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 $I=\int uvdx$ ,
$\Rightarrow I=u\int vdx-\int \left({\displaystyle \frac{du}{dx}}\int vdx\right)dx$
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
 $\Rightarrow I=u\int vdx-\int \left({\displaystyle \frac{du}{dx}}\int vdx\right)dx$
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
$$\begin{gathered} \Rightarrow I=u\int vdx-\int \left({\displaystyle \frac{du}{dx}}\int vdx\right)dx \\ \Rightarrow I=\cos 3x\int e^{3x}dx-\int \left({\displaystyle \frac{d\cos 3x}{dx}}\int e^{3x}dx\right)dx \\ \Rightarrow I=\cos 3x\left({\displaystyle \frac{e^{3x}}{3}}\right)-\int \left(-3\sin 3x\left({\displaystyle \frac{e^{3x}}{3}}\right)\right)dx \end{gathered}$$
$$\Rightarrow I=\cos 3x\left({\displaystyle \frac{e^{3x}}{3}}\right)+\int e^{3x}\sin 3xdx$$
$$\Rightarrow I=\cos 3x\left({\displaystyle \frac{e^{3x}}{3}}\right)+I_1$$- - - - - - - - - (3)
Here, $I_1=e^{3x}\sin 3x$
Now, we need to again use integration by parts to find this value. Therefore,
$$\begin{gathered} \Rightarrow I_1=\int e^{3x}\sin 3x \\ \Rightarrow I_1=u\int vdx-\int \left({\displaystyle \frac{du}{dx}}\int vdx\right)dx \\ \Rightarrow I_1=\sin 3x\int e^{3x}dx-\int \left({\displaystyle \frac{d\sin 3x}{dx}}\int e^{3x}dx\right)dx \\ \Rightarrow I_1=\sin 3x\left({\displaystyle \frac{e^{3x}}{3}}\right)-\int \left(3\cos 3x\left({\displaystyle \frac{e^{3x}}{3}}\right)\right)dx \end{gathered}$$
$$\Rightarrow I_1=\sin 3x\left({\displaystyle \frac{e^{3x}}{3}}\right)-\int e^{3x}\cos 3xdx$$- - - - - - - - - (4)
Now, our question was
 $\Rightarrow I=\int e^{3x}\cdot \cos 3xdx$
Therefore, equation (4) becomes
$$\Rightarrow I_1=\sin 3x\left({\displaystyle \frac{e^{3x}}{3}}\right)-I$$
Substituting this in equation (3), we get
$$\begin{gathered} \Rightarrow I=\cos 3x\left({\displaystyle \frac{e^{3x}}{3}}\right)+I_1 \\ \Rightarrow I=\cos 3x\left({\displaystyle \frac{e^{3x}}{3}}\right)+\sin 3x\left({\displaystyle \frac{e^{3x}}{3}}\right)-I \\ \Rightarrow 2I=\cos 3x\left({\displaystyle \frac{e^{3x}}{3}}\right)+\sin 3x\left({\displaystyle \frac{e^{3x}}{3}}\right) \\ \Rightarrow 2I={\displaystyle \frac{e^{3x}}{3}}\left(\cos 3x+\sin 3x\right) \\ \Rightarrow I={\displaystyle \frac{e^{3x}}{6}}\left(\cos 3x+\sin 3x\right)+c \end{gathered}$$
Where, c is integration constant.
Hence, we have found the integral of $\int e^{3x}\cdot \cos 3xdx$ .
So, the correct answer is “ $I={\displaystyle \frac{e^{3x}}{6}}\left(\cos 3x+\sin 3x\right)+c$ ”.

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