Question

How do you find the integral of $f\left( x \right) = {e^{2x}}\sin 3x$ using integration by parts?

Answer 1

Hint: Given the expression. We have to find the integral of the expression by applying integration by parts method. First, we will assign some variables to the given expression. Then, substitute the functions to the formula and apply the integration by parts. Then, find the integration of the expression. Then, we will obtain the same expression as the given expression on the right hand side. Combine the like terms and find the value of the integral.

Formula used:
Integration by parts is given by:
$\int {f\left( x \right)g\left( x \right)dx} = f\left( x \right)\int {g\left( x \right)dx} - \left( {\int {f'\left( x \right)} \int {g\left( x \right)dx} } \right)dx$

Complete step by step solution:
Let the integral be $I = \int {{e^{2x}}\sin 3x} dx$
Here, $f\left( x \right) = \sin 3x$ and $g\left( x \right) = {e^{2x}}$. Apply the integration by parts method.

$ \Rightarrow I = \sin 3x\int {{e^{2x}}dx} - \left( {\int {{{\left( {\sin 3x} \right)}^\prime }} \int {{e^{2x}}dx} } \right)dx$

Apply the chain rule of differentiation to the expression ${\left( {\sin 3x} \right)^\prime }$

$ \Rightarrow {\left( {\sin 3x} \right)^\prime } = 3\cos 3x$

$ \Rightarrow I = \sin 3x\int {{e^{2x}}dx} - \left( {\int {3\cos 3x} \int {{e^{2x}}dx} } \right)dx$

Apply the chain rule of integration to the expression, ${e^{2x}}dx$

$\int {{e^{2x}}dx} = \dfrac{1}{2}{e^{2x}}$

Substitute the value into the Integral I

$ \Rightarrow I = \sin 3x\left( {\dfrac{1}{2}{e^{2x}}} \right) - \left( {\int {3\cos 3x} \left( {\dfrac{1}{2}{e^{2x}}} \right)} \right)dx$

Move out the constant terms by applying the constant multiple rule of integration.

$ \Rightarrow I = \sin 3x\left( {\dfrac{1}{2}{e^{2x}}} \right) - \dfrac{3}{2}\int {\cos 3x} {e^{2x}}dx$

Again we will apply the integration by parts method to the right hand side expression.

Here, $f\left( x \right) = \cos 3x$ and $g\left( x \right) = {e^{2x}}$

$ \Rightarrow I = \sin 3x\left( {\dfrac{1}{2}{e^{2x}}} \right) - \dfrac{3}{2}\left[ {\cos 3x\int {{e^{2x}}dx} - \int {\left( {{{\left( {\cos 3x} \right)}^\prime }\int {{e^{2x}}dx} } \right)dx} } \right]$

Apply the chain rule of differentiation to the expression ${\left( {\cos 3x} \right)^\prime }$

$ \Rightarrow {\left( {\cos 3x} \right)^\prime } = - 3\sin 3x$

$ \Rightarrow I = \sin 3x\left( {\dfrac{1}{2}{e^{2x}}} \right) - \dfrac{3}{2}\left[ {\dfrac{{\cos 3x \cdot {e^{2x}}}}{2} - \int {\left( {\dfrac{{ - 3\sin 3x \cdot {e^{2x}}}}{2}} \right)dx} } \right]$

Further simplify the expression, we get:

\[ \Rightarrow I = \sin 3x\left( {\dfrac{1}{2}{e^{2x}}} \right) - \dfrac{3}{4}\cos 3x \cdot {e^{2x}} - \dfrac{3}{4}\int {\sin 3x \cdot {e^{2x}}dx} \]

Substitute $I = \int {{e^{2x}}\sin 3x} dx$ into the expression.

\[ \Rightarrow I = \sin 3x\left( {\dfrac{1}{2}{e^{2x}}} \right) - \dfrac{3}{4}\cos 3x \cdot {e^{2x}} - \dfrac{3}{4}I\]

Add \[\dfrac{3}{4}I\]on both sides, we get:

\[ \Rightarrow I + \dfrac{3}{4}I = \sin 3x\left( {\dfrac{1}{2}{e^{2x}}} \right) - \dfrac{3}{4}\cos 3x \cdot {e^{2x}}\]

Simplify the expression, we get:

\[ \Rightarrow \dfrac{7}{4}I = \dfrac{{2 \cdot {e^{2x}} \cdot \sin 3x - 3\cos 3x \cdot {e^{2x}}}}{4}\]

Multiply both sides by \[\dfrac{4}{7}\], we get:

\[ \Rightarrow \dfrac{7}{4}I \times \dfrac{4}{7} = \dfrac{{2 \cdot {e^{2x}} \cdot \sin 3x - 3\cos 3x \cdot {e^{2x}}}}{4} \times \dfrac{4}{7}\]

\[ \Rightarrow I = \dfrac{{{e^{2x}}}}{7}\left( {2\sin 3x - 3\cos 3x} \right)\]

Final answer: Hence the value of the expression, $\int {{e^{2x}}\sin 3x} dx$ is \[\dfrac{{{e^{2x}}}}{7}\left( {2\sin 3x - 3\cos 3x} \right)\]

Note: Please note that in such types of questions, the values of $f\left( x \right)$ and $g\left( x \right)$ must be carefully chosen so that we find the function of LHS in the right hand side and we can combine them to simplify the expression. Here are some important formulae that can be used in integration.

 $ \Rightarrow \int {{e^{ax}}} dx = \dfrac{1}{a}{e^{ax}}$
$ \Rightarrow \int {\sin x} dx = - \cos x + C$
$ \Rightarrow \int {\cos x} dx = \sin x + C$