Problem of the Week

Updated at Sep 2, 2013 4:27 PM

How can we find the integral of \(\cos^{2}x\)?

Below is the solution.



\[\int \cos^{2}x \, dx\]

1
Use Pythagorean Identities: \(\cos^{2}x=\frac{1}{2}+\frac{\cos{2x}}{2}\).
\[\int \frac{1}{2}+\frac{\cos{2x}}{2} \, dx\]

2
Use Sum Rule: \(\int f(x)+g(x) \, dx=\int f(x) \, dx+\int g(x) \, dx\).
\[\int \frac{1}{2} \, dx+\int \frac{\cos{2x}}{2} \, dx\]

3
Use this rule: \(\int a \, dx=ax+C\).
\[\frac{x}{2}+\int \frac{\cos{2x}}{2} \, dx\]

4
Use Constant Factor Rule: \(\int cf(x) \, dx=c\int f(x) \, dx\).
\[\frac{x}{2}+\frac{1}{2}\int \cos{2x} \, dx\]

5
Use Integration by Substitution on \(\int \cos{2x} \, dx\).
Let \(u=2x\), \(du=2 \, dx\), then \(dx=\frac{1}{2} \, du\)

6
Using \(u\) and \(du\) above, rewrite \(\int \cos{2x} \, dx\).
\[\int \frac{\cos{u}}{2} \, du\]

7
Use Constant Factor Rule: \(\int cf(x) \, dx=c\int f(x) \, dx\).
\[\frac{1}{2}\int \cos{u} \, du\]

8
Use Trigonometric Integration: the integral of \(\cos{u}\) is \(\sin{u}\).
\[\frac{\sin{u}}{2}\]

9
Substitute \(u=2x\) back into the original integral.
\[\frac{\sin{2x}}{2}\]

10
Rewrite the integral with the completed substitution.
\[\frac{x}{2}+\frac{\sin{2x}}{4}\]

11
Add constant.
\[\frac{x}{2}+\frac{\sin{2x}}{4}+C\]

Done
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