Evaluate : 1 ( )(3+2 ) d - Vidyadip

Question

Evaluate : $\int \frac{1}{(\sin \theta)(3+2 \cos \theta)} \cdot d \theta$

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Answer

$
\begin{aligned}
& \mathrm{I}=\int \frac{1}{(\sin \theta)(3+2 \cos \theta)} \cdot d \theta=\int \frac{\sin \theta}{\left(1-\cos ^2 \theta\right)(3+2 \cos \theta)} \cdot d \theta \\
& =\int \frac{\sin \theta}{(1-\cos \theta)(1+\cos \theta)(3+2 \cos \theta)} \cdot d \theta \\
& \text { put } \cos \theta=t \quad \therefore \quad-\sin \theta \cdot d \theta=1 \cdot d t \\
& \therefore \sin \theta \cdot d \theta=-1 \cdot d t \\
&
\end{aligned}
$
$
\text { Consider, } \begin{aligned}
\frac{-1}{(1-t)(1+t)(3+2 t)} & =\frac{A}{(1-t)}+\frac{B}{(1+t)}+\frac{C}{(3+2 t)} \\
& =\frac{A(1+t)(3+2 t)+B(1-t)(3+2 t)+C(1-t)(1+t)}{(1-t)(1+t)(3+2 t)}
\end{aligned}
$
$
\therefore \quad-1=A(1+t)(3+2 t)+B(1-t)(3+2 t)+C(1-t)(1+t)
$
at $t=1$,
$
\begin{aligned}
& -1=A(2)(5)+B(0)+C(0) \\
& -1=10 A \Rightarrow A=-\frac{1}{10}
\end{aligned}
$
at $t=-1$,
$
\begin{aligned}
& -1=A(0)+B(2)(1)+C(0) \\
& -1=2 B \quad \Rightarrow \quad B=-\frac{1}{2}
\end{aligned}
$
at $t=-\frac{3}{2}, \quad-1=A(0)+B(0)+C\left(+\frac{5}{2}\right)\left(-\frac{1}{2}\right)$
$
-1=-\frac{5}{4} C \quad \Rightarrow C=\frac{4}{5}
$
Thus,
$
\frac{-1}{(1-t)(1+t)(3+2 t)}=\frac{\left(-\frac{1}{10}\right)}{(1-t)}+\frac{\left(-\frac{1}{2}\right)}{(1+t)}+\frac{\left(\frac{4}{5}\right)}{(3+2 t)}
$
$
\begin{aligned}
\therefore \mathrm{I} & =\int\left[\frac{\left(-\frac{1}{10}\right)}{(1-t)}+\frac{\left(-\frac{1}{2}\right)}{(1+t)}+\frac{\left(\frac{4}{5}\right)}{(3+2 t)}\right] \cdot d t \\
& =-\frac{1}{10} \log (1-t) \cdot \frac{1}{(-1)}-\frac{1}{2} \log (1+t)+\frac{4}{5} \log (3+2 t) \cdot \frac{1}{2}+c \\
& =\frac{1}{10} \log (1-\cos \theta)-\frac{1}{2} \log (1+\cos \theta)+\frac{4}{10} \log (3+2 \cos \theta)+c \\
& =\frac{1}{10}\left(\log \frac{(1-\cos \theta)(3+2 \cos \theta)^4}{(1+\cos \theta)^5}\right)+c \quad \because \log a^m=m \cdot \log a
\end{aligned}
$

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