# Thread: Another Integration thing

1. ## Another Integration thing

Can you find out where I am going wrong?

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

When it gets substituted in it produces

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

due to the trigonometric identity of

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

it becomes

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

when multiplied out this goes to

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

when this is integrated it becomes

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

therefore the area should be

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

Which then gets to (almost there, I promise)

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

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

at last

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

2. Originally Posted by Beard
Can you find out where I am going wrong?

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

When it gets substituted in it produces

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

due to the trigonometric identity of

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

it becomes

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

when multiplied out this goes to

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

when this is integrated it becomes

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

therefore the area should be

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

Which then gets to (almost there, I promise)

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

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

at last

I believe that I went wrong when I insert the values but I don't know how to change them
Sorry but the notation is abominable and I'm not even going to bother correcting the numerous notational errors.

When you make the substitution you should also substitute for the integral terminals:

$x = 0 \Rightarrow s = 0$ and $x = \frac{\pi}{2} \Rightarrow s = 1$.

Then the integral becomes $\int_0^1 (1 - s^2)^2 \, {\color{red}ds} = \int_0^1 1 - 2 s^2 + s^4\, {\color{red}ds}$

which is much less work to solve.

3. $\int_{0}^{\frac{\pi}{2}} cos^{5}(x)dx$

There is a reduction formula for $\int cos^{n}(x)dx$

But we will do it.

You are on the right track, but the solution is not 203/480.

$\int_{0}^{\frac{\pi}{2}}\left(1-sin^{2}(x)\right)^{2}cos(x)dx$

$\int_{0}^{\frac{\pi}{2}}\left(cos(x)-2cos(x)sin^{2}(x)+cos(x)sin^{4}(x)\right)dx$

$\int_{0}^{\frac{\pi}{2}}cos(x)dx-2\int_{0}^{\frac{\pi}{2}}cos(x)sin^{2}(x)dx+\int_{ 0}^{\frac{\pi}{2}}cos(x)sin^{4}(x)dx$

It is obvious the first one is just sin(x), since the integral of cos(x) is sin(x).

$\int_{0}^{\frac{\pi}{2}}cos(x)sin^{2}(x)dx$

Let $u=sin(x), \;\ du=cos(x)dx$

$2\int u^{2}du=\frac{2}{3}u^{3}=\frac{2}{3}sin^{3}(x)$

$\int_{0}^{\frac{\pi}{2}}cos(x)sin^{4}(x)dx$

Let $u=sin(x), \;\ du=cos(x)dx$

$\int u^{4}du=\frac{1}{5}u^{5}=\frac{1}{5}sin^{5}(x)=$

Put it all together:

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

Using integration limits which we could have done back in the sub by changing to u=0 and 1, but we'll do it this way.

$\underbrace{sin(\frac{\pi}{2})}_{\text{1}}-\overbrace{\frac{2}{3}sin^{3}(\frac{\pi}{2})}^{\te xt{2/3}}+\underbrace{\frac{1}{5}sin^{5}(\frac{\pi}{2})} _{\text{1/5}}=\boxed{\frac{8}{15}}$

$1-\frac{2}{3}+\frac{1}{5}=\frac{8}{15}$

We can ignore the 0 limit because it evaluates to 0 anyway.

I felt unusually productive this go around.