Question

How do you find the integral of ${{e}^{x}}\sin x$?

Answer 1

As you can see that there are two functions in x and are written in the product form so we can use integration by parts to solve the integral of ${{e}^{x}}\sin x$ with respect to x. While solving this integral by integration by parts, we will take the first term as $\sin x$ and the second term as ${{e}^{x}}$. The formula for integration by parts for the following integral $\int{f\left( x \right)}g\left( x \right)dx$ is equal to $f\left( x \right)\int{g\left( x \right)}dx-\int{f'\left( x \right)}\int{g\left( x \right)}dx$ . In this integral, we have taken the first term as $f\left( x \right)$ and the second term as $g\left( x \right)$.

Complete step-by-step solution:
Let us find the integral of given expression,
$\int{\left( {{e}^{x}}\sin x \right)dx}$
Now, let us name the above integral as I.
I = $\int{\left( {{e}^{x}}\sin x \right)dx}$
We are going to integrate the above expression by using integration by parts. The formula of integration by parts for $\int{f\left( x \right)}g\left( x \right)dx$ is equal to:
$f\left( x \right)\int{g\left( x \right)}dx-\int{f'\left( x \right)}\int{g\left( x \right)}dx$
On applying integration by parts on the above integral in which the first term is $\sin x$ and the second term is ${{e}^{x}}$ we get,
$\begin{aligned} & I=\int{\left( \sin x{{e}^{x}} \right)}dx \\\ & =\sin x\int{{{e}^{x}}}dx-\int{\left( \sin x \right)'\int{{{e}^{x}}dx}} \\\ \end{aligned}$
We know that integration of ${{e}^{x}}$ with respect to x is ${{e}^{x}}$ and derivative of $\sin x$ with respect to x is $\cos x$ so using these values in the above expression we get,
$=\sin x{{e}^{x}}-\int{\cos x{{e}^{x}}}dx$
Now, we are going to integrate the above integral by using integration by parts by taking $\cos x$ as the first term and ${{e}^{x}}$ as the second term.
$I=\sin x{{e}^{x}}-\left[ \cos x\int{{{e}^{x}}dx}-\int{\left( \cos x \right)'\int{{{e}^{x}}dx}} \right]$
We know that derivative of $\cos x$ with respect to x is $-\sin x$ so substituting this relation in the above we get,
$\begin{aligned} & I=\sin x{{e}^{x}}-\left[ \cos x{{e}^{x}}+C-\int{-\sin x{{e}^{x}}dx} \right] \\\ & \Rightarrow I=\sin x{{e}^{x}}-\cos x{{e}^{x}}-C-\int{\sin x{{e}^{x}}dx} \\\ \end{aligned}$
In the above equation, “C” is the constant.
In the above, we have taken $\int{\left( {{e}^{x}}\sin x \right)dx}$ as I so we can substitute $\int{\sin x{{e}^{x}}dx}$ as I in the above equation and we get,
$\begin{aligned} & I=\sin x{{e}^{x}}-C-\cos x{{e}^{x}}-I \\\ & \Rightarrow 2I=\sin x{{e}^{x}}-\cos x{{e}^{x}}-C \\\ \end{aligned}$
Taking ${{e}^{x}}$ as common from the first two terms of the R.H.S of the above equation we get,
$2I={{e}^{x}}\left( \sin x-\cos x \right)-C$
Dividing 2 on both the sides we get,
$I=\dfrac{1}{2}\left( {{e}^{x}}\left( \sin x-\cos x \right) \right)-\dfrac{C}{2}$
In the above equation, we can write $-\dfrac{C}{2}$ as ${{C}_{1}}$ then the above equation becomes:
$I=\dfrac{1}{2}\left( {{e}^{x}}\left( \sin x-\cos x \right) \right)+{{C}_{1}}$
Hence, we got the integral of given expression as $\dfrac{1}{2}\left( {{e}^{x}}\left( \sin x-\cos x \right) \right)+{{C}_{1}}$.

Note: In the above solution, while solving the integral we have used integration by parts, and in which we have assigned some function as the first term and others as the second term so there is a priority trick for selecting the order of the functions.
The trick is as follows:
You should remember the word “ILATE”. In this word, each letter is the initial letter for mathematical expressions:
“I” represents “Inverse Trigonometric Functions”.
“L” represents “Logarithmic Functions”.
“A” represents “Algebraic Functions”.
“T” represents “Trigonometric Functions”.
“E” represents “Exponential Functions”.
Now, inverse trigonometric functions have a higher priority than logarithmic and all the functions are written after inverse trigonometric functions.
The integral given in the above problem is:
$\int{\left( {{e}^{x}}\sin x \right)dx}$
In the above integral, we have an exponential and trigonometric function and from the above trick, trigonometric function i.e. $\sin x$ has a higher priority than exponential function i.e. ${{e}^{x}}$. That’s why we have taken $\sin x$ as the first term and ${{e}^{x}}$ as the second.