Problems on Techniques of Integration

The trigonometric functions are nice functions since their derivatives and antiderivatives are easy to get. So for the integration by parts, these functions have the same behavior whether we differentiate them or take their antiderivatives. Therefore the focus should be on the other function involved in the integration. Since the derivative or antiderivative of $e^x$ is even easier, so in this case we may either differentiate or integrate, it will not make any major difference.

Set

\begin{displaymath}\left\{\begin{array}{lll}
u &=& \cos(ax)\\
dv &=& e^{bx} dx\;.
\end{array}\right.\end{displaymath}

Then

\begin{displaymath}\left\{\begin{array}{lll}
du &=& -a \sin(ax) dx\\
v &=& \displaystyle \frac{1}{b} e^{bx}\;.
\end{array}\right.\end{displaymath}

Since

\begin{displaymath}\int u dv = u v - \int v du\;,\end{displaymath}

we get

\begin{displaymath}\begin{array}{lll}
\displaystyle \int \cos(ax)e^{bx} dx &=&\d...
...os(ax)e^{bx} + \frac{a}{b}\int \sin(ax)e^{bx} dx
\end{array}\;.\end{displaymath}

In order to integrate the function $\sin(a x)e^{b x}$, we will need to do another integration by parts. This is very common that an integration by parts may lead to another one or may be more integration by parts. Set

\begin{displaymath}\left\{\begin{array}{lll}
u &=&\sin(ax)\\
dv &=& e^{bx}dx\;.
\end{array}\right.\end{displaymath}

Then

\begin{displaymath}\left\{\begin{array}{lll}
du &=&a \cos(ax)dx\\
v &=&\displaystyle \frac{1}{b} e^{bx}\;.
\end{array}\right.\end{displaymath}

Since

\begin{displaymath}\int u dv = u v - \int v du\;,\end{displaymath}

we get

\begin{displaymath}\begin{array}{lll}
\displaystyle \int \sin(ax)e^{bx} dx &=&\d...
... x)e^{b x} - \frac{a}{b}\int \cos(a x)e^{b x} dx
\end{array}\;.\end{displaymath}

Hence

\begin{displaymath}\int \cos(a x)e^{b x} dx = \frac{1}{b}\cos(a x)e^{b x} + \fra...
...in(a x)e^{b x} - \frac{a}{b}\int \cos(a x)e^{b x} dx\right] \;.\end{displaymath}

which implies

\begin{displaymath}\left(1 + \frac{a^2 }{b^2 }\right) \int \cos(x)e^x dx = \frac{1}{b }\cos(a x)e^{b x} + \frac{a }{b^2 }\sin(a x)e^{b x} \end{displaymath}

or

\begin{displaymath}\int \cos(ax)e^{bx} dx = \frac{b}{a^2 + b^2 } \cos(a x)e^{b x} + \frac{a}{a^2 + b^2 } \sin(a x)e^{b x} + C\;.\end{displaymath}

It may be surprising that the constant $C$ showed up at the end and not before. In fact, it has to do with the formula

\begin{displaymath}\int u dv = u v - \int v du\;.\end{displaymath}

Indeed, when obtaining it, we had to take the antiderivative of $(uv)'$ which is not $uv$ but $uv + C$. In any case, one should not forget the constant $C$ when dealing with antiderivatives or indefinite integrals.


If you prefer to jump to the next problem, click on Next Problem below.

[Next Problem] [Matrix Algebra]
[Trigonometry] [Calculus]
[Geometry] [Algebra]
[Differential Equations] [Complex Variables]

S.O.S MATHematics home page

Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.

Mohamed A. Khamsi

Copyright 1999-2017 MathMedics, LLC. All rights reserved.
Contact us
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA
users online during the last hour