# Which is the easiest way to evaluate $\int \limits_{0}^{\pi/2} (\sqrt{\tan x} +\sqrt{\cot x})$?

Which is the easiest way to evaluate $\int \limits_{0}^{\pi/2} (\sqrt{\tan x} +\sqrt{\cot x})$?

I have reduced this problem to $$2\int_0^{\pi/2} \sqrt{\tan x} \ dx$$

but now, evaluating this integral is giving me some problems, simply substituting $u=\tan(x)$ and then $\mathrm{d}u=\sec^2(x)\mathrm{d}x \Rightarrow \frac{\mathrm{d}u}{1+u^2}=\mathrm{d}x$ and which in turn gives something a bit ugly, I was wondering which is the most elegant way to evaluate this?

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$${\int_0^{\frac{\pi}{2}} \sqrt{\tan x}dx + \sqrt{\cot x}dx}$$ $$={\int_0^{\frac{\pi}{2}}\frac{\sin x + \cos x}{\sqrt{\sin x \cos x}}dx = \int_0^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\frac{\sqrt{2\sin{x}\cos{x}}}{\sqrt{2}}}dx = \sqrt{2}\int_0^{\frac{\pi}{2}} \frac{ \sin{x} + \cos{x}}{\sqrt{1 - (1 - 2 \sin{x} \cos{x})}}dx}$$ $${=\sqrt{2}\int_0^{\frac{\pi}{2}} \frac{ \sin{x} + \cos{x}}{\sqrt{1 - (\sin{x} - \cos{x})^2}}dx}$$

Let ${t = \sin{x} - \cos{x}}$, $\Large {{\small{dx}} = \frac{dt}{\sin{x} + \cos{x}}}$ $${x \to \frac{\pi}{2} \implies t = (\sin{x} - \cos{x}) \to 1}$$ $${x \to 0 \implies t = (\sin{x} - \cos{x}) \to -1}$$

$$\sqrt{2}\int_{-1}^{1} \frac{1}{\sqrt{1 - t^2}}dt = \sqrt{2}\left[\sin^{-1}{t}\right]_{-1}^{1} = \sqrt{2}\left[\frac{\pi}{2} - \left(- \frac{\pi}{2} \right) \right] = \sqrt{2} \pi$$

I think this might be the simplest approach.

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Brilliant substitution. Clever Manipulation of the integral! +1 – user21436 Jan 22 '12 at 14:08

I would argue the easiest way is to use the Gamma function. Notice that by making the change $x=\sin^2(u)$ we get that $$\int_0^1 x^{-\frac{1}{4}}(1-x)^{-\frac{3}{4}}dx=\int_0^{\pi/2}\sqrt{\tan(x)}dx$$ Then this is $$B\left(\frac{1}{4},\frac{3}{4}\right)=\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{3}{4}\right)=\frac{\pi}{\sin\left(\frac{\pi}{4}\right)}=\frac{\pi}{\sqrt{2}}.$$

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Let $u=\sqrt{\tan(x)}$. Then $u^2 = \tan(x)$ and $2 u \mathrm{d} u = (1+ \tan^2(x)) \mathrm{d} x$. Thus $$\int_0^{\frac{\pi}{2}} \sqrt{\tan(x)} \mathrm{d} x = \int_0^\infty \frac{2u^2}{1+u^4} \mathrm{d} u$$ Since $1+u^4 = (1 + \sqrt{2} u + u^2)( 1- \sqrt{2} u + u^2)$, partial fraction decomposition applies: $$\frac{2u^2}{1+u^4} = \frac{1}{\sqrt{2}} \left( \frac{u}{u^2-\sqrt{2} u+1}-\frac{u}{u^2+\sqrt{2} u+1} \right)$$ Hence $$\begin{eqnarray} \int \frac{2u^2}{1+u^4} \mathrm{d} u &=& \frac{1}{2 \sqrt{2}} \log \left(\frac{u^2-\sqrt{2} u+1}{u^2+\sqrt{2} u+1}\right) + \\ &\phantom{=}& \frac{\tan ^{-1}\left(\sqrt{2} u+1\right) -\tan ^{-1}\left(1-\sqrt{2} u\right) }{\sqrt{2}} \end{eqnarray}$$ Applying the fundamental theorem of calculus: $$\int_0^{\pi/2} \sqrt{\tan(x)} \mathrm{d} x = \frac{\pi}{\sqrt{2}}$$

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Hint: subtituting $u=\sin^2 x$ you will get the beta function, you will also need some basic properties of beta and gamma functions

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+1, I see you beat me too it. – Eric Naslund Jan 18 '12 at 21:52
However this question was in my school, and we haven't learnt gamma function yet! so that's not the easiest way! – CuriousSciDude Dec 3 '15 at 10:54

