Mathematics

Solve:
$$\displaystyle \int{\sin 3x\cos 4x dx}$$


SOLUTION

We have,

$$\int{\sin 3x\cos 4x dx}$$

On multiply and divide by 2 and we get,

$$ \int{\dfrac{2}{2}\sin 3x\cos 4x}dx $$

$$ \Rightarrow \dfrac{1}{2}\int{2\sin 3x\cos 4x dx} $$


Now using formula,

$$2\sin A\cos B=\sin \left( A+B \right)+\sin \left( A-B \right)$$


So,

$$ \Rightarrow \dfrac{1}{2}\int{\left[ \sin \left( 3x+4x \right)+\sin \left( 3x-4x \right) \right]}dx $$

$$ \Rightarrow \dfrac{1}{2}\int{\left[ \sin 7x-\sin x \right]}dx $$

$$ \Rightarrow \dfrac{1}{2}\int{\sin 7xdx-\dfrac{1}{2}\int{\sin xdx}} $$

$$ \Rightarrow \dfrac{1}{2}\left( -\dfrac{\cos 7x}{7} \right)-\dfrac{1}{2}\left( -\cos x \right)+C $$

$$ \Rightarrow -\dfrac{\cos 7x}{14}+\dfrac{1}{2}\cos x+C $$


Hence, this is the answer.

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Subjective Medium Published on 17th 09, 2020
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