Integrate the following expression $\int{2\sin 3x\sin 6xdx}$.

Hint: At first we will use identity that $2\sin A\sin B=\cos \left( A-B \right)-\cos \left( A+B \right)$ and then take A as $6x$ and B as $3x$ . After breaking into two terms use the formula,
$\int{\cos ax}dx=\dfrac{\sin ax}{a}+c$ Where a , c is any constant.

Complete step-by-step answer:
In the question we are asked to integrate the function $2\sin 3x\sin 6x$ with respect $'x'$ and find the function in terms of x.
At first we will understand what is in definite integral. It is also known by the names antiderivative, inverse derivative, primitive function. It is said that the indefinite integral of a function F is a differentiable function F whose derivative is equal to the original function f. This can be stated symbolically as $F'=f$ . The process of solving antiderivatives is called anti differentiation (or definite integration) and its opposite operation is called differentiation, which is the operation of finding derivatives.
Anti derivative are related to definite integrals through the fundamental theorem of calculus; the definite integral of a function over an interval is equal to the difference between the values of an anti derivative evaluated at the end point of the interval is equal to the difference between the values of an anti derivative evaluated at the end points of the interval.
For example: let $f\left( x \right)=\dfrac{{{x}^{3}}}{3}$ then $f\left( x \right)={{x}^{2}}$ as it’s a derivative of $f\left( x \right)=\dfrac{{{x}^{3}}}{3}$ . As the derivative of constant is $'0'$ so ${{x}^{2}}$ will have infinite number of anti derivative such as $\dfrac{{{x}^{3}}}{3},\dfrac{{{x}^{3}}}{3}+1$ , $\dfrac{{{x}^{3}}}{3}-2$ etc.
Thus $f\left( x \right)$ can be represented as $\dfrac{{{x}^{3}}}{3}+c$ where c is any constant.
So, we have to integrate the function, $2\sin 3x\sin 6x$ .
Here at first we will use the identity, $2\sin A\sin B=\cos \left( A-B \right)-\cos \left( A+B \right)$ .
Now here instead of A taking $6x$ and instead B taking $3x$ we get,
$2\sin 6x\sin 3x=\cos \left( 6x-3x \right)-\cos \left( 6x+3x \right)$ or, $2\sin 3x\sin 6x=\cos \left( 3x \right)-cox\left( 9x \right)$ .
Now we can integrate $\cos 3x-\cos 9x$ instead of $2\sin 3x\sin 6x$ as both the values are the same.
So, $\int{2\sin 3x\sin 6xdx}$ or, $\int{\left( \cos 3x-\cos 9x \right)dx}$ .
Now as we know that $\int{\left\{ f\left( x \right)-g\left( x \right) \right\}dx}$ .
$=\int{f\left( x \right)dx-\int{g\left( x \right)dx}}$ .
Now we can write,
$\int{\cos 3xdx-\int{\cos 6xdx}}$ .
So, now we will apply identity that,
$\int{\cos axdx=\dfrac{\sin ax}{\text{a}}}+c$ .
Where a , c is any constant.
Now we can write,
$\int{\cos 3xdx-\int{\cos 6xdx}}$ as $\dfrac{\text{sin3x}}{3}+{{c}_{1}}-\dfrac{\sin 6x}{6}+{{c}_{2}}$ .
Here ${{c}_{1}}$ and ${{c}_{2}}$ are constant so their sum would also be considered as constant c.
Hence the expression will be $\dfrac{\sin 3x}{3}-\dfrac{\text{sin6x}}{6}+c$ .

Note: We cab first simplify expression as by substituting sin6x as $2\sin 3x\cos 3x$ and then writing expression as $4{{\sin }^{2}}3x\cos 3x$ .
After that taking $\sin 3x$ as t. So, $3\cos 3x$ as $\dfrac{dt}{dx}$ or $4\cos 3xdx$ as $\dfrac{4dt}{3}$ .
So the whole expression will be changed from $\int{2\sin 3x\sin 6xdx}$as $\dfrac{4}{3}\int{{{t}^{2}}dt}$ and thus solving it and finally substituting t as $\sin 3x$