To integrate the trigonometric function we will use the trigonometric identity:
\(\sin x \cos y = \frac{1}{2} [\sin (x+y) + \sin (x-y)]\)Form this identity\(\sin 4x \cos 3x = \frac{1}{2} (\sin 7x + \sin x)\)
Therefore, \(\int (\sin4x \cos3x)dx = \int \frac{1}{2} (\sin7x + \sin x)dx\)
From the above equation we have:
\(\int \frac{1}{2}(\sin7x + \sin x)dx = \frac{1}{2} \int (\sin7x + \sin x)dx\) …………(ii)According to the properties of integration, the integral of the sum of two functions is equal to the sum of integrals of the given functions, i.e.,
\(\int [f(x)+g(x)]dx = \int f(x).dx + \int g(x).dx\)Therefore equation 2 can be rewritten as:
\(\frac{1}{2} \int (\sin7x ) + \frac{1}{2}\int (\sin x)dx\)= \(\frac{-\cos 7x}{14} + \frac{-\cos x}{2} + C\)
This gives us the required integration of the given function.
