Question

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

Answer 1

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 \\\$