Question

How do you find the integral of $\left( {{e}^{x}} \right)\left( \cos x \right)dx$ ?

Answer 1

Hint: The given equation is an integral equation. We need to expand the equation. Now we must solve the real parts of the equation. After solving the real part, the integral must be evaluated by using u-substitution. Now use Euler’s identity and take out the real part from the resultant equation.

Complete step-by-step solution:
Here $\left( {{e}^{x}} \right)\left( \cos x \right)dx$ Can be solved by the complexifying integral method.
Here in the above, the $\cos x$ is the real part of ${{e}^{ix}}$.
By the Euler’s rule identity that is ${{e}^{i\theta }}=\cos \left( \theta \right)+i\sin \left( \theta \right)$.
Now we can rewrite the integral as:
$\int{{{e}^{x}}\cos \left( x \right)dx=\int{{{e}^{x}}.\operatorname{Re}\left( {{e}^{ix}} \right)dx=}}$
Here the given equation is complex. In this type of equation ${{e}^{x}}$ is a real factor.
Real factor outside or inside the real part functions doesn’t vary the total value of the equation.
So now we can write a real part function and the integral part inside it.
Now the equation changes as:
$\begin{align}
  & \Rightarrow \operatorname{Re}\left( \int{{{e}^{x}}{{e}^{^{ix}}}dx} \right) \\
 & \Rightarrow \operatorname{Re}\left( \int{{{e}^{x+ix}}dx} \right) \\
 & \Rightarrow \operatorname{Re}\left( \int{{{e}^{\left( i+1 \right)x}}} \right)dx \\
\end{align}$
By using u-substitution we can evaluate the integral:
$\Rightarrow \operatorname{Re}\left( \dfrac{{{e}^{\left( i+1 \right)x}}}{i+1}+c \right)$
Now let’s take the common exponent out of the equation.
$\Rightarrow \operatorname{Re}\left( \dfrac{{{e}^{x}}{{e}^{ix}}}{i+1} \right)+c$
$\Rightarrow {{e}^{x}}.\operatorname{Re}\left( \dfrac{{{e}^{ix}}}{i+1} \right)+c$
To expand the above equation we can use Euler’s rule identity.
$\Rightarrow {{e}^{x}}.\operatorname{Re}\left( \dfrac{i-1}{\left( i+1 \right)\left( i-1 \right)}\left( \cos \left( x \right)+i\sin \left( x \right) \right) \right)+c$
Now let’s calculate the denominator.
 $\Rightarrow {{e}^{x}}.\operatorname{Re}\left( \dfrac{i-1}{-1-1}\left( \cos \left( x \right)+i\sin \left( x \right) \right) \right)+c$
$\Rightarrow {{e}^{x}}.\operatorname{Re}\left( \dfrac{i-1}{-2}\left( \cos \left( x \right)+i\sin \left( x \right) \right) \right)+c$
$\Rightarrow {{e}^{x}}.\operatorname{Re}\left( \left( \dfrac{i}{-2}-\dfrac{1}{-2} \right)\left( \cos \left( x \right)+i\sin \left( x \right) \right) \right)+c$
Now we should multiply both the terms inside the bracket.
$\Rightarrow {{e}^{x}}.\operatorname{Re}\left( -\dfrac{i}{2}\cos \left( x \right)+\dfrac{1}{2}\sin \left( x \right)+\dfrac{1}{2}\cos \left( x \right)+\dfrac{i}{2}\sin \left( x \right) \right)+c$
Now we can separate the real parts easily.
$\Rightarrow {{e}^{x}}\left( \dfrac{1}{2}\sin \left( x \right)+\dfrac{1}{2}\cos (x) \right)+c$
$\Rightarrow \dfrac{{{e}^{x}}}{2}\left( \sin \left( x \right)+\cos \left( x \right) \right)+c$
Hence the integral of $\left( {{e}^{x}} \right)\left( \cos x \right)dx$ is $\dfrac{{{e}^{x}}}{2}\left( \sin \left( x \right)+\cos \left( x \right) \right)+c$

Note: Euler’s formulae are used for integrated equations. Leonhard Euler is the mathematics scientist who evolved the Euler’s formula. This Euler’s formula is used for the complex numbers.
Integral equation which has trigonometric functions can be evaluated by using Euler's formula.
The integrand can be simplified by using the half-angle formula.