# What is the integral of int sin^4 (x) dx?

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.