Question

Find:$\displaystyle \int \dfrac{dx}{\sin x + \sin 2x}$

Answer 1

We are given:

$\displaystyle \int \dfrac{dx}{\sin x + \sin 2x}$

Step 1: Use the identity:

$\sin 2x = 2 \sin x \cos x$

So the denominator becomes:

$\sin x + \sin 2x = \sin x + 2 \sin x \cos x = \sin x (1 + 2 \cos x)$

Now the integral becomes:

$\displaystyle \int \dfrac{dx}{\sin x (1 + 2 \cos x)}$

Step 2: Use substitution:

Let $u = \cos x \Rightarrow du = -\sin x \, dx \Rightarrow -du = \sin x \, dx$

Substitute into the integral:

$\displaystyle \int \dfrac{dx}{\sin x (1 + 2 \cos x)} = \int \dfrac{-du}{(1 + 2u)}$

Step 3: Integrate:

$\displaystyle \int \dfrac{-du}{1 + 2u} = -\int \dfrac{du}{1 + 2u}$

Use the substitution $v = 1 + 2u \Rightarrow dv = 2 \, du \Rightarrow du = \dfrac{dv}{2}$

$\displaystyle -\int \dfrac{1}{v} \cdot \dfrac{dv}{2} = -\dfrac{1}{2} \int \dfrac{dv}{v} = -\dfrac{1}{2} \ln |v| + C$

Back-substitute:

$v = 1 + 2u = 1 + 2 \cos x$

Final Answer:

$\boxed{ -\dfrac{1}{2} \ln |1 + 2 \cos x| + C }$