Question

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

Answer 1

In the above given question, we are given an expression which is the product of the cosine function and the exponential function, that is written as $\left( {\cos x} \right)\left( {{e^x}} \right)$ . We have to find the antiderivative of the above given expression, which means that we have to integrate the given expression. In order to approach the solution, we have to use the method of integration by parts.

Complete step by step answer:
Given expression is, $\left( {\cos x} \right)\left( {{e^x}} \right)$. We have to find the integration of $\left( {\cos x} \right)\left( {{e^x}} \right)$ i.e. $\int {\left( {\cos x} \right)\left( {{e^x}} \right)} dx$. Let us suppose that,
$\Rightarrow I = \int {\left( {\cos x} \right)\left( {{e^x}} \right)} dx$
Now, let $u = \cos x$ be the first function and $v = {e^x}$ be the second function,
So that, we have their derivatives as $\dfrac{{du}}{{dx}} = - \sin x$ and $\dfrac{{dv}}{{dx}} = {e^x}$ respectively.

Now applying the method of integration by parts, we can write
$\Rightarrow \int {uvdx = u\int v } - \int {\left( {\int v \dfrac{{du}}{{dx}}} \right)dx}$
Substituting their above values, we have the equation as,
$\Rightarrow I = \int {\left( {\cos x} \right)\left( {{e^x}} \right)} dx = \cos x\int {{e^x}} - \int {\left( {\int {{e^x}} \cdot \dfrac{d}{{dx}}\cos x} \right)dx}$
That gives us the equation,
$\Rightarrow I = {e^x}\cos x - \int {{e^x} \cdot \left({ - \sin x} \right)dx}$
That is,
$\Rightarrow I = {e^x}\cos x + \int {{e^x}\sin xdx}$ ...(1)

Now since we have another integral $\int {{e^x}\sin xdx}$ , hence we have to apply the method of integration by parts one more time. This time, let $u = \sin x$ be the first function and $v = {e^x}$ be the second function. So, we have their derivatives as and $\dfrac{{dv}}{{dx}} = {e^x}$ respectively.
That gives us the following integration,
$\Rightarrow \int {\sin x \cdot {e^x}} dx = \sin x\int {{e^x}} - \int {\left( {\int {{e^x}} \cdot \dfrac{d}{{dx}}\sin x} \right)dx}$
That gives us the equation,
$\Rightarrow \int {\sin x \cdot {e^x}} dx = {e^x}\sin x - \int {\left( {{e^x}\cos x} \right)dx}$
But we have $I = \int {\left( {\cos x} \right)\left( {{e^x}} \right)} dx$ , therefore we can write it as,
$\Rightarrow \int {\sin x \cdot {e^x}} dx = {e^x}\sin x - I$ ...(2)

Now substituting the equation (2) in the equation (1), we have the equation,
$\Rightarrow I = {e^x}\cos x + {e^x}\sin x - I$
That gives us,
$\Rightarrow 2I = {e^x}\cos x + {e^x}\sin x$
$\Rightarrow I = \dfrac{{{e^x}\cos x + {e^x}\sin x}}{2}$
That is,
$\Rightarrow I = \dfrac{1}{2}{e^x}\left( {\sin x + \cos x} \right)$
Hence,
$\therefore \int {\left( {\cos x} \right)\left( {{e^x}} \right)} dx = \dfrac{1}{2}{e^x}\left( {\sin x + \cos x} \right)$
That is the required value of the integral $\int {\left( {\cos x} \right)\left( {{e^x}} \right)} dx$.

Therefore, the anti derivative of $\int {\left( {\cos x} \right)\left( {{e^x}} \right)} dx$ is $\dfrac{1}{2}{e^x}\left( {\sin x + \cos x} \right)$.

Note: There are various methods to find the antiderivative or the integration of different functions. Different methods are used to solve different types of integrals as per the convenience of the calculation and for the suitable functions. Some of the most used integration methods are:
-Integration by Substitution.
-Integration by Parts.
-Integration Using Trigonometric Identities.
-Integration of Some particular function.
-Integration by Partial Fraction.