How do you integrate $\sin 3x\cos 3xdx$?

Hint: Integration is a method of adding or summing up the parts to find the whole. It is studied under the branch of calculus. It is the process of finding the antiderivatives of a function. It is similar to adding the slices to make it whole. The integration is the inverse process of differentiation. Here we integrate the trigonometric values. For that we use some trigonometric identities and then we solve with integration by substitution and integration by parts. Then we solve by formula and complete step by step explanation.
Formulas used:
$\sin 2x = 2\sin x\cos x$
$\int {\sin nx} dx = - \dfrac{{\cos nx}}{n} + c$

Complete step by step answer:Let us integrate $\sin 3x\cos 3xdx$
\[\sin 3x\cos 3xdx = \dfrac{{\sin 6x}}{2}\] $\int {\sin 3x\cos 3xdx} $
Now by using formula we convert$\sin 3x\cos 3x$ into $\sin 6x$
$\sin 2x = 2\sin x\cos x$
$\sin 6x = 2\sin 3x\cos 3x$
We bring 2 to the left hand side, we get
$ \Rightarrow \dfrac{{\sin 6x}}{2} = \sin 3x\cos 3x$
Now we integrate
\[ \Rightarrow \sin 3x\cos 3xdx = \dfrac{{\sin 6x}}{2}\]
\[ \Rightarrow \int {\dfrac{{\sin 6x}}{2}dx} \]
Now bring constant term, out of integral we get,
\[ \Rightarrow \dfrac{1}{2}\int {\sin 6xdx} \]
Here we use integral formula mentioned in formula used, we get
\[ \Rightarrow \dfrac{1}{2}\left( {\dfrac{{ - \cos 6x}}{6}} \right) + C\]
Now multiplying constant terms, we get
\[ \Rightarrow - \dfrac{1}{{12}}\cos 6x + C\]
Hence,
$ \Rightarrow \int {\sin 3x\cos 3xdx = - \dfrac{1}{{12}}\cos 6x + C} $
Hence we get the required answer.

Note:
In applied Maths, there are lots of problems involving integration of functions. The integration is more of an art in comparison with any other process in mathematics. There are various methods or techniques of integration. The methods are Integration by parts, Integration by t-substitution, Integration by trigonometric substitution, Integration by and partial fraction. Here we used Integration by trigonometric substitution. If you’re able to calculate the indefinite integrals, then definite integrals can be easily done. Integration and differentiation is also a pair of inverse functions similar to addition – subtraction and multiplication-division.