# Help with a trig integral

• Mar 2nd 2010, 11:25 AM
bobhudy
Help with a trig integral
OK I'm studying for a calculus exam and I ran across a problem where I thought I did everything right, but I can't know for sure because it's an even problem and I don't have an answer.

Problem:

$\displaystyle \int cos^5 (x) dx =$

What I did:

$\displaystyle \int cos^5 (x) dx =$

Splitting into squares:
$\displaystyle \int [cos^2 (x)]^2 cos (x) dx$

Using cos^2 + sin^2 = 1:
$\displaystyle \int [1 - sin^2 (x)]^2 cos (x) dx$

Substituting:
$\displaystyle u = \sin (x)$
$\displaystyle du = \cos (x) dx$

$\displaystyle \int (1 - u^2)^2 dx$

Foiling:
$\displaystyle \int (1 - 2u^2 - u^4) dx$

Integrating:
$\displaystyle u - \frac{2}{3}u^3 - \frac{1}{5}u^5$

Unsubstituting:
$\displaystyle sin (x) - \frac{2}{3}sin^3 (x) - \frac{1}{5}sin^5 (x)$

$\displaystyle sin (x) - \frac{2}{3}sin^3 (x) - \frac{1}{5}sin^5 (x) + C$

I typed the problem into Wolfram Alpha and it gave me this:

$\displaystyle \frac{5}{8}sin (x) - \frac{5}{48}sin (3x) - \frac{1}{80}sin (5x) + C$

This looks alot different than my answer, and they did it using something called a 'reduction formula', which seems like something I would never remember. Is my answer wrong, or did they just do it differently? Or is Wolfram Alpha not the best place to be checking my answers?

Thanks
• Mar 2nd 2010, 11:50 AM
General
Quote:

Originally Posted by bobhudy
OK I'm studying for a calculus exam and I ran across a problem where I thought I did everything right, but I can't know for sure because it's an even problem and I don't have an answer.

Problem:

$\displaystyle \int cos^5 (x) dx =$

What I did:

$\displaystyle \int cos^5 (x) dx =$

Splitting into squares:
$\displaystyle \int [cos^2 (x)]^2 cos (x) dx$

Using cos^2 + sin^2 = 1:
$\displaystyle \int [1 - sin^2 (x)]^2 cos (x) dx$

Substituting:
$\displaystyle u = \sin (x)$
$\displaystyle du = \cos (x) dx$

$\displaystyle \int (1 - u^2)^2 dx$

Foiling:
$\displaystyle \int (1 - 2u^2 - u^4) dx$

Integrating:
$\displaystyle u - \frac{2}{3}u^3 - \frac{1}{5}u^5$

Unsubstituting:
$\displaystyle sin (x) - \frac{2}{3}sin^3 (x) - \frac{1}{5}sin^5 (x)$

$\displaystyle sin (x) - \frac{2}{3}sin^3 (x) - \frac{1}{5}sin^5 (x) + C$

I typed the problem into Wolfram Alpha and it gave me this:

$\displaystyle \frac{5}{8}sin (x) - \frac{5}{48}sin (3x) - \frac{1}{80}sin (5x) + C$

This looks alot different than my answer, and they did it using something called a 'reduction formula', which seems like something I would never remember. Is my answer wrong, or did they just do it differently? Or is Wolfram Alpha not the best place to be checking my answers?

Thanks

You did not make any mistake.
Since: $\displaystyle sin (x) - \frac{2}{3}sin^3 (x) - \frac{1}{5}sin^5 (x) \, =\, \frac{5}{8}sin (x) - \frac{5}{48}sin (3x) - \frac{1}{80}sin (5x)$.
• Mar 2nd 2010, 12:43 PM
bobhudy
Quote:

Originally Posted by General
You did not make any mistake.
Since: $\displaystyle sin (x) - \frac{2}{3}sin^3 (x) - \frac{1}{5}sin^5 (x) \, =\, \frac{5}{8}sin (x) - \frac{5}{48}sin (3x) - \frac{1}{80}sin (5x)$.

thanks. i now realise i shouldve just subbed in a number and seen if theyre the same -_-