Mathematics

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


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

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

$$=\dfrac{1}{2}\displaystyle\int{\left(\sin{7x}+\sin{x}\right)dx}$$ ........  using  $$2\sin{A}\cos{B}=\sin{\left(A+B\right)}+\sin{\left(A-B\right)}$$

$$=\dfrac{1}{2}\left[\dfrac{-\cos{7x}}{7}-\cos{x}\right]+c$$

$$=-\dfrac{1}{14}\cos{7x}-\dfrac{1}{2}\cos{x}+c$$
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Subjective Medium Published on 17th 09, 2020
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