Question

Integral of sin 4x cos3x dx

Answer 1

Answer

The required integral is (1/14)cos7x + (1/2)cosx

Solution

We know that ,

$\sf\bullet \: \: 2\sin A\cos B = \sin(A+B) + sin(A-B)\\\\ \sf\implies \sin A\cos B = \dfrac{1}{2} \{ \sin(A+B) + \cos(A-B) \}$

Thus , we will have :

$\sf\implies \sin4x\cos3x = \dfrac{1}{2} \{ \sin(4x + 3x) + \sin(4x -3x) \} \\\\ \sf\implies \sin4x\cos3x = \dfrac{1}{2}\sin7x + \dfrac{1}{2}\sin x$

Thus ,on integrating we have :

$\sf\implies \int \sin 4x \cos3x = \int\dfrac{1}{2}\sin 7x dx +\dfrac{1}{2} \int \sin x dx \\\\ \sf\implies \int \sin 4x \cos3x = \dfrac{1}{14}\cos7x + \dfrac{1}{2} \cos x$