Question

Find : $\int \frac{\cos \theta }{(4+{\sin }^2\theta )(5-4{\cos }^2\theta )}d\theta$

Answer 1

Let $I=\int \frac{\cos \theta }{(4+{\sin }^2\theta )(5-4{\cos }^2\theta )}d\theta$
$=\int \frac{\cos \theta d\theta }{(4+{\sin }^2\theta )(5-4(1-{\sin }^2\theta ))}$
$=\int \frac{\cos \theta d\theta }{(4+{\sin }^2\theta )(1+4{\sin }^2\theta )}$
Put $\sin \theta =t\Rightarrow \cos \theta d\theta =dt$
$I=\int \frac{dt}{(4+t^2)(1+4t^2)}$
For partial fraction,
$\frac{1}{(4+t^2)(1+4t^2)}=\frac{At+B}{4+t^2}+\frac{Ct+D}{1+4t^2}$
$1=(At+B)(1+4t^2)+(Ct+D)(4+t^2)$
Expanding the equation
$1=(B+4D)+(A+4C)t+(4B+D)t^2+(4A+C)t^3$

Put $t=0$, we have $B+4D=1$ ...(i)
Equating the coefficient of $t$ on both sides, we have

$A+4C=0$ ...(ii)

Equating the coeff of $t^2$ and $t^3$ respectively, we obtain

$4B+D=0$ ...(iii)

and $4A+C=0$ ...(iv)

Solving $\left(i\right),\left(ii\right),\left(iii\right)\text{and}\left(iv\right)$, we obtain

$A=C=0\text{and}B=\frac{-1}{15}\text{and}D=\frac{4}{15}\frac{1}{(4+t^2)(1+4t^2)}=\frac{\frac{-1}{15}}{4+t^2}+\frac{\frac{4}{15}}{1+4t^2}=\frac{-1}{15}\times \frac{1}{4+t^2}+\frac{4}{15}\times \frac{1}{4}\frac{1}{(\frac{1}{4}+t^2)}\therefore I=\frac{-1}{15}\int \frac{1}{{\left(2\right)}^2+t^2}dt+\frac{1}{15}\int \frac{1}{{\left(\frac{1}{2}\right)}^2+t^2}dt=\frac{-1}{15}\times \frac{1}{2}{\tan }^{-1}\frac{t}{2}+\frac{1}{15}\times 2{\tan }^{-1}2t+C$

$=\frac{-1}{30}{\tan }^{-1}\left(\frac{\sin \theta }{2}\right)+\frac{2}{15}{\tan }^{-1}2\left(\sin \theta \right)$ + $C$