Question

Find the integral of the function
$\sin 3x.\cos 4x$

Answer 1

Hint: In the above type of integration question first of all we will have to convert them by using trigonometric formulae in that form in which we can easily integrate them, so, we have to remember the sine and cosine sum angle formulae.

Complete step by step answer:
In the above question we have to find the integral of $\sin 3x.\cos 4x$ which is in the multiplication form and we don’t know the integration of this kind. So, we will try to split it as the sum/difference of sine and cosine.
The formulae of trigonometry that we use to split the given trigonometric expression as sum/difference of sine and cosine are as shown below;
$\begin{aligned} & \sin (A+B)=\sin A\cos B+\cos A\sin B \\\ & \sin (A-B)=\sin A\cos B-\cos A\sin B \\\ \end{aligned}$
So, by using the above formulae we can write the given expression is as follow;

& \sin 3x.\cos 4x=\dfrac{\sin (3x+4x)+\sin (3x-4x)}{2} \\\ & \Rightarrow \sin 3x.\cos 4x=\dfrac{\sin (7x)+\sin (-x)}{2}=\dfrac{\sin (7x)-\sin (x)}{2} \\\ \end{aligned}$$ Now, we will integrate the above expression which is shown below; $$\begin{aligned} & \Rightarrow \int{\sin 3x\cos 4xdx=\int{\dfrac{\sin 7x-\sin x}{2}}}dx \\\ & \Rightarrow \int{\dfrac{\sin 7x}{2}}dx-\int{\dfrac{\sin x}{2}}dx \\\ & \Rightarrow \dfrac{-\cos 7x}{2\times 7}+\dfrac{\cos x}{2}+c \\\ \end{aligned}$$ Hence, the answer will be $$~-\dfrac{\cos 7x}{14}~~+\dfrac{\cos x}{2}+c$$, where c is any arbitrary constant. Therefore, the integral of the given function in the above question will be $$~-\dfrac{\cos 7x}{14}~~+\dfrac{\cos x}{2}+c$$. NOTE: Be careful while doing integration because there is a chance that you might make some mistake and your final answer will be wrong. Also take care of the sign during the calculation. Remember the sine and cosine sum angle formulae which i have already mentioned above in the solution.Also go through the other trigonometric formulae because it helps you while solving the above type of integral.