Question

Solve:
 $\int \sqrt{\cot x}dx$

Answer 1

Hint: If we try to replace $\cot x$ with simpler terms like $\sin x$ and $\cos x$ , then we will end up complicating it. It is better to solve it using $\cot x$ itself and by using the required formulas to solve the integral.

Complete step-by-step answer:
Let us consider $$\sqrt{\cot x}=t$$, then $\cot x=t^2$
On differentiating both sides of $\cot x=t^2$ with respect to $x$ , we get,
 $$\begin{gathered} -\cos e{c}^{2}x.dx=2t.dt \\ \Rightarrow -\cos e{c}^{2}x=2t{\displaystyle \frac{dt}{dx}} \\ \Rightarrow \cos e{c}^{2}x=-2t{\displaystyle \frac{dt}{dx}} \end{gathered}$$
Using the formula, $1+{\cot }^{2}x=\cos e{c}^{2}x$ , in the above equation we get,
 $1+{\cot }^{2}x=-2t.{\displaystyle \frac{dt}{dx}}$
Since, $\cot x=t^2$ , on substituting in the above equation, we get,
 $1+t^4=-2t.{\displaystyle \frac{dt}{dx}}$
On rearranging the terms, the equation would be,
 $dx={\displaystyle \frac{-2t}{1+t^4}}dt$
Substituting the values of $\sqrt{\cot x}$ and $dx$ in $\int \sqrt{\cot x}dx$ , we get,
 $$\begin{gathered} I=\int t.{\displaystyle \frac{-2t}{1+t^4}}dt \\ \Rightarrow I=\int {\displaystyle \frac{-2t^2}{1+t^4}}dt \end{gathered}$$
We can take the negative sign out of the integral,
 $I=-\int {\displaystyle \frac{2t^2}{1+t^4}}dt$
To solve this integral, we add and subtract the numerator with $1$ :
$$\begin{gathered} I=-\int {\displaystyle \frac{2t^2+1-1}{t^4+1}}dt \\ =-\int {\displaystyle \frac{t^2+1+t^2-1}{t^4+1}}dt \\ =-\int \left({\displaystyle \frac{t^2+1}{t^4+1}}+{\displaystyle \frac{t^2-1}{t^4+1}}\right)dt \end{gathered}$$
Now, we split the integral into two different integrals and solve them separately.
The two integrals are:
 $I_1=\int {\displaystyle \frac{t^2+1}{t^4+1}}dt$ and $I_2=\int {\displaystyle \frac{t^2-1}{t^4+1}}dt$

On dividing the numerator and the denominator of the first integral with $t^2$ ,
$$I_1=\int {\displaystyle \frac{1+{\displaystyle \frac{1}{t^2}}}{t^2+{\displaystyle \frac{1}{t^2}}}}dt$$
The denominator can be written as $t^2+{\displaystyle \frac{1}{t^2}}={\left(t-{\displaystyle \frac{1}{t}}\right)}^2+2$
On replacing this value in the denominator, we get,
$$I_1=\int {\displaystyle \frac{1+{\displaystyle \frac{1}{t^2}}}{{\left(t-{\displaystyle \frac{1}{t}}\right)}^2+2}}dt$$
Let us take $t-{\displaystyle \frac{1}{t}}=u$
Differentiating the above equation with respect to $t$ , we get,
 $\left(1+{\displaystyle \frac{1}{t^2}}\right)dt=du$
On substituting these values in $I_1$ we get,
$$I_1=\int {\displaystyle \frac{du}{u^2+2}}$$
Applying the formula, $\int {\displaystyle \frac{dx}{a^2+x^2}}={\displaystyle \frac{1}{a}}{\tan }^{-1}\left({\displaystyle \frac{x}{a}}\right)$ in $I_1$ :
 $I_1={\displaystyle \frac{1}{\sqrt{2}}}{\tan }^{-1}\left({\displaystyle \frac{u}{\sqrt{2}}}\right)$
Now we replace $u$ with the value of $t$ as we know $u=t-{\displaystyle \frac{1}{t}}$
 $$\begin{gathered} I_1={\displaystyle \frac{1}{\sqrt{2}}}{\tan }^{-1}\left({\displaystyle \frac{t-{\displaystyle \frac{1}{t}}}{\sqrt{2}}}\right) \\ \Rightarrow I_1={\displaystyle \frac{1}{\sqrt{2}}}{\tan }^{-1}\left({\displaystyle \frac{t^2-1}{\sqrt{2}t}}\right) \end{gathered}$$
Now we replace $t$ as we know $t=\sqrt{\cot x}$
 $I_1={\displaystyle \frac{1}{\sqrt{2}}}{\tan }^{-1}\left({\displaystyle \frac{\cot x-1}{\sqrt{2\cot x}}}\right)$

