Mathematics

Solve $$\int {\cos 4x\cos xdx} $$


SOLUTION

Consider the given integral.

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

 

We know that

$$\cos A\cos B=\dfrac{1}{2}\left[ \cos \left( A+B \right)+\cos \left( A-B \right) \right]$$

 

Therefore,

$$ I=\int{\dfrac{1}{2}\left[ \cos \left( 4x+x \right)+\cos \left( 4x-x \right) \right]dx} $$

$$ I=\dfrac{1}{2}\int{\left[ \cos \left( 5x \right)+\cos \left( 3x \right) \right]dx} $$

$$ I=\dfrac{1}{2}\left[ \dfrac{\sin 5x}{5}+\dfrac{\sin 3x}{3} \right]+C $$

$$ I=\dfrac{\sin 5x}{10}+\dfrac{\sin 3x}{6}+C $$

 

Hence, this is the answer.

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Subjective Medium Published on 17th 09, 2020
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