# ∫ √ Cot θ D θ - Mathematics

Sum
$\int\sqrt{\cot \text{θ} d } \text{ θ}$

#### Solution

$\text{ We have,}$

$I = \int\sqrt{\cot \theta} d \text{ θ}$

$\text{ Putting cot} \text{ θ} = t^2$

$\Rightarrow - {cosec}^2 \text{ θ dθ}= 2t \text{ dt }$

$\Rightarrow d\theta = - \frac{2t \text{ dt }}{\cos e c^2 \text{ θ }}$

$\Rightarrow d\theta = \frac{- 2t \text{ dt}}{1 + co t^2 \text{ θ}}$

$\Rightarrow d\theta = \frac{- 2t \text{ dt}}{1 + t^4}$

$\therefore I = \int t\left( \frac{- 2t \text{ dt }}{1 + t^4} \right)$

$= - \int\left( \frac{2 t^2}{1 + t^4} \right)dt$

$= - \int\left( \frac{t^2 + 1 + t^2 - 1}{t^4 + 1} \right)dt$

$= - \int\left( \frac{t^2 + 1}{t^4 + 1} \right)dt - \int\frac{\left( t^2 - 1 \right)dt}{t^4 + 1}$

 \text{Dividing numerator and denominator by} \text{  t}^2

$I = - \int\left( \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} \right)dt - \int\left( \frac{1 - \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} \right)dt$

$= - \int\frac{\left( 1 + \frac{1}{t^2} \right)dt}{t^2 + \frac{1}{t^2} - 2 + 2} - \int\frac{\left( 1 - \frac{1}{t^2} \right)dt}{t^2 + \frac{1}{t^2} + 2 - 2}$

$= - \int\frac{\left( 1 + \frac{1}{t^2} \right)dt}{\left( t - \frac{1}{t} \right)^2 + \left( \sqrt{2} \right)^2} - \int\frac{\left( 1 - \frac{1}{t^2} \right)dt}{\left( t + \frac{1}{t} \right)^2 - \left( \sqrt{2} \right)^2}$

$\text{ Putting t} - \frac{1}{t} = p$

$\Rightarrow \left( 1 + \frac{1}{t^2} \right)dt = dp$

$\text{ Putting}\ t + \frac{1}{t} = q$

$\Rightarrow \left( 1 - \frac{1}{t^2} \right)dt = dq$

$I = - \int \frac{dp}{p^2 + \left( \sqrt{2} \right)^2} - \int\frac{dq}{q^2 - \left( \sqrt{2} \right)^2}$

$= - \frac{1}{\sqrt{2}} \tan^{- 1} \left( \frac{p}{\sqrt{2}} \right) - \frac{1}{2\sqrt{2}}\text{ log} \left| \frac{q - \sqrt{2}}{q + \sqrt{2}} \right| + C$

$= - \frac{1}{\sqrt{2}} \tan^{- 1} \left( \frac{t - \frac{1}{t}}{\sqrt{2}} \right) - \frac{1}{2\sqrt{2}}\text{ log }\left| \frac{t + \frac{1}{t} - \sqrt{2}}{1 + \frac{1}{t} + \sqrt{2}} \right| + C$

$= - \frac{1}{\sqrt{2}} \text{ tan}^{- 1} \left( \frac{t^2 - 1}{\sqrt{2} t} \right) - \frac{1}{2\sqrt{2}}\text{ log} \left| \frac{t^2 + 1 - \sqrt{2}t}{t^2 + 1 + \sqrt{2}t} \right| + C$

$= - \frac{1}{\sqrt{2}} \tan^{- 1} \left( \frac{\cot \theta - 1}{2\sqrt{\cot \theta}} \right) - \frac{1}{2\sqrt{2}}\text{ log } \left| \frac{\cot \theta + 1 - \sqrt{2 \cot \theta}}{\cot \theta + 1 + \sqrt{2 \cot \theta}} \right| + C$

Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Q 2 | Page 190