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How do you find the integral of $ {e^{3x}} \cdot \cos 3xdx $ ?

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Last updated date: 13th Jul 2024
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Answer
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Hint: To find the integral of $ {e^{3x}} \cdot \cos 3xdx $ , we are going to use integration by parts. The formula for integration by parts is
For $ I = \int {uvdx} $ ,
$ \Rightarrow I = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)} } dx $
Here, the values of u and v are selected on the basis of ILATE rule. We have explained the integration by parts method in detail below.

Complete step by step solution:
In this question, we are given an expression and we need to find its integral. First of all, let this given integral be equal to $ I $.
The given expression is: $ I = {e^{3x}} \cdot \cos 3xdx $ - - - - - - - - - - - - - - - (1)
Integrating equation (1), we get
 $ \Rightarrow I = \int {{e^{3x}} \cdot \cos 3xdx} $ - - - - - - - - - - (2)
Here, we can see that the expression in equation (2) is in the form $ I = \int {uvdx} $ . That is we are going to use Bernoulli's rule for integration by parts to find the integral of equation (2).
Bernoulli’s rule for integration by parts for $ I = \int {uvdx} $ is
 $ \Rightarrow I = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)} } dx $
Here, we have to decide the value of u and v based on their order using the ILATE rule. ILATE stands for
I – Inverse trigonometric Functions
L – Log functions
A – Algebraic Functions
T – Trigonometric functions
E – Exponential functions
So, here
 $ u = \cos 3x $ and $ v = {e^{3x}} $
Therefore, we get
\[
   \Rightarrow I = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)} } dx \\
   \Rightarrow I = \cos 3x\int {{e^{3x}}dx - \int {\left( {\dfrac{{d\cos 3x}}{{dx}}\int {{e^{3x}}dx} } \right)} } dx \\
   \Rightarrow I = \cos 3x\left( {\dfrac{{{e^{3x}}}}{3}} \right) - \int {\left( { - 3\sin 3x\left( {\dfrac{{{e^{3x}}}}{3}} \right)} \right)} dx \;
 \]
\[ \Rightarrow I = \cos 3x\left( {\dfrac{{{e^{3x}}}}{3}} \right) + \int {{e^{3x}}\sin 3xdx} \]
\[ \Rightarrow I = \cos 3x\left( {\dfrac{{{e^{3x}}}}{3}} \right) + {I_1}\]- - - - - - - - - (3)
Here, $ {I_1} = {e^{3x}}\sin 3x $
Now, we need to again use integration by parts to find this value. Therefore,
\[
   \Rightarrow {I_1} = \int {{e^{3x}}\sin 3x} \\
   \Rightarrow {I_1} = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)} } dx \\
   \Rightarrow {I_1} = \sin 3x\int {{e^{3x}}dx - \int {\left( {\dfrac{{d\sin 3x}}{{dx}}\int {{e^{3x}}dx} } \right)} } dx \\
   \Rightarrow {I_1} = \sin 3x\left( {\dfrac{{{e^{3x}}}}{3}} \right) - \int {\left( {3\cos 3x\left( {\dfrac{{{e^{3x}}}}{3}} \right)} \right)} dx \;
 \]
\[ \Rightarrow {I_1} = \sin 3x\left( {\dfrac{{{e^{3x}}}}{3}} \right) - \int {{e^{3x}}\cos 3xdx} \]- - - - - - - - - (4)
Now, our question was
 $ \Rightarrow I = \int {{e^{3x}} \cdot \cos 3xdx} $
Therefore, equation (4) becomes
\[ \Rightarrow {I_1} = \sin 3x\left( {\dfrac{{{e^{3x}}}}{3}} \right) - I\]
Substituting this in equation (3), we get
\[
   \Rightarrow I = \cos 3x\left( {\dfrac{{{e^{3x}}}}{3}} \right) + {I_1} \\
   \Rightarrow I = \cos 3x\left( {\dfrac{{{e^{3x}}}}{3}} \right) + \sin 3x\left( {\dfrac{{{e^{3x}}}}{3}} \right) - I \\
   \Rightarrow 2I = \cos 3x\left( {\dfrac{{{e^{3x}}}}{3}} \right) + \sin 3x\left( {\dfrac{{{e^{3x}}}}{3}} \right) \\
   \Rightarrow 2I = \dfrac{{{e^{3x}}}}{3}\left( {\cos 3x + \sin 3x} \right) \\
   \Rightarrow I = \dfrac{{{e^{3x}}}}{6}\left( {\cos 3x + \sin 3x} \right) + c \;
 \]
Where, c is integration constant.
Hence, we have found the integral of $ \int {{e^{3x}} \cdot \cos 3xdx} $ .
So, the correct answer is “ $I = \dfrac{{{e^{3x}}}}{6}\left( {\cos 3x + \sin 3x} \right) + c $ ”.

Note: There is also a shortcut method for solving this question. We have a direct formula that is
 $ \Rightarrow \int {{e^{ax}}\cos bxdx = \dfrac{{{e^{ax}}}}{{{a^2} + {b^2}}}\left( {a\cos bx + b\sin bx} \right)} + c $
Therefore,
 $
   \Rightarrow \int {{e^{3x}}\cos 3xdx = \dfrac{{{e^{3x}}}}{{{3^2} + {3^2}}}\left( {3\cos 3x + 3\sin 3x} \right)} + c \\
   \Rightarrow \int {{e^{3x}}\cos 3xdx = } \dfrac{{{e^{3x}}}}{{18}} \times 3\left( {\cos 3x + \sin 3x} \right) + c \\
   \Rightarrow \int {{e^{3x}}\cos 3xdx = } \dfrac{{{e^{3x}}}}{6}\left( {\cos 3x + \sin 3x} \right) + c \\
  $