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Solve:
 $ \int {\sqrt {\cot x} dx} $

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Last updated date: 03rd Mar 2024
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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,
 $
   - \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.
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