Question

How do you integrate $\int {{e^{2x}}\sin \left( {3x} \right)dx} $ by integration by parts method?

Answer 1

Hint: We integrate the given number by parts. First we categorize these two functions into $u$ and $v$ on the basis of the ILATE rule. Then we use the formula to integrate and solve further. After integrating twice, we substitute the recurring function with $I$ and then solve it further.

Formula used: The formula for integration by parts method is given as follows:
$uv = u\int {vdx} - \int {\left\{ {\dfrac{{du}}{{dx}}\int v dx} \right\}dx} $
Where $u$ represents the first function and $v$ represents the second function. We determine the first function and the second function using the ILATE method. Here in ILATE method, the trigonometric form comes first, hence the trigonometric function will be the first function $u$ and the exponential function will be the second function $v$

Complete step by step answer:
Let $I = \int {{e^{2x}}\sin \left( {3x} \right)dx} $
Now we have to take, $\sin \left( {3x} \right) = u$ and ${e^{2x}} = v$ .
Applying the formula for integration by parts, we get:
$ \Rightarrow I = \sin \left( {3x} \right)\int {{e^{2x}}dx - \int {\left\{ {\dfrac{d}{{dx}}\left( {\sin \left( {3x} \right)} \right)\int {{e^{2x}}dx} } \right\}dx} } $
On integrating and differentiating the respective functions, we get:
\[ \Rightarrow I = \dfrac{{\sin \left( {3x} \right){e^{2x}}}}{2} - \int {\left\{ {\dfrac{{3\cos \left( {3x} \right){e^{2x}}}}{2}} \right\}dx} \]
On taking out the constant outside the integral symbol, we get:
\[ \Rightarrow I = \dfrac{{\sin \left( {3x} \right){e^{2x}}}}{2} - \dfrac{3}{2}\int {\left\{ {\cos \left( {3x} \right){e^{2x}}} \right\}dx} \]
We find that \[\cos \left( {3x} \right){e^{2x}}\] can be integrated by parts again. Hence to solve the sum further, we integrate this by parts:
\[ \Rightarrow I = \dfrac{{\sin \left( {3x} \right){e^{2x}}}}{2} - \dfrac{3}{2}\left[ {\dfrac{{\cos \left( {3x} \right){e^{2x}}}}{2} - \int {\left\{ {\dfrac{d}{{dx}}\left( {\cos 3x} \right)\int {{e^{2x}}dx} } \right\}} dx} \right]\]
\[ \Rightarrow I = \dfrac{{\sin \left( {3x} \right){e^{2x}}}}{2} - \dfrac{3}{2}\left[ {\dfrac{{\cos \left( {3x} \right){e^{2x}}}}{2} - \int {\left\{ {\dfrac{{ - 3\sin \left( {3x} \right){e^{2x}}}}{2}} \right\}dx} } \right]\]
Taking out the constant term again:
\[ \Rightarrow I = \dfrac{{\sin \left( {3x} \right){e^{2x}}}}{2} - \dfrac{3}{2}\left[ {\dfrac{{\cos \left( {3x} \right){e^{2x}}}}{2} + \dfrac{3}{2}\int {\sin \left( {3x} \right){e^{2x}}dx} } \right]\]
Now as we already know that $I = \int {{e^{2x}}\sin \left( {3x} \right)dx} $, therefore let substitute this in the latter half of the equation.
Therefore, \[I = \dfrac{{\sin \left( {3x} \right){e^{2x}}}}{2} - \dfrac{3}{2}\left[ {\dfrac{{\cos \left( {3x} \right){e^{2x}}}}{2} + \dfrac{3}{2}I} \right]\]
On opening the brackets and further simplifying, we get:
\[ \Rightarrow I = \dfrac{{\sin \left( {3x} \right){e^{2x}}}}{2} - \dfrac{3}{4}\cos \left( {3x} \right){e^{2x}} - \dfrac{9}{4}I + C\]
On adding $ + \dfrac{9}{4}I$ to both sides, we get:
\[ \Rightarrow I + \dfrac{9}{4}I = \dfrac{{\sin \left( {3x} \right){e^{2x}}}}{2} - \dfrac{3}{4}\cos \left( {3x} \right){e^{2x}} + C\]
On solving it further, we get:
\[ \Rightarrow \dfrac{{13}}{4}I = \dfrac{{\sin \left( {3x} \right){e^{2x}}}}{2} - \dfrac{3}{4}\cos \left( {3x} \right){e^{2x}} + C\]
On dividing both the sides with $\dfrac{4}{{13}}$ , we get:
\[ \Rightarrow I = \dfrac{4}{{13}}\left[ {\dfrac{{\sin \left( {3x} \right){e^{2x}}}}{2} - \dfrac{3}{4}\cos \left( {3x} \right){e^{2x}}} \right] + C\]
On simplifying it further:
\[ \Rightarrow I = \dfrac{2}{{13}}\sin \left( {3x} \right){e^{2x}} - \dfrac{3}{{13}}\cos \left( {3x} \right){e^{2x}} + C\]
So, the correct answer is \[ I = \dfrac{2}{{13}}\sin \left( {3x} \right){e^{2x}} - \dfrac{3}{{13}}\cos \left( {3x} \right){e^{2x}} + C\]

Note: In integration by part, we follow the ILATE rule, which stands for
I – INVERSE TRIGONOMETRIC FUNCTIONS
L-LOGARITHMIC FUNCTIONS
A-ALGEBRAIC FUNCTIONS LIKE POYNOMINALS ETC
T-TRIGONOMETRIC FUNCTIONS
E-EXPONENTIAL FUNCTIONS
This rule helps us to determine which function to be considered as the first function and which one the second while using integration by parts method.