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 \\

$

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