Question

Solve the following integral: $\int{\sin 3x.\cos 2xdx}$

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 given by $\sin (A-B)=\sin A\cos B-\cos A\sin B$ and $\sin (A+B)=\sin A\cos B+\cos A\sin B$.

In the above question, we have to find the integral of $\sin 3x.\cos 2x$ 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.
\begin{align} & \sin (A+B)=\sin A\cos B+\cos A\sin B \\ & \sin (A-B)=\sin A\cos B-\cos A\sin B \\ \end{align}
\begin{align} & \sin 3x.\cos 2x\text{ }=\text{ }\dfrac{\sin (3x+2x)+\sin (3x-2x)}{2} \\ & \Rightarrow \sin 3x.\cos 2x\text{ }=\text{ }\dfrac{\sin (5x)+\sin (x)}{2}=\dfrac{\sin (5x)+\sin (x)}{2} \\ \end{align}
\begin{align} & \Rightarrow \int{\sin 3x\cos 2xdx=\int{\dfrac{\sin 5x+\sin x}{2}}}dx \\ & \Rightarrow \int{\dfrac{\sin 5x}{2}}dx+\int{\dfrac{\sin x}{2}}dx \\ & \because \int{\sin xdx=-\cos x} \\ & \Rightarrow \dfrac{-\cos 5x}{2\times 5}-\dfrac{\cos x}{2}+c \\ \end{align}
Hence, the answer will be $~-\dfrac{\cos 5x}{10}~~-\dfrac{\cos x}{2}+c$, where c is any arbitrary constant.
Therefore, the integral of the given function in the above question will be $~-\dfrac{\cos 5x}{10}~~-\dfrac{\cos x}{2}+c$.