Integrate the following function: \[x\sin 3x\]

Hint: Here we will be proceeding by letting the given function as \[h\left( x \right)\]. Then use integration by parts methods by knowing its first function and second function by using the ILATE rule. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Let the given function be \[h\left( x \right) = x\sin 3x\]
This function can be integrated by using the method of integration by parts.
By ILATE rule we take \[x\] as the first function (algebraic function) and \[\sin 3x\] (trigonometric function) as second function for integration by parts.
We know that \[\int {f\left( x \right)} g\left( x \right)dx = f\left( x \right)\int {g\left( x \right)dx} - \int {f'\left( x \right)} \int {g\left( x \right)dx} + c\] where \[f\left( x \right)\] is first function and \[g\left( x \right)\] is second function in integrating by parts.
Applying integrating by parts to the function \[h\left( x \right)\], we have
\[
   \Rightarrow \int {h\left( x \right) = \int {\left( x \right)\left( {\sin 3x} \right)dx} } \\
   \Rightarrow \int {h\left( x \right)} = x\int {\left( {\sin 3x} \right)dx} - \int {x'\int {\left( {\sin 3x} \right)dx} } \\
\]
By using the formula, \[\int {\sin axdx} = \dfrac{{ - \cos ax}}{a} + c\] we have
\[
   \Rightarrow \int {h\left( x \right)} = \left( x \right)\left( {\dfrac{{ - \cos 3x}}{3}} \right) - \int {\left( 1 \right)\left( {\dfrac{{ - \cos 3x}}{3}} \right)dx} \\
   \Rightarrow \int {h\left( x \right)} = \dfrac{{ - x\cos 3x}}{3} - \int {\dfrac{{ - \cos 3x}}{3}dx} \\
   \Rightarrow \int {h\left( x \right)} = \dfrac{{ - x\cos 3x}}{3} + \dfrac{1}{3}\int {\cos 3xdx} \\
\]
By using the formula, \[\int {\cos ax} = \dfrac{{\sin ax}}{a} + c\] we have
\[
   \Rightarrow \int {h\left( x \right)} = \dfrac{{ - x\cos 3x}}{3} + \dfrac{1}{3}\left( {\dfrac{{\sin 3x}}{3}} \right) + c \\
   \Rightarrow \int {h\left( x \right)} = \dfrac{{ - x\cos 3x}}{3} + \dfrac{{\sin 3x}}{9} + c \\
  \therefore \int {x\sin 3x} = \dfrac{{ - x\cos 3x}}{3} + \dfrac{{\sin 3x}}{9} + c \\
\]
Thus, the integral of \[x\sin 3x\] is \[\dfrac{{ - x\cos 3x}}{3} + \dfrac{{\sin 3x}}{9} + c\].

Note: A constant namely integrating constant that is added to the function obtained by evaluating the indefinite integral of a given function, indicating that all indefinite integrals of the given function differ by, at most, a constant. So, it is necessary to add integrating constants after completion of the integral.