# ∫ Cot X + Cot 3 X 1 + Cot 3 X Dx - Mathematics

Sum
$\int\frac{\cot x + \cot^3 x}{1 + \cot^3 x} \text{ dx}$

#### Solution

$\text{We have},$
$I = \int\left( \frac{\cot x + \cot^3 x}{1 + \cot^3 x} \right) dx$
$= \int\left[ \frac{\cot x \left( 1 + \cot^2 x \right)}{1 + \cot^3 x} \right]dx$
$= \int\left( \frac{\cot x {cosec}^2 x}{1 + \cot^3 x} \right) dx$
$\text{Putting} \cot x = t$
$\Rightarrow - \text{ cosec}^2 x\ dx = dt$
$\Rightarrow \text{cosec}^2 x\ dx = - dt$
$\therefore I = - \int\frac{\text{ t dt}}{1 + t^3}$
$= - \int\frac{\text{ t dt}}{\left( 1 + t \right) \left( t^2 - t + 1 \right)}$
$\text{ Let } \frac{t}{\left( 1 + t \right) \left( t^2 - t + 1 \right)} = \frac{A}{t + 1} + \frac{Bt + C}{t^2 - t + 1}$
$\Rightarrow \frac{t}{\left( 1 + t \right) \left( t^2 - t + 1 \right)} = \frac{A \left( t^2 - t + 1 \right) + \left( Bt + C \right) \left( t + 1 \right)}{\left( t + 1 \right) \left( t^2 - t + 1 \right)}$
$\Rightarrow t = A \left( t^2 - t + 1 \right) + B t^2 + Bt + Ct + C$
$\Rightarrow t = \left( A + B \right) t^2 + \left( B + C - A \right) t + A + C$
$\text{Equating Coefficients of like terms}$
$A + B = 0 . . . . . \left( 1 \right)$
$B + C - A = 1 . . . . . \left( 2 \right)$
$A + C = 0 . . . . . \left( 3 \right)$
$\text{Solving} \left( 1 \right), \left( 2 \right) \text{and} \left( 3 \right), \text{we get}$
$A = - \frac{1}{3}$
$B = \frac{1}{3}$
$C = \frac{1}{3}$
$\therefore \frac{t}{\left( 1 + t \right) \left( t^2 - t + 1 \right)} = - \frac{1}{3 \left( t + 1 \right)} + \frac{1}{3} \left( \frac{t + 1}{t^2 - t + 1} \right)$
$\Rightarrow \frac{t}{\left( 1 + t \right) \left( t^2 - t + 1 \right)} = - \frac{1}{3 \left( t + 1 \right)} + \frac{1}{6} \left[ \frac{2t + 2}{t^2 - t + 1} \right]$
$\Rightarrow \frac{t}{\left( 1 + t \right) \left( t^2 - t + 1 \right)} = - \frac{1}{3 \left( t + 1 \right)} + \frac{1}{6} \left[ \frac{2t - 1 + 3}{t^2 - t + 1} \right]$
$\therefore I = - \left[ - \frac{1}{3}\int\frac{dt}{t + 1} + \frac{1}{6}\int\left( \frac{2t - 1}{t^2 - t + 1} \right) dt + \frac{1}{2}\int\frac{dt}{t^2 - t + 1} \right]$
$= + \frac{1}{3}\int\frac{dt}{t + 1} - \frac{1}{6}\int\left( \frac{2t - 1}{t^2 - t + 1} \right) dt - \frac{1}{2}\int\frac{dt}{t^2 - t + \frac{1}{4} - \frac{1}{4} + 1}$
$= \frac{1}{3}\int\frac{dt}{t + 1} - \frac{1}{6}\int\frac{\left( 2t - 1 \right) dt}{\left( t^2 - t + 1 \right)} - \frac{1}{2}\int\frac{dt}{\left( t - \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}$
$\text{ let t}^2 - t + 1 = p$
$\Rightarrow \left( 2t - 1 \right) dt = dp$
$\therefore I = \frac{1}{3}\int\frac{dt}{t + 1} - \frac{1}{6}\int\frac{dp}{p} - \frac{1}{2}\int\frac{dt}{\left( t - \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}$
$= \frac{1}{3} \text{ log} \left| t + 1 \right| - \frac{1}{6} \text{ log} \left| p \right| - \frac{1}{2} \times \frac{2}{\sqrt{3}} \text{ tan}^{- 1} \left( \frac{t - \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) + C$
$= \frac{1}{3} \text{ log }\left| t + 1 \right| - \frac{1}{6} \text{ log }\left| p \right| - \frac{1}{\sqrt{3}} \text{ tan}^{- 1} \left( \frac{2t - 1}{\sqrt{3}} \right) + C$
$= \frac{1}{3} \text{ log }\left| \cot x + 1 \right| - \frac{1}{6} \text{ log }\left| \cot^2 x - \cot x + 1 \right| - \frac{1}{\sqrt{3}} \text{ tan}^{- 1} \left( \frac{\text{ 2 cot x} - 1}{\sqrt{3}} \right) + C$

Concept: Indefinite Integral Problems
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Revision Excercise | Q 130 | Page 205