Problems on Techniques of Integration

The trigonometric functions $\cos$ and $\sin$ are nice functions since their derivatives and antiderivatives are easy to obtain. 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. In this case, we must differentiate $f(x) = x$ because its derivative gives the constant 1.


u &=&x\\
dv &=& \cos(x)dx\;.


du &=&dx\\
v &=& \sin(x)\;.


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

we get

\begin{displaymath}\int x \cos(x) dx = x \sin(x) - \int sin(x) dx \end{displaymath}

which implies

\begin{displaymath}\int x \cos(x) dx = x \sin(x) - [-\cos(x)] + C = x \sin(x) + \cos(x) + C\;.\end{displaymath}

It is a common mistake to forget the constant $C$.

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