Question

What is the value of integral \[\int {{e^{ax}}\cos \left( {bx} \right)\,dx} \] ?

Answer 1

Hint: In the given question, we are required to find the value of an integral provided to us in the problem itself. So, we will evaluate the integral by using the integration by parts method. So, we consider the given integral as a new variable. We must know the integrals of exponential function and trigonometric functions in order to solve the given problem.

Complete step by step answer:
Consider $I = \int {{e^{ax}}\cos \left( {bx} \right)\,dx} $
In integration by parts method, we integrate a function which is a product of two functions using a formula:
$\int {f\left( x \right)g\left( x \right)dx = f\left( x \right)\int {g\left( x \right)dx - \int {\left[ {\dfrac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right]dx} } } $
Hence, using integration by parts method and considering ${e^{ax}}$ as first function and $\cos \left( {bx} \right)$ as second function, we get
\[ \Rightarrow I = \left[ {{e^{ax}}\int {\cos \left( {bx} \right)dx} } \right] - \int {\left[ {\dfrac{d}{{dx}}\left( {{e^{ax}}} \right) \times \int {\cos \left( {bx} \right)dx} } \right]} dx\]

Now, we know that the derivative of \[\left( {{e^{ax}}} \right)\] with respect to x is \[a{e^{ax}}\] using the chain rule of differentiation. Also, we know that the integral of $\cos \left( {bx} \right)$ with respect to x is $\dfrac{{\sin \left( {bx} \right)}}{b}$.
\[ \Rightarrow I = \left[ {{e^{ax}} \times \dfrac{{\sin \left( {bx} \right)}}{b}} \right] - \int {\left[ {a{e^{ax}} \times \dfrac{{\sin \left( {bx} \right)}}{b}} \right]} dx\]
Simplifying the expression, we get,
\[ \Rightarrow I = \left[ {\dfrac{{{e^{ax}}\sin \left( {bx} \right)}}{b}} \right] - \int {\left[ {\dfrac{{a{e^{ax}}\sin \left( {bx} \right)}}{b}} \right]} dx\]
Taking the constant term outside of the integral, we get,
\[ \Rightarrow I = \left( {\dfrac{{{e^{ax}}\sin \left( {bx} \right)}}{b}} \right) - \dfrac{a}{b}\int {{e^{ax}}\sin \left( {bx} \right)\,} dx\]

Now, we do the integration of \[\int {{e^{ax}}\sin \left( {bx} \right)\,} \] using the integration by parts method again. So, we get,
\[ \Rightarrow I = \left( {\dfrac{{{e^{ax}}\sin \left( {bx} \right)}}{b}} \right) - \dfrac{a}{b}\left[ {{e^{ax}}\int {\sin \left( {bx} \right)dx} - \int {\left[ {\dfrac{d}{{dx}}\left( {{e^{ax}}} \right) \times \int {\sin \left( {bx} \right)dx} } \right]} dx} \right]\]
Now, we know that the integral of $\sin x$ with respect to x is $ - \cos x$ and derivative of \[\left( {{e^{ax}}} \right)\] with respect to x is \[a{e^{ax}}\] using the chain rule of differentiation. Hence, substituting the values of the known derivatives and integral, we get the value of original integral as:
\[ \Rightarrow I = \left( {\dfrac{{{e^{ax}}\sin \left( {bx} \right)}}{b}} \right) - \dfrac{a}{b}\left[ {{e^{ax}}\left( {\dfrac{{ - \cos \left( {bx} \right)}}{b}} \right) - \int {\left[ {a{e^{ax}} \times \left( {\dfrac{{ - \cos \left( {bx} \right)}}{b}} \right)} \right]} dx} \right]\]

Simplifying the integral, we get,
\[ \Rightarrow I = \left( {\dfrac{{{e^{ax}}\sin \left( {bx} \right)}}{b}} \right) - \dfrac{a}{b}\left[ {{e^{ax}}\left( {\dfrac{{ - \cos \left( {bx} \right)}}{b}} \right) + \int {\dfrac{a}{b}{e^{ax}}\cos \left( {bx} \right)} dx} \right]\]
Taking constants out of the integral, we get,
\[ \Rightarrow I = \left( {\dfrac{{{e^{ax}}\sin \left( {bx} \right)}}{b}} \right) - \dfrac{a}{b}\left[ {\dfrac{{ - {e^{ax}}\cos \left( {bx} \right)}}{b} + \dfrac{a}{b}\int {{e^{ax}}\cos \left( {bx} \right)} dx} \right]\]
Opening brackets and substituting the integral \[\int {{e^{ax}}\cos \left( {bx} \right)} dx\] as $I$, we get,
\[ \Rightarrow I = \left( {\dfrac{{{e^{ax}}\sin \left( {bx} \right)}}{b}} \right) + \dfrac{a}{{{b^2}}}{e^{ax}}\cos \left( {bx} \right) - \dfrac{{{a^2}}}{{{b^2}}}\int {{e^{ax}}\cos \left( {bx} \right)} dx\] \[ \Rightarrow I = \left( {\dfrac{{{e^{ax}}\sin \left( {bx} \right)}}{b}} \right) + \dfrac{a}{{{b^2}}}{e^{ax}}\cos \left( {bx} \right) - \dfrac{{{a^2}}}{{{b^2}}}I\]

Now, we have to find the value of integral I from the equation. So, we get,
\[ \Rightarrow I + \dfrac{{{a^2}}}{{{b^2}}}I = \left( {\dfrac{{{e^{ax}}\sin \left( {bx} \right)}}{b}} \right) + \dfrac{a}{{{b^2}}}{e^{ax}}\cos \left( {bx} \right)\]
\[ \Rightarrow \left( {\dfrac{{{b^2} + {a^2}}}{{{b^2}}}} \right)I = \dfrac{1}{b}{e^{ax}}\sin \left( {bx} \right) + \dfrac{a}{{{b^2}}}{e^{ax}}\cos \left( {bx} \right)\]
\[ \Rightarrow \left( {\dfrac{{{b^2} + {a^2}}}{{{b^2}}}} \right)I = \dfrac{a}{{{b^2}}}{e^{ax}}\left[ {\dfrac{b}{a}\sin \left( {bx} \right) + \cos \left( {bx} \right)} \right]\]
Cross multiplying the terms of the equation, we get,
\[\therefore I = \left( {\dfrac{a}{{{b^2} + {a^2}}}} \right)\left( {{e^{ax}}} \right)\left[ {\dfrac{b}{a}\sin \left( {bx} \right) + \cos \left( {bx} \right)} \right] + C\], where $C$ is any arbitrary constant.

So, the value of integral \[\int {{e^{ax}}\cos \left( {bx} \right)\,dx} \] is \[\left( {\dfrac{a}{{{b^2} + {a^2}}}} \right)\left( {{e^{ax}}} \right)\left[ {\dfrac{b}{a}\sin \left( {bx} \right) + \cos \left( {bx} \right)} \right] + C\], where $C$ is any arbitrary constant.

Note:Integration by parts method can be used to solve integrals of various complex functions involving the product of functions within itself easily. In the given question, integration by parts method has been used twice to make the process much easier and organized. We may have to apply the integration by parts method multiple times in order to reach the final answer in some questions. One must know the integral and derivatives of some basic functions like exponential function to get to deal with such questions.