We will employing the substitution $u=\sqrt{\tan x}$: $$u'= \frac{1+\tan^2 x}{2 \sqrt{\tan x}}$$ and $$2\int_0^{\pi/2} \sqrt{\tan x}\,dx = 4 \int_0^\infty \frac{u^2}{1+u^4} du= 2 \int_{-\infty}^\infty \frac{u^2}{1+u^4} du.$$ The last integral has two poles ($u_1 = e^{i\pi/4}$, $u_2=e^{i3\pi/4}$) in the upper complex half-plane. The corresponding residue are $$\text{Res}_{u=u_1} \frac{u^2}{1+u^4} = -\frac{u_2}{4} \qquad\qquad \text{Res}_{u=u_1} \frac{u^2}{1+u^4} = -\frac{u_1}{4}.$$

Thus the value of the integral is $$2\int_0^{\pi/2} \sqrt{\tan x}\,dx =- \pi i (u_1+u_2)=\sqrt{2}\pi$$

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This is high-school level problem, so hopefully that's not the suitable way to do it. – VelvetThunder Jan 18 '12 at 21:20
@MaxX With substitution $u = \sqrt{\tan(x)}$ you get $\int_0^{\pi/2} \sqrt{\tan(x)} \mathrm{d} x = \int_0^\infty \frac{2u^2}{1+u^4} \mathrm{d} u$. The integrand has an elementary anti-derivative, so one can apply the fundamental theorem of calculus. – Sasha Jan 18 '12 at 21:24
@MarxX: of course one can also solve it using some arctan and logs. However, you were asking for the easiest solution and there is nothing easier then residue theorem :-) – Fabian Jan 18 '12 at 21:31
@Fabian assuming you know Residue theorem.... – Pureferret Jan 19 '12 at 0:56
In the sense that the question involves "the most elegant" way, I do not understand the downvote – Fabian Jan 19 '12 at 7:27

$$\int_0^{\frac{\pi}{2}} \sqrt{\tan(x)} \mathrm{d} x = \int_0^\infty \frac{2u^2}{1+u^4} \mathrm{d} u$$

$$= \int^{\infty}_0 \frac{u^2+1}{1+u^4} + \frac{u^2-1}{1+u^4} \mathrm{d} u$$

$$= \int^{\infty}_0 \frac{\mathrm{d} (u-1/u)}{ (u-1/u)^2 +2 } + \int^{\infty}_0 \frac{\mathrm{d} (u+1/u)}{ (u+1/u)^2 -2 }$$

and these have simple primitives in terms of arctan and logs. I like the Beta function approach or residues better, but this is something a high schooler can do.

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The text passage contains a mathematical question about evaluating a specific integral and multiple methods to solve it, including the use of the Beta and Gamma functions, substitution, and complex analysis (residue theorem). The passage also provides answers to the integral using these methods.
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# Which is the easiest way to evaluate $\int \limits_{0}^{\pi/2} (\sqrt{\tan x} +\sqrt{\cot x})$?

Which is the easiest way to evaluate $\int \limits_{0}^{\pi/2} (\sqrt{\tan x} +\sqrt{\cot x})$?

I have reduced this problem to $$2\int_0^{\pi/2} \sqrt{\tan x} \ dx$$

but now, evaluating this integral is giving me some problems, simply substituting $u=\tan(x)$ and then $\mathrm{d}u=\sec^2(x)\mathrm{d}x \Rightarrow \frac{\mathrm{d}u}{1+u^2}=\mathrm{d}x$ and which in turn gives something a bit ugly, I was wondering which is the most elegant way to evaluate this?

-

$${\int_0^{\frac{\pi}{2}} \sqrt{\tan x}dx + \sqrt{\cot x}dx}$$ $$={\int_0^{\frac{\pi}{2}}\frac{\sin x + \cos x}{\sqrt{\sin x \cos x}}dx = \int_0^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\frac{\sqrt{2\sin{x}\cos{x}}}{\sqrt{2}}}dx = \sqrt{2}\int_0^{\frac{\pi}{2}} \frac{ \sin{x} + \cos{x}}{\sqrt{1 - (1 - 2 \sin{x} \cos{x})}}dx}$$ $${=\sqrt{2}\int_0^{\frac{\pi}{2}} \frac{ \sin{x} + \cos{x}}{\sqrt{1 - (\sin{x} - \cos{x})^2}}dx}$$

Let ${t = \sin{x} - \cos{x}}$, $\Large {{\small{dx}} = \frac{dt}{\sin{x} + \cos{x}}}$ $${x \to \frac{\pi}{2} \implies t = (\sin{x} - \cos{x}) \to 1}$$ $${x \to 0 \implies t = (\sin{x} - \cos{x}) \to -1}$$

$$\sqrt{2}\int_{-1}^{1} \frac{1}{\sqrt{1 - t^2}}dt = \sqrt{2}\left[\sin^{-1}{t}\right]_{-1}^{1} = \sqrt{2}\left[\frac{\pi}{2} - \left(- \frac{\pi}{2} \right) \right] = \sqrt{2} \pi$$

I think this might be the simplest approach.

