# Evaluate the following integral: \int \sin 4x \cos 5x \ dx

## Question:

Evaluate the following integral: {eq}\int \sin 4x \cos 5x \ dx {/eq}

## Indefinite Integral:

Whenever we come across the trigonometric function that is to be integrated, then the best way to integrate is by first simplifying the integrand and then applying the integration formulas that are available.

Here the indefinite integral is:

{eq}\int \sin 4x \cos 5x \ dx\\ {/eq}

So here we will simplify the integrand using the identity:

{eq}\cos \left(t\right)\sin \left(s\right)=\frac{\sin \left(s+t\right)+\sin \left(s-t\right)}{2}\\ {/eq}

So this will give us:

{eq}\displaystyle \int \sin \left(4x\right)\cos \left(5x\right)dx=\int \frac{\sin \left(4x+5x\right)+\sin \left(4x-5x\right)}{2}dx\\ \displaystyle =\frac{1}{2}\left(\int \sin \left(4x+5x\right)dx+\int \sin \left(4x-5x\right)dx\right)\\ \displaystyle =\frac{1}{2}\left(\int \sin \left(9x\right)dx+\int \sin \left(-x\right)dx\right)\\ \displaystyle =\frac{1}{2}\left(-\frac{1}{9}\cos \left(9x\right)+\cos \left(x\right)\right)+C~~~~~~~~~~~~~\left [ \because \int \sin \left(u\right)du=-\cos \left(u\right)+c, \right ] {/eq}