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Calculus Level 2

$\int_{0}^{\pi/2} \frac{\sqrt{\cot(x)}}{\sqrt{\cot(x)} + \sqrt{\tan(x)}}dx$

The value of the integral above can be written as $\dfrac{a\pi^{c}}{b}$ , where $a$ and $b$ are positive coprime integers. Enter $a+b+c$ .

8 4 6 5

5 solutions

Chris Lewis
Jun 12, 2019

(Almost) no trig needed! Let the integral be $I$ . With the substitution $y=\frac{\pi}{2}-x$ we find

$I=\int_0^{\frac{\pi}{2}} \frac{\sqrt{\tan{x}}}{\sqrt{\cot{x}}+\sqrt{\tan{x}}} dx = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\cot{y}}}{\sqrt{\cot{y}}+\sqrt{\tan{y}}} dy$

Adding the two forms of $I$ we get

$2I=\int_0^{\frac{\pi}{2}} \frac{\sqrt{\cot{x}}+\sqrt{\tan{x}}}{\sqrt{\cot{x}}+\sqrt{\tan{x}}} dx = \int_0^{\frac{\pi}{2}} dx = \frac{\pi}{2}$

and so $I=\frac{\pi}{4}$ . Writing this in the required form we have $(a,b,c)=(1,4,1)$ and $a+b+c=\boxed6$ .

Of course, it can be done the hard way.

$\int \frac{\sqrt{\cot (x)}}{\sqrt{\tan (x)}+\sqrt{\cot (x)}} \, dx \Rightarrow \\ \frac{1}{4} \left(2 \tan ^{-1}(\tan (x))-\log \left(1-\sqrt{\tan (x)}\right)- \log \left(\sqrt{\tan (x)}+1\right)+\log (\tan (x)+1)+ \\ \sqrt{\tan (x)} \sqrt{\cot (x)} \left(\log (1-\tan (x))+\log (\tan (x)+1)-\log \left(\sec ^2(x)\right)\right)\right)$

You need to take the limit of the expression to get the lower limit of $0$ .

You need to take the limit of the expression from below to get the upper limit of $\frac{\pi}{4}$ .

This give the answer of 6 per given formula.

A Former Brilliant Member - 1 year, 12 months ago

How did you derive this indefinite integral?

Pi Han Goh - 1 year, 11 months ago

$\int \frac{\sqrt{\cot(x)}}{\sqrt{\tan(x)}+\sqrt{\cot(x)}} \, dx$

$\int \frac{\sqrt{\tan(x)} \sqrt{\cot(x)}-\cot(x)}{\tan(x) -\cot(x)} \, dx$

$\int \left(\frac{\sqrt{\tan(x)} \sqrt{\cot(x)}}{\tan(x)-\cot(x)}-\frac{\cot(x)}{\tan(x)-\cot(x)}\right) \, dx$

$\int \frac{\sqrt{\tan(x)} \sqrt{\cot(x)}}{\tan(x)-\cot(x)} \, dx-\int \frac{\cot(x)}{\tan(x)-\cot(x)} \, dx$

$\int \frac{\sqrt{\tan(x)} \sqrt{\cot(x)}}{\tan(x)-\cot(x)} \times \frac{(\tan(x)) \left(\sec ^2(x)\right)}{(\tan(x)) \left(\sec ^2(x)\right)}\, dx-\int \frac{\cot(x)}{\tan(x)-\cot(x)} \, dx$

$u=x \tan(x)$ and $du=\sec^2(x)\,dx$

$\int \frac{u}{u^4-1} \, du-\int \frac{\cot(x)}{\tan(x)-\cot(x)} \, dx$

$s=u^2$ and $ds=2u\,du$

$\frac{1}{2} \int \frac{1}{s^2-1} \, ds-\int \frac{\cot(x)}{\tan(x)-\cot(x)} \, dx$

$-\frac{1}{2} \int \frac{1}{1-s^2} \, ds-\int \frac{\cot(x)}{\tan(x)-\cot(x)} \, dx$

$-\frac{1}{2}\tanh(s)-\int \frac{\cot(x)}{\tan(x)-\cot(x)} \, dx$

$-\frac{1}{2}\tanh(s)-\int \frac{\sec ^2(x)}{-\left(\sec ^2(x)\right)+\left(\sec ^2(x)\right) \left(\tan ^2(x)\right)} \, dx$

$-\frac{1}{2}\tanh(s)-\int \frac{\sec ^2(x)}{-\left(\sec ^2(x)\right)+\left(\sec ^2(x)\right) \left(\tan ^2(x)\right)} \frac{(\tan(x)) \left(\sec ^2(x)\right)}{(\tan(x)) \left(\sec ^2(x)\right)} \, dx$

$-\frac{1}{2}\tanh(s)-\int \frac{\sec ^2(x)}{-\left(\sec ^2(x)\right)+\left(\sec ^2(x)\right) \left(\tan ^2(x)\right)} \, dx$

Using $\sec ^2(x)=\tan ^2(x)+1$

$-\frac{1}{2}\tanh(s)-\int \frac{\sec ^2(x)}{-1+\tan ^4(x)} \, dx$

$p=\tan(x)$ and $dp=\sec(x)\,dx$

$-\frac{1}{2}\tanh(s)-\int \frac{1}{p^4-1} \, dp$

$-\frac{1}{2}\tanh(s)-\int \left(-\frac{1}{2 \left(p^2+1\right)}-\frac{1}{4 (p+1)}+\frac{1}{4 (p-1)}\right) \, dp$

