Question

Find the value of $\int {\sin 3x.\cos 4x} dx$.

Answer 1

This is a question of indefinite integration. Here we have to integrate $\sin 3x\cos 4x$ with respect to $dx$. Our basic aim is to write the term $\sin 3x\cos 4x$ as the sum of $\sin nx$ and $\cos nx$, so that its integration becomes simple. We have studied a trigonometric formula $\sin \left( {A + B} \right) + \sin \left( {A - B} \right) = 2\sin A\cos B$. Using this formula convert $\sin 3x\cos 4x$ into addition of two terms then integrate the individual terms with respect to $dx$ and then add them to get the required result.

Complete step-by-step answer:
Given: find the value of $\int {\sin 3x.\cos 4x} dx$.
Firstly, we have to convert $\sin 3x\cos 4x$ as the addition of two terms.
We have studied a famous trigonometric formula $\sin \left( {A + B} \right) + \sin \left( {A - B} \right) = 2\sin A\cos B$.
Putting $A = 3x$ and $B = 4x$ we get,
$\Rightarrow 2\sin 3x\cos 4x = \sin \left( {3x + 4x} \right) + \sin \left( {3x - 4x} \right) \\\ \Rightarrow \sin 3x\cos 4x = \dfrac{1}{2}\left( {\sin 7x + \sin \left( { - x} \right)} \right) \\\$
We know that $\sin \left( { - \theta } \right) = - \sin \theta$ applying this. we get,
$\Rightarrow \sin 3x\cos 4x = \dfrac{1}{2}\left( {\sin 7x - \sin x} \right) \\\ \Rightarrow \sin 3x\cos 4x = \dfrac{1}{2}\sin 7x - \dfrac{1}{2}\sin x \\\$
Now, $\int {\sin 3x.\cos 4x} dx$ can be written as $\int {\left( {\dfrac{1}{2}\sin 7x - \dfrac{1}{2}\sin x} \right)dx}$.
$= \int {\dfrac{1}{2}\sin 7xdx} - \int {\dfrac{1}{2}\sin xdx}$
Since $\dfrac{1}{2}$ is a constant term so it can be taken out of both the integration.
$= \dfrac{1}{2}\left( {\int {\sin 7xdx} - \int {\sin xdx} } \right)$- - - - - - - - - - - - - - - - - - - - - (1)
Now, take $7x = u$ then differentiating both side we get,
$7dx = du$
Now, by putting $7x = u$ and $dx = \dfrac{{du}}{7}$, we can write $\int {\sin 7xdx}$ as $\int {\sin u\left( {\dfrac{{du}}{7}} \right) = \dfrac{1}{7}\int {\sin udu} }$.
We know that integration of $\sin x$ with respect to $dx$ gives $- \cos x$. That is $\int {\sin xdx = - \cos x} + c$.
Now, $\dfrac{1}{7}\int {\sin udu = \dfrac{1}{7}\left( { - \cos u} \right)} + c$
Putting $u = 7x$.
we get the value of $\int {\sin 7xdx}$ is $\dfrac{{ - 1}}{7}\cos 7x + c$.
And the value of $\int {\sin xdx}$ is $- \cos x + c$.
Putting these values in equation (1) we get,

$$= \dfrac{1}{2}\left( {\int {\sin 7xdx} - \int {\sin xdx} } \right) \\\ = \dfrac{1}{2}\left( { - \dfrac{1}{7}\cos 7x + c - \cos x + c} \right) \\\ = - \dfrac{1}{{14}}\left( {\cos 7x + 7\cos x} \right) + c \\\$$

Thus, the required value of $\int {\sin 3x.\cos 4x} dx$ is $- \dfrac{1}{{14}}\left( {\cos 7x + 7\cos x} \right) + c$.

Note:
If the value of this integration is to be calculated over limits them we have to simply put the upper limit in $- \dfrac{1}{{14}}\left( {\cos 7x + 7\cos x} \right)\,$ and then lower limits and the difference of these two gives the required results.