Hi,

How can you use it to evaluate I₈ for I_n=ò₀^p/2 cosⁿ x dx?
I was not sure how to evaluate it. Hope someone can show me how. Then I shall go and do another example myself.

Hal.

You have the correct reduction formula. But you also have limits. So what is cos^n-1 x sin x when evaluated at the limits? Zero. So in fact the reduction formula is:

n I_n = (n-1)I_n-2

Does this make it easier?

Yes! Thank you! It works out much neater to evaluate!

Hi again,

When I got down to the term I₁, I said, let I₁=òcos x dx=[sinp -sin0]=0
So when I multiplied the other terms the I₉ became = 0. Where have I gone wrong?

Up to the term I₃ , I got it to be 128/315. Is there a restriction for n? Is that why its goes wrong? Could someone just confirm this for me. I think it is because of our original nI_n = (n-1)I_n-2 ?

The answer is correct. The point is cosx=sin(p/2+x), so if we make a change of variable, we see that we are integrating an odd function (sin⁹ y) with symmetric limits (-p/2 to p/2), hence zero.

Kerwin

Thanks Kerwin.

Hello there,

I am having trouble with this question. Some hints would be nice.

I_n = [(2^n-2 )/(n-1)]3^1/2 + [(n-2)/(n-1)]I_n-2

Hence evaluate I₇ .

How do I show what is required? What is the best way to break it up? Thanks.

Hal.

Try writing it as sec² x x sec^n-2 x, then using parts to obtain the reduction formula.

Hi,

I tried that previously but obviously I made a mess of it. I thought there might be a different way to do it. OK, I'll try it again and get back to you.

Hal

Hi,

How would you go about doing this type:

These things are often easier to manipulate in trigonometric form. Particularly if the integral is between 0 and 1 or something nice like that. So try substituting something like x = sin t. You should find this reduces it to one of your previous reduction problems. Did you have any luck with the secⁿ x reduction formula?

Hi Michael,

I just tried the Secⁿ x reduction again. I got it to the general form (n-1)I_n = tanx.sec^n-2 x + (n-2)I_n-2 form. Is that correct? I am going to put in the boundary conditions to see whether I get it in the form that they require.

I haven't tried the sub you suggested yet. But I will and will let you know how it goes. Thanks also for the general tip MIchael.

Hal.

Hi Michael,

x=sin² t sounds good to me. I amassuming the integralis from 0 to 1,so t goes from 0 to p/2. We get:

I_n=ò₀^p/2sinⁿ tcost×2sintcost dt=2ò₀^p/2 sinⁿ⁺¹ tcost×cost dt.

because sin and cos are positivein this range.

This now lends itselfto integration by parts, with the separation occurring at the ×.

I_n=2sinⁿ⁺²t×cost/(n+2)+2ò₀^p/2sinⁿ⁺³ t dt/(n+2)

The first bit vanishes when the limits are put in. So:

(n+2)I_n=2ò₀^p/2 sinⁿ⁺³ t dt

And:

(n) I_n-2-(n+2)I_n=2ò₀^p/2sinⁿ⁺¹ tcos² t dt=I_n

and so n I_n-2=(n+3)I_n as required.

You probably can do this directly without making the trig substitution but I find trig expressions much easier to manipulate.