$\int \frac{sin 2 x}{sin 5 x sin 3 x} d x$ .

Open in App

Solution

Given : $\int \frac{sin 2 x d x}{sin 5 x sin 3 x}$
$\int \frac{sin 2 x d x}{sin 5 x sin 3 x} = \int \frac{(sin 5 x cos 3 x - cos 5 x sin 3 x)}{sin 5 x sin 3 x} d x = \int (cot 3 x - cot 5 x) d x = \frac{1}{3} log | sin 3 x | - \frac{1}{5} log | sin 5 x | + C$
Hence the correct answer is $\frac{1}{3} log | sin 3 x | - \frac{1}{5} log | sin 5 x | + C$

Suggest Corrections

Similar questions

\int \frac{s i n 2 x}{s i n 5 x s i n 3 x} d x

\int \frac{sin 2 x}{sin 5 x sin 3 x} d x

\int \frac{sin 2 x}{sin 5 x sin 3 x} d x

\int \frac{sin 2 x}{sin 5 x sin 3 x} d x = \frac{1}{3} log | sin 3 x | - f (x)

f (x) =

\int \frac{sin 2 x}{sin 5 x sin 3 x} d x = \frac{1}{3} log | sin 3 x | - \frac{1}{5} log | f (x) | + C

f (x) =

Join BYJU'S Learning Program