Evaluate : 1 2 x+ 2 x d x

Question

Evaluate : $\int \frac{1}{2 \cos x+\sin 2 x} \cdot d x$

✓

Answer

$
\begin{aligned}
& \mathrm{I}=\int \frac{1}{2 \cos x+\sin 2 x} \cdot d x \quad=\int \frac{1}{2 \cos x+2 \sin x \cdot \cos x} \cdot d x=\int \frac{1}{2(\cos x)(1+\sin x)} \cdot d x \\
& =\frac{1}{2} \cdot \int \frac{\cos x}{\cos ^2 x(1+\sin x)} \cdot d x=\frac{1}{2} \cdot \int \frac{\cos x}{\left(1-\sin ^2 x\right)(1+\sin x)} \cdot d x \\
&
\end{aligned}
$
put $\sin x=t$
$
\therefore \quad \cos x \cdot d x=1 \cdot d t
$
$
=\frac{1}{2} \cdot \int \frac{1}{\left(1-t^2\right)(1+t)} \cdot d t=\frac{1}{2} \cdot \int \frac{1}{(1-t)(1+t)(1+t)} \cdot d t=\frac{1}{2} \cdot \int \frac{1}{(1-t)(1+t)^2} \cdot d t
$
Consider, $\quad \frac{1}{(1-t)(1+t)^2}=\frac{A}{(1-t)}+\frac{B}{(1+t)}+\frac{C}{(1+t)^2}=\frac{A(1+t)^2+B(1-t)(1+t)+C(1-t)}{(1-t)(1+t)^2}$
$
\therefore \quad 1=A(1+t)^2+B(1-t)(1+t)+C(1-t)
$
at $t=1$,
$1=A(2)^2+B(0)+C(0)$
$1=4 A \quad \Rightarrow A=\frac{1}{4}$
at $t=-1, \quad 1=A(0)+B(0)+C(2)$
$1=2 C \quad \Rightarrow C=\frac{1}{2}$
at $t=0, \quad 1=A(1)^2+B(1)(1)+C(1)$
$1=A+B+C$
$1=\frac{1}{4}+B+\frac{1}{2} \Rightarrow B=\frac{1}{4}$
Thus, $\quad \frac{1}{(1-t)(1+t)^2}=\frac{\left(\frac{1}{4}\right)}{(1-t)}+\frac{\left(\frac{1}{4}\right)}{(1+t)}+\frac{\left(\frac{1}{2}\right)}{(1+t)^2}$
$\therefore \mathrm{I}=\int\left[\frac{\left(\frac{1}{4}\right)}{(1-t)}+\frac{\left(\frac{1}{4}\right)}{(1+t)}+\frac{\left(\frac{1}{2}\right)}{(1+t)^2}\right] \cdot d t=\frac{1}{2}\left[\frac{1}{4} \log (1-t) \cdot \frac{1}{(-1)}+\frac{1}{4} \log (1+t)+\frac{1}{2} \cdot \frac{(-1)}{(1+t)}\right]+c$
$=\frac{1}{2}\left[\frac{1}{4} \log (1-t) \cdot \frac{1}{(-1)}+\frac{1}{4} \log (1+t)+\frac{1}{2} \cdot \frac{-1}{1+t}\right]+c$
$=\frac{1}{8}\left[-\log (1-\sin x)+\log (1+\sin x)-\frac{2}{1+\sin x}\right]+c=\frac{1}{8}\left[\log \left(\frac{1+\sin x}{1-\sin x}\right)-\frac{2}{1+\sin x}\right]+c$
$\therefore \quad \int \frac{1}{2 \cos x+\sin 2 x} \cdot d x=\frac{1}{8}\left[\log \left(\frac{1+\sin x}{1-\sin x}\right)-\frac{2}{1+\sin x}\right]+c$