# Integrate the following integral:

\[\int {\sin mx\cos nx dx} ,m \ne n\].

Answer

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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.

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.

Last updated date: 20th Sep 2023

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