$-\frac{1}{2}\tanh(s)+\frac{1}{2}\int \frac{1}{p^2+1} \, dp+\frac{1}{4}\int \frac{1}{p+1} \, dp-\frac{1}{4}\int \frac{1}{p-1} \, dp$

$-\frac{1}{2}\tanh(s)+\frac{1}{2}\tan^{-1}(p) +\frac{1}{4}\log(p+1)-\frac{1}{4}\log(p-1)$

$-\frac{1}{2}\tanh(tan(x)^2)+\frac{1}{2}\tan^{-1}(tan(x)) +\frac{1}{4}\log(tan(x)+1)-\frac{1}{4}\log(tan(x)-1)$

A Former Brilliant Member - 1 year, 11 months ago

คลุง แจ็ค
Jun 17, 2019

Here is another way of doing it.

$\displaystyle \int \frac{\sqrt{\cot{x}}}{\sqrt{\cot{x}}+\sqrt{\tan{x}}} dx =\int \frac{\cos{x}}{\cos{x}+\sin{x}} dx=\frac{1}{\sqrt{2}} \int \frac{\cos{x}}{\sin{(x+\pi/4)}} dx = \frac{1}{\sqrt{2}} \int \frac{\cos{(x+\pi/4-\pi/4)}}{\sin{(x+\pi/4)}} dx = \frac{1}{2} \int \frac{\cos{(x+\pi/4)}+\sin{(x+\pi/4)}}{\sin{(x+\pi/4)}} dx$

This integral is easy enough to integrate directly.

$\displaystyle \frac{1}{2} \times \left (\ln{(\sin{(x+\pi/4)})} +x \right )$

Substitute the limit of integration in and you get the answer pretty quickly.

Manifold M
Jun 12, 2019

$I= \int_{0}^{\pi/2} \frac{\sqrt{\cot(x)}}{\sqrt{\cot(x)} + \sqrt{\tan(x)}}dx = -\int_{\pi/2}^{0} \frac{\sqrt{\cot(\pi/2-t)}}{\sqrt{\cot(\pi/2-t)} + \sqrt{\tan(\pi/2-t)}}dt = \int_{0}^{\pi/2} \frac{\sqrt{\tan(t)}}{\sqrt{\tan(t)} + \sqrt{\cot(t)}}dx = I$

$2I = \int_{0}^{\pi/2} (\frac{\sqrt{\cot(x)}}{\sqrt{\cot(x)} + \sqrt{\tan(x)}} + \frac{\sqrt{\tan(x)}}{\sqrt{\cot(x)} + \sqrt{\tan(x)}})dx =\int_{0}^{\pi/2} \frac{\sqrt{\cot(x)} + \sqrt{\tan(x)}}{\sqrt{\cot(x)} + \sqrt{\tan(x)}}dx = \int_{0}^{\pi/2} dt = \frac{\pi}{2}$

Hence $I=\frac{\pi}{4}$ . So 1+4+1=6

Chew-Seong Cheong
Jun 13, 2019

$\begin{aligned} I & = \int_0^\frac \pi 2 \frac {\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}} dx & \small \color{#3D99F6} \text{Using } \int_a^b f(x)\ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_0^\frac \pi 2 \left( \frac {\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}} + \frac {\sqrt{\cot \left(\frac \pi 2- x\right)}}{\sqrt{\cot \left(\frac \pi 2- x\right)}+\sqrt{\tan \left(\frac \pi 2- x\right)}} \right) dx \\ & = \frac 12 \int_0^\frac \pi 2 \left( \frac {\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}} + \frac {\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{\cot x}} \right) dx \\ & = \frac 12 \int_0^\frac \pi 2 dx = \frac \pi 4 \end{aligned}$

Therefore, $a+b+c = 1+1+4 = \boxed 6$ .

@Moshe M , you have to mention: $a$ and $b$ are coprime integers, if not $(a,b) = (1,4), (2,8), (3,12), ...$ are all solutions.

Chew-Seong Cheong - 1 year, 12 months ago

Of course! Sorry : )

Manifold M - 1 year, 12 months ago

Naren Bhandari
Jun 13, 2019

$\int_0^{\frac{1}{2}\pi} \dfrac{\sqrt {\cot x} }{\sqrt{\cot x }+\sqrt{\tan x}}\,dx = I =\int_0^{\frac{1}{2}\pi} \dfrac{\cos x}{\cos x+ \sin x}\,dx \\ = \dfrac{\pi}{2} - \int_0^{\frac{1}{2}\pi}\dfrac{\sin x }{\cos x+\sin x}=\dfrac{\pi}{2} -\int_0^{\frac{1}{2}\pi}\dfrac{\cos \left(\frac{\pi}{2} -x\right)}{\sin\left(\frac{\pi}{2}-x\right)+\cos\left(\frac{\pi}{2}-x\right)}\,dx= \frac{\pi}{2} -I \\ 2I= \dfrac{\pi}{2}\implies I =\dfrac{\pi}{4}$ thus $a+b+c=6$ . We use the identity $\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$

Let's get trigy

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