$\int \frac{sin 2 x}{(1 + sin x) (2 + sin x)} d x = - 2 log | 1 + sin x | + m log | 2 + sin x | + c$ .Find $m$

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Solution

$\int \frac{sin 2 x}{(1 + sin x) (2 + sin x)} d x = - 2 l n | 1 + sin x | + m l n | 2 + sin x | + c$ find m
$\int \frac{2 sin x cos x}{(1 + sin x) (2 + sin x)} d x$
put $2 sin x = t$
$cos x d x = d t .$
$\int \frac{2 (t - 2)}{(t - 1) t} d t$
$2 \int \frac{t - 1 - 1}{t (t - 1)} d t$
$2 \int \frac{t - 1}{t (t - 1)} d t - 2 \int \frac{1}{t (t - 1)} d t$
$2 \int \frac{1}{t} d t - 2 \int \frac{t - (t - 1)}{t (t - 1)} d t$
$2 l n | t | - 2 \int \frac{t}{t (t - 1)} d t + 2 \int \frac{t - 1}{t (t - 1)} d t$
$2 l n | t | - 2 l n | t - 1 | + 2 l n | t | + c$
$4 l n | t | - 2 l n | t - 1 | + c$
$4 l n | 2 + sin x | - 2 l n | 1 + sin x | + c = 2 l n | 1 + sin x | + m l n | 2 + sin x | + c$
Companying L.H.S, R.H.S.
$m = 4$

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