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Brilliant substitution. Clever Manipulation of the integral! +1 – user21436 Jan 22 '12 at 14:08

I would argue the easiest way is to use the Gamma function. Notice that by making the change $x=\sin^2(u)$ we get that $$\int_0^1 x^{-\frac{1}{4}}(1-x)^{-\frac{3}{4}}dx=\int_0^{\pi/2}\sqrt{\tan(x)}dx$$ Then this is $$B\left(\frac{1}{4},\frac{3}{4}\right)=\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{3}{4}\right)=\frac{\pi}{\sin\left(\frac{\pi}{4}\right)}=\frac{\pi}{\sqrt{2}}.$$

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Let $u=\sqrt{\tan(x)}$. Then $u^2 = \tan(x)$ and $2 u \mathrm{d} u = (1+ \tan^2(x)) \mathrm{d} x$. Thus $$\int_0^{\frac{\pi}{2}} \sqrt{\tan(x)} \mathrm{d} x = \int_0^\infty \frac{2u^2}{1+u^4} \mathrm{d} u$$ Since $1+u^4 = (1 + \sqrt{2} u + u^2)( 1- \sqrt{2} u + u^2)$, partial fraction decomposition applies: $$\frac{2u^2}{1+u^4} = \frac{1}{\sqrt{2}} \left( \frac{u}{u^2-\sqrt{2} u+1}-\frac{u}{u^2+\sqrt{2} u+1} \right)$$ Hence $$\begin{eqnarray} \int \frac{2u^2}{1+u^4} \mathrm{d} u &=& \frac{1}{2 \sqrt{2}} \log \left(\frac{u^2-\sqrt{2} u+1}{u^2+\sqrt{2} u+1}\right) + \\ &\phantom{=}& \frac{\tan ^{-1}\left(\sqrt{2} u+1\right) -\tan ^{-1}\left(1-\sqrt{2} u\right) }{\sqrt{2}} \end{eqnarray}$$ Applying the fundamental theorem of calculus: $$\int_0^{\pi/2} \sqrt{\tan(x)} \mathrm{d} x = \frac{\pi}{\sqrt{2}}$$

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Hint: subtituting $u=\sin^2 x$ you will get the beta function, you will also need some basic properties of beta and gamma functions

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+1, I see you beat me too it. – Eric Naslund Jan 18 '12 at 21:52
However this question was in my school, and we haven't learnt gamma function yet! so that's not the easiest way! – CuriousSciDude Dec 3 '15 at 10:54

We will employing the substitution $u=\sqrt{\tan x}$: $$u'= \frac{1+\tan^2 x}{2 \sqrt{\tan x}}$$ and $$2\int_0^{\pi/2} \sqrt{\tan x}\,dx = 4 \int_0^\infty \frac{u^2}{1+u^4} du= 2 \int_{-\infty}^\infty \frac{u^2}{1+u^4} du.$$ The last integral has two poles ($u_1 = e^{i\pi/4}$, $u_2=e^{i3\pi/4}$) in the upper complex half-plane. The corresponding residue are $$\text{Res}_{u=u_1} \frac{u^2}{1+u^4} = -\frac{u_2}{4} \qquad\qquad \text{Res}_{u=u_1} \frac{u^2}{1+u^4} = -\frac{u_1}{4}.$$

Thus the value of the integral is $$2\int_0^{\pi/2} \sqrt{\tan x}\,dx =- \pi i (u_1+u_2)=\sqrt{2}\pi$$

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This is high-school level problem, so hopefully that's not the suitable way to do it. – VelvetThunder Jan 18 '12 at 21:20
@MaxX With substitution $u = \sqrt{\tan(x)}$ you get $\int_0^{\pi/2} \sqrt{\tan(x)} \mathrm{d} x = \int_0^\infty \frac{2u^2}{1+u^4} \mathrm{d} u$. The integrand has an elementary anti-derivative, so one can apply the fundamental theorem of calculus. – Sasha Jan 18 '12 at 21:24
@MarxX: of course one can also solve it using some arctan and logs. However, you were asking for the easiest solution and there is nothing easier then residue theorem :-) – Fabian Jan 18 '12 at 21:31
@Fabian assuming you know Residue theorem.... – Pureferret Jan 19 '12 at 0:56
In the sense that the question involves "the most elegant" way, I do not understand the downvote – Fabian Jan 19 '12 at 7:27

$$\int_0^{\frac{\pi}{2}} \sqrt{\tan(x)} \mathrm{d} x = \int_0^\infty \frac{2u^2}{1+u^4} \mathrm{d} u$$

$$= \int^{\infty}_0 \frac{u^2+1}{1+u^4} + \frac{u^2-1}{1+u^4} \mathrm{d} u$$

$$= \int^{\infty}_0 \frac{\mathrm{d} (u-1/u)}{ (u-1/u)^2 +2 } + \int^{\infty}_0 \frac{\mathrm{d} (u+1/u)}{ (u+1/u)^2 -2 }$$

and these have simple primitives in terms of arctan and logs. I like the Beta function approach or residues better, but this is something a high schooler can do.

-
user
The text passage contains a mathematical question about evaluating a specific integral and multiple methods to solve it, including the use of the Beta and Gamma functions, substitution, and complex analysis (residue theorem). The passage also provides answers to the integral using these methods.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS