1. ## Integrating pi x/2

Hi,

I'm looking for help with what to do when you have a division over pi x.
I need to intergrate the following between 1 and 0

cos (pi x/2) dx

I'm fine with intergrating it as cos pi x. But I'm stuck with what to do when adding in the /2 under the pi x.

any help would be greatly appreciated! thanks!

2. Hello, Gill8752!

It's a simple substitution . . .

$\displaystyle \int^1_0\cos\left(\frac{\pi}{2}\,x\right)\,dx$

Let: $\displaystyle u \:=\:\frac{\pi}{2}\,x \quad\Rightarrow\quad du \:=\:\frac{\pi}{2}\,dx \quad\Rightarrow\quad dx \:=\:\frac{2}{\pi}\,du$

Substitute: .$\displaystyle \int \cos u \left(\frac{2}{\pi}\,du\right) \;=\;\frac{2}{\pi}\int \cos u\,du$

Got it?

3. I'm not really sure I do following you. I haven't done any substitution in lesson's. I've only been shown chain rules for standard integrals

eg cos(ax+b) = 1/a sin(ax+b)

4. Originally Posted by Gill8752
I'm not really sure I do following you. I haven't done any substitution in lesson's. I've only been shown chain rules for standard integrals

eg cos(ax+b) = 1/a sin(ax+b)

Put $\displaystyle a= \frac{\pi}{2}$ and b=0 then.

5. So would that would make it

cos pi x/2 = 1/pi/2 sin (pi x +b)

That doesnt seem right to me?

p.s how do you get your equations to look right? what program do you use?

6. I've tried it another way - does this make sense or is it totally wrong!??

$\displaystyle \int^1_0\cos\left(\frac{\pi x}{2}\right)\,dx$

into

$\displaystyle \int^1_0\sin\left(\frac{\pi x^2}{2x}\right)\,dx$

There should be a 1/2 infront of the pi within the bracket but i couldnt figure out how to do that.

7. Originally Posted by Gill8752
I've tried it another way - does this make sense or is it totally wrong!??

$\displaystyle \int^1_0\cos\left(\frac{\pi x}{2}\right)\,dx$

into

$\displaystyle \int^1_0\sin\left(\frac{\pi x^2}{2x}\right)\,dx$

There should be a 1/2 infront of the pi within the bracket but i couldnt figure out how to do that.
Totally wrong.

You need to use the rule that you yourself more or less stated (see post #3):

$\displaystyle \int \! \cos (ax+b) \, dx = \frac{1}{a} \, \sin (ax+b)$ (I have ignored the arbitrary constant of integration since you're using this result to get a definite integral).

In your case $\displaystyle a = \frac{\pi}{2}$ and $\displaystyle b = 0$.

8. So am i right in thinking it would be this...

$\displaystyle \int\frac{\pi }{2}\, \sin ( \frac{\pi}{2} x +b)$

9. Originally Posted by Gill8752
So am i right in thinking it would be this...

$\displaystyle \int\frac{\pi }{2}\, \sin ( \frac{\pi}{2} x +b)$
Sorry but you seem to have several fundamental misunderstandings and have made several mistakes.

1. There should be no integral sign as you've already done the integration.

2. There should be no $\displaystyle b$ since if you look at what I posted you ought to realise that $\displaystyle b = 0$ in your case. The formula I gave you was a general case.

3. $\displaystyle a = \frac{\pi}{2}$, so how can $\displaystyle \frac{1}{a} =\frac{\pi}{2} ?$ It doesn't.