Question

Find : $\int \frac{\sin \theta d\theta }{(4+{\cos }^2\theta )(2-{\sin }^2\theta )}$

Answer 1

Let $I=\int \frac{\sin \theta d\theta }{(4+{\cos }^2\theta )(2-(1-{\cos }^2\theta ))}$
= $\int \frac{\sin \theta d\theta }{(4+{\cos }^2\theta )(1+{\cos }^2\theta )}$
= $\int \frac{\sin \theta d\theta }{(4+{\cos }^2\theta )(1+{\cos }^2\theta )}$
Put $\cos \theta$
$\Rightarrow -\sin \theta d\theta =dt\Rightarrow \sin \theta d\theta =-dt$
$\Rightarrow I=-\int \frac{dt}{(4+t^2)(1+t^2)}$
For Partial Fraction, we have Let
$\frac{1}{(4+t^2)(1+t^2)}=\frac{At+B}{4+t^2}+\frac{Ct+D}{1+t^2}$
$1=(At+B)(1+t^2)+(Ct+D)(4+t^2)$
$1=(B+4D)+(A+4C)t+(B+D)t^2+(A+C)t^3$

$\text{Put t = 0, we have B + 4D = 1 ...(i)}$
$\text{Equating the coeff. of t on both sides, we have A + 4 C = 0...(ii)}$
$\text{Equating the coeff. of}{t}^2\text{and}{t}^3\text{respectively,}$
$\text{we obtain B+D = 0...(iii) and A + C = 0 ...(iv)}$
$\text{Solving (i), (ii),(iii) and (iv), we obtain}$
$\text{A = C = 0, B =}-\frac{1}{3}andD=\frac{1}{3}$
$\therefore \frac{1}{(4+t^2)(1+t^2)}=\frac{-\frac{1}{3}}{4+t^2}+\frac{\frac{1}{3}}{1+t^2}$
$I=-[\frac{-1}{3}\int \frac{1}{4+t^2}dt+\frac{1}{3}\int \frac{1}{1+t^2}dt]$
$=-[\frac{-1}{6}{\tan }^{-1}\frac{t}{2}+\frac{1}{3}{\tan }^{-1}t]+C$
$=\frac{1}{6}{\tan }^{-1}\left(\frac{\cos \theta }{2}\right)-\frac{1}{3}{\tan }^{-1}\left(\cos \theta \right)$ + C