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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 $\sin (A+B)=\sin A\cos B+\cos A\sin B$.

Complete step-by-step answer:

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.

The formulae of trigonometry that we will use to split the given trigonometric expression as sum/difference of sine and cosine are as shown below;

$\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}$

So, by using the above formulae we can write the given expression as follows;

\[\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}\]

Now, we will integrate the above expression which is shown below;

\[\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 \[~-\dfrac{\cos 5x}{10}~~-\dfrac{\cos x}{2}+c\].

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. Do not forget to add the arbitrary constant c, as it is an indefinite integral. Remember the sine and cosine sum angle formulae which are already mentioned above in the solution. Also, go through the other trigonometric formulae because it helps you while solving the above type of integral.

Complete step-by-step answer:

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.

The formulae of trigonometry that we will use to split the given trigonometric expression as sum/difference of sine and cosine are as shown below;

$\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}$

So, by using the above formulae we can write the given expression as follows;

\[\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}\]

Now, we will integrate the above expression which is shown below;

\[\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 \[~-\dfrac{\cos 5x}{10}~~-\dfrac{\cos x}{2}+c\].

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. Do not forget to add the arbitrary constant c, as it is an indefinite integral. Remember the sine and cosine sum angle formulae which are already mentioned above in the solution. Also, go through the other trigonometric formulae because it helps you while solving the above type of integral.

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