Mathematics

Find the integral of the function
$$\sin 4x\sin 8x$$


SOLUTION

Consider the given function.

$$I=\int{\sin 4x\sin 8xdx}$$

$$ I=\int{\sin 4x\sin 2\left( 4x \right)dx} $$

$$ I=\int{\sin 4x\left( 2\sin 4x\cos 4x \right)dx} $$

$$ I=2\int{{{\sin }^{2}}4x\cos 4xdx} $$

 

Let $$t=\sin 4x$$

$$ \dfrac{dt}{dx}=4\cos 4x $$

$$ \dfrac{dt}{4}=\cos 4xdx $$

 

Therefore,

$$ I=\dfrac{2}{4}\int{{{t}^{2}}dt} $$

$$ I=\dfrac{1}{2}\left[ \dfrac{{{t}^{3}}}{3} \right]+C $$

$$ I=\left[ \dfrac{{{t}^{3}}}{6} \right]+C $$

 

On putting the value of $$t$$, we get

$$I=\dfrac{{{\sin }^{3}}4x}{6}+C$$

 

Hence, this is the answer.

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