Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Solve:  $\int {\sqrt {\cot x} dx}$

Last updated date: 21st Jul 2024
Total views: 350.4k
Views today: 4.50k
Verified
350.4k+ views
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.

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,
$- \cos e{c^2}x.dx = 2t.dt \\ \Rightarrow - \cos e{c^2}x = 2t\dfrac{{dt}}{{dx}} \\ \Rightarrow \cos e{c^2}x = - 2t\dfrac{{dt}}{{dx}} \\$
Using the formula, $1 + {\cot ^2}x = \cos e{c^2}x$ , in the above equation we get,
$1 + {\cot ^2}x = - 2t.\dfrac{{dt}}{{dx}}$
Since, $\cot x = {t^2}$ , on substituting in the above equation, we get,
$1 + {t^4} = - 2t.\dfrac{{dt}}{{dx}}$
On rearranging the terms, the equation would be,
$dx = \dfrac{{ - 2t}}{{1 + {t^4}}}dt$
Substituting the values of $\sqrt {\cot x}$ and $dx$ in $\int {\sqrt {\cot x} dx}$ , we get,
$I = \int {t.\dfrac{{ - 2t}}{{1 + {t^4}}}dt} \\ \Rightarrow I = \int {\dfrac{{ - 2{t^2}}}{{1 + {t^4}}}dt} \\$
We can take the negative sign out of the integral,
$I = - \int {\dfrac{{2{t^2}}}{{1 + {t^4}}}dt}$
To solve this integral, we add and subtract the numerator with $1$ :
$I = - \int {\dfrac{{2{t^2} + 1 - 1}}{{{t^4} + 1}}d} t \\ = - \int {\dfrac{{{t^2} + 1 + {t^2} - 1}}{{{t^4} + 1}}d} t \\ = - \int {\left( {\dfrac{{{t^2} + 1}}{{{t^4} + 1}} + \dfrac{{{t^2} - 1}}{{{t^4} + 1}}} \right)d} t \\$
Now, we split the integral into two different integrals and solve them separately.
The two integrals are:
${I_1} = \int {\dfrac{{{t^2} + 1}}{{{t^4} + 1}}} dt$ and ${I_2} = \int {\dfrac{{{t^2} - 1}}{{{t^4} + 1}}} dt$

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

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

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.