# How to find $\int_0 ^ \frac{\pi}{2} \frac{\cot x}{\cot x + \csc x}\,dx \,$?

How to find $$\int_0 ^ \frac{\pi}{2} \frac{\cot x}{\cot x + \csc x}\,dx \,\,?$$

The integrand $$\frac{\cot x}{\cot x + \csc x}$$ is not defined at $$x =0$$. But the function is bounded on $$(0 , \frac{\pi}{2}]$$. $$\lim _{x \to 0} \frac{\cot x}{\cot x + \csc x} = 0$$ So this is not an improper integral.

My Attempt : $$\int_0 ^ \frac{\pi}{2} \frac{\cot x}{\cot x + \csc x} = \lim_{t \to 0} \int_t ^ \frac{\pi}{2} \frac{\cot x}{\cot x + \csc x} = \lim_{t \to 0} \left[(\frac{\pi}{2} - 1)+ (\tan{\frac{t}{2} - t})\right] = \left(\frac{\pi}{2} - 1\right)$$.

I know how to find the anti-derivative of the integrand. I first found out the anti-derivative of the integrand in $$[t , \frac{\pi}{2}]$$ , where $$0 < t < \frac{\pi}{2}$$. Let's say it is $$\phi(t)$$. Then I find $$\lim_{t \to 0} \phi(t)$$. I am not sure if this is a right way. Can anyone please check it?

• Your attempt looks correct to me. Commented Jul 15, 2020 at 6:31
• Since the integral is proper you don't need to take limits. Rather try to find a function $F$ which is continuous on $[0,\pi/2]$ and the derivative $F'$ equals the integrand on $(0,\pi/2)$. The Fundamental Theorem of Calculus then says that the integral is $F(\pi/2)-F(0)$. Commented Jul 15, 2020 at 11:37
• See math.stackexchange.com/a/3754943/72031 for the statement of FTC. Commented Jul 15, 2020 at 11:38
• Can you please give me the proof of this link's theorem? I am quite curious to see how it is being possible..@ParamanandSingh Commented Jul 20, 2020 at 11:26

$$\int_0 ^ \frac{\pi}{2} \frac{\cot x}{\cot x + cosec x} dx=\int_{0}^{\pi/2} \frac{\cos x}{1+\cos x} dx=\pi/2-\int_{0}^{\pi/2} \frac{dx}{1+\cos x} dx$$ $$=\pi/2-\frac{1}{2}\int_{0}^{\pi/2} \sec^2(x/2)~dx=\pi/2-\tan x |_{0}^{\pi/2}=\pi/2-1.$$

• I know how to solve it. I was asking about the point $0$ where the integrand is not defined. This is also not an improper integral. Is my attempt fine? 2Dr Zafar Commented Jul 15, 2020 at 7:20
• But the function in the first place is not defined at $x = 0$. Commented Jul 15, 2020 at 7:30
• I do not think You can change the function for your convenience in this way. Commented Jul 15, 2020 at 7:31
• @sam for the integral to exist either $f(0)$ or $\lim_{x \to 0} f(x)$ should be finite. So in your case $f(x)=\frac{\cot x}{\cot x+\csc x}$, $f(0)=\frac{\infty}{\infty}$ (undefined), but $\lim_{x \to 0} f(x)=1$, by L-Hospital's rule, or just by simplifying $f(x)$. Commented Jul 15, 2020 at 7:38
• @sani: it does not matter if the integrand is not defined at a finite number of points. At those points you can define the function in any manner. However what is more important that the function remains bounded and the interval of integration is also bounded. Otherwise we need to use the idea of improper integrals. Commented Jul 15, 2020 at 11:34

my solution $$\int_0 ^ \frac{\pi}{2} \frac{\cot x}{\cot x + \csc x}\mathrm{d}x$$ $$=\int_0 ^ \frac{\pi}{2} \frac{\cot x(\csc x-\cot x)}{\csc^2 x-\cot^2x}\mathrm{d}x$$ $$=\int_0 ^ \frac{\pi}{2} \csc x\cot x-\cot^2 x)\mathrm{d}x$$

$$=\int_0 ^ \frac{\pi}{2} \csc x\cot x-\csc^2 x+1)\mathrm{d}x$$

$$=(-\csc x+\cot x+x)_0^{\pi/2}$$ $$=\frac{\pi}{2}-1$$

• I know how to solve it. I was asking about the point $0$ where the integrand is not defined. This is also not an improper integral. Is my attempt fine? @Jyee Fischer Commented Jul 15, 2020 at 7:19

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \begin{align} \int_{0}^{\pi/2}{\cot\pars{x} \over \cot\pars{x} + \csc\pars{x}}\,\dd x & = \int_{0}^{\pi/2}{\cos\pars{x} \over \cos\pars{x} + 1}\,\dd x = \int_{0}^{\pi/2}{2\cos^{2}\pars{x/2} - 1 \over 2\cos^{2}\pars{x/2}}\,\dd x \\[5mm] & = \int_{0}^{\pi/2}\bracks{1 - {1 \over 2}\sec^{2}\pars{x \over 2}}\,\dd x = \left.x - \tan\pars{x \over 2}\right\vert_{\ 0}^{\ \pi/2} \\[5mm] & = {\pi \over 2} - \tan\pars{\pi \over 4} = \bbox[15px,#ffd,border:1px solid navy]{{\pi \over 2} - 1}\ \approx\ 0.5708 \end{align}

• I know how to solve it. I was asking about the point $0$ where the integrand is not defined. This is also not an improper integral. Is my attempt fine? @Felix Marin Commented Jul 15, 2020 at 7:20

HINT: Take , $$\cot x- \csc x=t$$

then, $$\cot x+ \csc x=\frac{1}{t}$$

and $$\cot x=\frac{1}{2}\left[t+\frac{1}{t}\right]$$

I think you can proceed from here.

• I know how to solve it. I was asking about the point $0$ where the integrand is not defined. This is also not an improper integral. Is my attempt fine? Commented Jul 15, 2020 at 7:20
• @sam That is why I reshuffled (simplified) the integrand which is finite everywhere in $[0,\pi/2]$. For example $frac{x}{x+x^2}=\frac{1}{1+x}$ is finite at $x=0$. Commented Jul 15, 2020 at 7:28
• At $x = 0$ , what is $t = ?$@Venkat Amith Commented Jul 15, 2020 at 7:33
• At $x=0$ then $t=$, $cotx- cosecx=t$ , $\frac{cosx-1}{sinx}$, $t=-Tan(\frac{x}{2})$ then $t=0$ at $x=0$. @sani Commented Jul 15, 2020 at 9:15