Now, we shall solve $I_2$
 $I_2=\int {\displaystyle \frac{t^2-1}{t^4+1}}dt$
On dividing the numerator and the denominator of the second integral with $t^2$ ,
$$I_2=\int {\displaystyle \frac{1-{\displaystyle \frac{1}{t^2}}}{t^2+{\displaystyle \frac{1}{t^2}}}}dt$$
The denominator can be written as $t^2+{\displaystyle \frac{1}{t^2}}={\left(t+{\displaystyle \frac{1}{t}}\right)}^2-2$
$$I_2=\int {\displaystyle \frac{1-{\displaystyle \frac{1}{t^2}}}{{\left(t+{\displaystyle \frac{1}{t}}\right)}^2-2}}dt$$
Let us take $v=t+{\displaystyle \frac{1}{t}}$
Differentiating the above equation with respect to $t$ , we get,
 $dv=\left(1-{\displaystyle \frac{1}{t^2}}\right)dt$
On substituting these values in $I_2$ we get,
$$I_2=\int {\displaystyle \frac{dv}{v^2-2}}$$
Applying the formula, $\int {\displaystyle \frac{dx}{x^2-a^2}}={\displaystyle \frac{1}{2a}}\ln \left({\displaystyle \frac{x-a}{x+a}}\right)$ in $I_2$ :
$$I_2={\displaystyle \frac{1}{2\sqrt{2}}}\ln \left({\displaystyle \frac{v-\sqrt{2}}{v+\sqrt{2}}}\right)$$
Now we replace $v$ with the value of $t$ as we know $v=t+{\displaystyle \frac{1}{t}}$
$$\begin{gathered} I_2={\displaystyle \frac{1}{2\sqrt{2}}}\ln \left({\displaystyle \frac{t+{\displaystyle \frac{1}{t}}-\sqrt{2}}{t+{\displaystyle \frac{1}{t}}+\sqrt{2}}}\right) \\ \Rightarrow I_2={\displaystyle \frac{1}{2\sqrt{2}}}\ln \left({\displaystyle \frac{t^2+1-\sqrt{2}t}{t^2+1+\sqrt{2}t}}\right) \end{gathered}$$
Now we replace $t$ as we know that $t^2=\cot x$
$$I_2={\displaystyle \frac{1}{2\sqrt{2}}}\ln \left({\displaystyle \frac{\cot x+1-\sqrt{2\cot x}}{\cot x+1+\sqrt{2\cot x}}}\right)$$
Now we have to combine both the integrals, $I=-\left(I_1+I_2\right)$
 $I=-\left[{\displaystyle \frac{1}{\sqrt{2}}}{\tan }^{-1}\left({\displaystyle \frac{\cot x-1}{\sqrt{2\cot x}}}\right)+{\displaystyle \frac{1}{2\sqrt{2}}}\ln \left({\displaystyle \frac{\cot x+1-\sqrt{2\cot x}}{\cot x+1+\sqrt{2\cot x}}}\right)\right]+C$
Therefore, $\int \sqrt{\cot x}dx=-{\displaystyle \frac{1}{\sqrt{2}}}{\tan }^{-1}\left({\displaystyle \frac{\cot x-1}{\sqrt{2\cot x}}}\right)-{\displaystyle \frac{1}{2\sqrt{2}}}\ln \left({\displaystyle \frac{\cot x+1-\sqrt{2\cot x}}{\cot x+1+\sqrt{2\cot x}}}\right)+C$
Where $C$ represents the constant of integration.
So, the correct answer is “ $-{\displaystyle \frac{1}{\sqrt{2}}}{\tan }^{-1}\left({\displaystyle \frac{\cot x-1}{\sqrt{2\cot x}}}\right)-{\displaystyle \frac{1}{2\sqrt{2}}}\ln \left({\displaystyle \frac{\cot x+1-\sqrt{2\cot x}}{\cot x+1+\sqrt{2\cot x}}}\right)+C$ ”.

Note: Integrals are widely used in a variety of fields. Integrals, for example, are used in probability theory to calculate the probability of a random variable falling within a given range. In the substitution method in the final answer do not forget to put the original function back.