Question

# Integrate the following integral:$\int {\sin mx\cos nx dx} ,m \ne n$.

Hint:Break the given integral into two parts using trigonometric identities

We have the given integral as
$\int {\sin mx\cos nxdx}$
Now, divide and multiply by 2,
$= \dfrac{1}{2}\int {2\sin mx\cos nxdx}$
We know the identity
$2\sin A\cos B = \sin (A + B) + \sin (A - B)$
Using this identity, such that $A = mx$and $B = nx$ we get,
$= \dfrac{1}{2}\int {\sin (mx + nx) + \sin (mx - nx)} dx$
$= \dfrac{1}{2}\int {\sin (m + n)x + \sin (m - n)x} dx$
We know that $\int {\sin ax} dx = \dfrac{{ - \cos ax}}{a}$
Therefore, we get,
$= \dfrac{1}{2}\left( {\dfrac{{ - \cos (m + n)}}{{m + n}} - \dfrac{{\cos (m - n)}}{{m - n}}} \right) + c$
Where $c$ is an integration constant.

Note: To solve these types of questions, we must have an adequate knowledge of various integration properties and identities, evaluating the integral within such parameters, will lead towards the required solution.