Question

Evaluate:

$\int {x\,\sin 3x\,dx} $
A. $ - \dfrac{1}{3}x\cos 3x + \dfrac{1}{9}\sin 3x + c$
B. $ - \dfrac{1}{3}x\cos 3x + \dfrac{1}{3}\sin 3x + c$
C. $\dfrac{1}{3}x\sin 3x - \dfrac{1}{3}\cos 3x + c$
D. None of these

Answer 1

Hint: At first we learn about the formula integration by parts i.e.
$ \Rightarrow \int {f(x)\,g(x)\,dx = f(x)\,\int {g(x)\,dx\,} - \int {f'(x)} } \left( {\int {g(x)\,dx\,} } \right)\,dx + c$, Using this formula we’ll simply the given integration to get the required answer to match the given options.

Complete step by step answer:

Given data: The expression $\int {x\,\sin 3x\,dx} $
We know if two functions of the same independent variable are multiplied together and we have to find the integration of that product then we use the formula named as integration by parts.
Let say the functions be $f(x)$ and $g(x)$ then
$ \Rightarrow \int {f(x)\,g(x)\,dx = f(x)\,\int {g(x)\,dx\,} - \int {f'(x)} } \left( {\int {g(x)\,dx\,} } \right)\,dx + c$
Now using integration by parts in the given expression
i.e. $\int {x\,\sin 3x\,dx} = x\int {\sin 3x\,dx} - \int {1\left( {\int {\sin 3x\,dx} } \right)} \,dx + c$
Using $\int {\sin \,nx\,dx} = - \dfrac{{\cos nx}}{n}$ , we get,
$ = x\left( {\dfrac{{ - \cos 3x}}{3}} \right) - \int {1\left( {\dfrac{{ - \cos 3x}}{3}} \right)} \,dx + c$
Taking $\dfrac{1}{3}$ common from both the terms, we get,
$ = \dfrac{1}{3}\left[ { - x\cos 3x - \int { - \cos 3x\,dx} } \right] + c$
On simplifying, we get,
$ = \dfrac{1}{3}\left[ { - x\cos 3x + \int {\cos 3x\,dx} } \right] + c$
Using $\int {\cos nx\,dx} = \dfrac{{\sin nx}}{n}$, we get,
$ = \dfrac{1}{3}\left[ { - x\cos 3x + \dfrac{{\sin 3x}}{3}} \right] + c$
On simplifying the brackets,
$ = - \dfrac{{x\cos 3x}}{3} + \dfrac{{\sin 3x}}{9} + c$
Therefore the solution of the given integral is $ - \dfrac{{x\cos 3x}}{3} + \dfrac{{\sin 3x}}{9} + c$
Option(A) is correct.

Note: In this solution we compared the function $f(x)$ as ‘x’ and $g(x)$ as ‘sin3x’, but if we have done the opposite the answer might have gotten complex as on the integration of ‘x’ we will a higher degree value. So we are always supposed to assume $f(x)$ which on differentiation will give us a lower degree value, this will result in a simpler expression to simplify.