Can you find out where I am going wrong?

$\displaystyle \int_{0}^{\frac{\pi}{2}}\cos^5(x) .dx\ where\ s = \sin(x) $ and therefore $\displaystyle \frac{ds}{dx} = cos(x)$

When it gets substituted in it produces

$\displaystyle \int_{0}^{\frac{\pi}{2}}\cos^4(x) .ds$

due to the trigonometric identity of

$\displaystyle \cos^2(x) = 1 - \sin^2(x)$

it becomes

$\displaystyle \int_{0}^{\frac{\pi}{2}}(1 - \sin^2(x))(1 - \sin^2(x))$

when multiplied out this goes to

$\displaystyle \int_{0}^{\frac{\pi}{2}}1 - 2s^2 + s^4 (using\ earlier\ substitution\ of\ s = \sin(x)\ to\ simplify\ it)$

when this is integrated it becomes

$\displaystyle \int_{0}^{\frac{\pi}{2}}s - \frac{2s^3}{3} + \frac{s^5}{5}$

therefore the area should be

$\displaystyle \left[\sin(x) - \frac{2\sin^3(x)}{3} + \frac{\sin^5(x)}{5}\right]_{0}^{\frac{\pi}{2}}$

Which then gets to (almost there, I promise)

$\displaystyle \left[\sin(\frac{\pi}{2}) - \frac{2sin^3(\frac{\pi}{2})}{3} + \frac{sin^5(\frac{\pi}{2})}{5}\right]$ and because sin(0) is 0 I left the second half out.

$\displaystyle \frac{1}{2} - \frac{1}{12} + \frac{1}{160} all\ equalling\ \frac{203}{480}$

at last

I believe that I went wrong when I insert the values but I don't know how to change them