Integration of Exponential Functions
The basic formulaes:\[ \int e^x \, dx = e^x + C \] and \[\int a^x dx = \frac{a^x}{\ln(a)} +C\]
Find the indefinite integral \[\int (3e^x+2^x) dx,\] using \(C\) as the constant of integration.
We have \[\begin{align} \int (3e^x+2^x) dx &=3\int e^x dx+\int 2^x dx \\ &=3e^x+\frac{2^x}{\ln 2}+C, \end{align}\] where \(C\) is the constant of integration.
Find the indefinite integral \[\int e^{x+2} dx,\] using \(C\) as the constant of integration.
We have \[\begin{align} \int e^{x+2} dx &=\int e^x e^2 dx \\ &=e^2\int e^x dx \\ &=e^2 e^x +C \\ &=e^{x+2}+C, \end{align}\] where \(C\) is the constant of integration.
Typical Cases
Case 1: Suppose we have an exponential functions clubbed as: \(\displaystyle \int e^x(f(x) + f'(x)) dx\). In this case, the integral is \(e^xf(x) + C\)
Find the indefinite integral \[\int e^x(\sin(x) + \cos(x))dx\] using \(C\) as the constant of integration.
We have the integral in the form of \(\displaystyle \int e^x(f(x) + f'(x)) dx\).
where \(f(x) = \sin(x)\). So our integral is \[e^x\sin(x) + C\]
Case 2: Suppose we have the integrate of the form \[I = \int e^{ax}\cos(bx+c)\].
It's integral is
\[I=\frac{e^{ax}(a\cos(bx+c)+b\sin(bx+c))}{a^2 + b^2}\] .
We'll integrate the above using integration by parts as:
\[I = \int e^{ax}\cos(bx+c)\]
\[= \cos(bx+c)\frac{e^{ax}}{a} + \frac{b}{a} \int e^{ax}\sin(bx+c)\]
\[= \cos(bx+c)\frac{e^{ax}}{a} + \frac{b}{a}\left(\frac{e^{ax}}{a}\sin(bx+c) - \frac{b}{a} \int e^{ax}\cos(bx+c)\right)\]
\[=\frac{e^{ax}(a\cos(bx+c)+b\sin(bx+c))}{a^2} - \frac{b^2}{a^2}I\]
So \[I(1+\frac{b^2}{a^2}) = \frac{e^{ax}(a\cos(bx+c)+b\sin(bx+c))}{a^2}\]
Therefore \[I=\frac{e^{ax}(a\cos(bx+c)+b\sin(bx+c))}{a^2 + b^2}\]
Note: The above example is also applicable for the form \[I = \int e^{ax}\sin(bx+c)\]
Find the indefinite integral \[\int e^{2x}\cos(5x+3) dx,\] using \(C\) as the constant of integration.
So, by the above result, we obtain our answer as \[\frac{e^{2x}(2\cos(5x+3)+5\sin(5x+3))}{29} + C\]
Case 3: If the integrate is of the form:
\[\int \frac{ae^x + be^{-x}}{pe^x + qe^{-x}} dx\]
In this case, express NUM = \(\alpha (DEN) + \beta \frac{d}{dx}(DEN)\) and then integrate as usual.
Note: NUM means the Numerator of the integrand, and DEN means the Denominator of the integrand.
Find the indefinite integral
\[\int \frac{2e^x + 3e^{-x}}{e^x - 5e^{-x}} dx\]
using C as the constant of integration.
We can write \(2e^x + 3e^{-x} = \alpha(e^x - 5e^{-x}) + \beta(e^x + 5e^{-x})\)
So comparing the coefficients of \(e^x \text{ and } e^{-x}\), we obtain \(\alpha + \beta = 2\) and \(\alpha - \beta = -\frac{3}{5}\).
So we obtain \(\alpha = \frac7{10}, \beta=\frac{13}{10}\). So we have
\[\int \frac{2e^x + 3e^{-x}}{e^x - 5e^{-x}} dx\]
\[=\alpha \int dx + \beta \int \frac{e^x + 5e^{-x}}{e^x - 5e^{-x}} dx\]
Put \(e^x - 5e^{-x} = t\) => \((e^x + 5e^{-x})dx = dt\)
\[=\frac7{10} \int dx +\frac{13}{10}\int \frac{dt}{t}\]
\[=\frac{7x}{10} +\frac{13}{10}\ln |t| + C \]
\[=\frac{7x}{10} +\frac{13}{10}\ln |e^x - 5e^{-x}| + C \]