Find (4+ ^2 ) (5-4 ^2 ) d .

Question

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

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Answer

According to the question, $I=\int \frac{\cos \theta}{\left.\left(4+\sin ^2 \theta\right)\left(5-4 \cos ^2 \theta\right)\right)} d \theta$
$=\int \frac{\cos \theta}{\left(4+\sin ^2 \theta\right)\left(5-4\left(1-\sin ^2\theta\right)\right)} d \theta\left(\therefore \cos ^2 \theta=1-\sin ^2 \theta\right)$
$=\int \frac{\cos \theta}{\left(4+\sin ^2 \theta\right)\left(5-4+4 \sin ^2 \theta\right)} d \theta$
$=\int \frac{\cos \theta}{\left(4+\sin ^2 \theta\right)\left(1+4 \sin ^2 \theta\right)} d \theta$
Let $\sin \theta=t $
$\Rightarrow \cos \theta \ d \ \theta=d t$
Then, $I=\int \frac{d t}{\left(4+t^2\right)\left(1+4 t^2\right)}$
let, $\frac{1}{\left(4+t^2\right)\left(1+4 t^2\right)}=\frac{A}{4+t^2}+\frac{B}{1+4 t^2}$
using partial fractions
At $t =0, \frac{A}{4}+\frac{B}{1}=\frac{1}{4 \times 1} $
$\Rightarrow A+4 B=1$
At $t=1, \frac{A}{5}+\frac{B}{5}=\frac{1}{5 \times 5}$
$ \Rightarrow 5 A+5 B=1$
On solving Equations $(i)$ and $(ii),$ we get
$A=\frac{-1}{15} $ and $ B=\frac{4}{15}$
$\frac{1}{\left(4+t^2\right)\left(1+4 t^2\right)}=\frac{-\frac{1}{15}}{4+t^2}+\frac{\frac{4}{15}}{1+4 t^2}$
$\Rightarrow \frac{1}{\left(4+t^2\right)\left(+4 t^2\right)}$
$=\frac{-1}{15\left(4+t^2\right)}+\frac{4}{15\left(1+4 t^2\right)}$
Integrating both sides $\text{w.r.t.} \ t,$
$\Rightarrow \int \frac{1}{\left(4+t^2\right)\left(1+4 t^2\right)} d t=\frac{-1}{15} \int \frac{1}{4+t^2} d t+\frac{4}{15} \int \frac{1}{1+4 t^2} d t$
$=\frac{-1}{15} \int \frac{1}{2^2+t^2}+\frac{4}{15 \times 4} \int \frac{1}{\left(\frac{1}{2}\right)^2+t^2} d t$
$=\frac{-1}{15} \cdot \frac{1}{2} \tan ^{-1} \frac{t}{2}+\frac{1}{15} \cdot \frac{1}{1 / 2} \tan ^{-1} \frac{t}{1 / 2}+C\left[\because \int \frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c\right]$
put $ t=\sin \theta$
$=\frac{-1}{30} \tan ^{-1} \frac{\sin \theta}{2}+\frac{2}{15} \tan ^{-1} 2 \sin \theta+C$