd like to see a different solution, because I feel there's a simpler one maybe using the symmetry of the integrand: $$\int_0^{\pi/2} \frac{\sqrt{\tan x}}{\sin x(\cos x+\sin x)}\mathrm{d}x=2\int_0^{\pi/4} \frac{\sqrt{\tan x}}{\sin x(\cos x+\sin x)}\mathrm{d}x$$

• What could be more simpler than finding a simple antiderivative for your integral? Quit while you are happy, ahead, and with the most efficient solution. Cheers. Commented Aug 1, 2023 at 10:13
• @NinadMunshi Thank you for the comment. In fact, I'm not sure if there's a simpler solution, maybe it's just my gut feeling Commented Aug 1, 2023 at 10:15
• The integral is equivalent to $\int_{-\pi/4}^{\pi/4} \sec x\sqrt{\sec(2x)}\,dx = \sqrt{2}\int_{0}^{\pi/2}[\cos x(1+\cos x)]^{-1/2}dx$, though I'm not sure how that helps. Commented Aug 1, 2023 at 10:27
• Simpler is a subjective term. What is simple for me may or may not be simple for you. Commented Aug 1, 2023 at 13:12
• @MathStackexchangeIsNotSoBad that's not always true Commented Aug 1, 2023 at 13:17

Substitute $$t=\sqrt{\tan x}-\sqrt{\cot x}$$. Then $$dt =\frac{\sqrt{\cot x}+\sqrt{\tan x}}{2\sin x \cos x}dx,\>\>\>\>\> t^2+4=\frac{(\sin x + \cos x)^2}{\sin x \cos x}$$ and \begin{align} &\int_0^{\pi/2} \frac{\sqrt{\tan x}}{\sin x(\cos x+\sin x)}{d}x\\ =&\int_0^{\pi/2} \frac{\sqrt{\tan x}+\sqrt{\cot x}}{(\sin x +\cos x)^2}dx=\int_{-\infty}^\infty\frac{2}{t^2+4}dt=\pi \end{align}

Use the substitution $$t=\sqrt{\tan x}$$. When we see $$\sqrt{\tan x}$$ in the integrand, this substitution will often rationalize the integrand for us. $$u^2=\tan x, \space dx=\frac{2u\space du}{1+u^4}$$ $$\int_0^{\pi/2}\frac{\sqrt{\tan x}}{\sin x (\cos x +\sin x)}dx=\int_0^\infty\frac{u}{\frac{u^2}{\sqrt{u^4+1}}(\frac{1}{\sqrt{u^4+1}}+\frac{u^2}{\sqrt{u^4+1}})}\frac{2u\space du}{1+u^4}=\int_0^\infty\frac{2du}{1+u^2}=\pi$$ We can also find the indefinite integral. Using properties of arctan, we can find another substitution that works for fun $$\int \frac{\sqrt{\tan x}}{\sin x (\cos x +\sin x)}dx=2\arctan (\sqrt{\tan x})+C=\arctan\left(\frac{2}{\sqrt{\cot x}-\sqrt{\tan x}}\right)+C$$ This is why $$t=\sqrt{\cot x}-\sqrt{\tan x}$$ is also fine

It's not simpler than the other suggestions, but just for the sake of completing eyeballfrog's comment:

Shifting the interval of integration by substituting $$x\to x+\dfrac\pi4$$, we can reduce

$$I=\int_0^\tfrac\pi2 \frac{\sqrt{\tan x}}{\sin x(\cos x+\sin x)} \, dx = \int_0^\tfrac\pi4 \frac2{\cos x\sqrt{1-2\sin^2x}} \, dx$$

by making use of the identity

$$\tan\left(x+\frac\pi4\right) = \frac{1-\tan x}{1+\tan x}$$

and replacing $$x\to-x$$ in the negative half of $$\left[-\dfrac\pi4,\dfrac\pi4\right]$$.

Now continue with $$\sin x=\sin\dfrac\pi4\sin y$$ to transform and evaluate:

$$I = \int_0^\tfrac\pi2 \frac{2\sqrt2}{1+\cos^2y} \, dy = \int_0^\tfrac\pi2 \frac{2\sqrt2 \, \sec^2y}{\sec^2y+1} \, dy = \int_0^\tfrac\pi2 \frac{2}{\left(\frac{\tan y}{\sqrt2}\right)^2+1} \, d\frac{\tan y}{\sqrt2}$$