Reduction question (1 Viewer)

cutemouse · Apr 13, 2009

Hello,

I can do the showing parts, just don't know how to find the integral...

Could someone please help me?

Thanks

azureus88 · Apr 13, 2009

let x=cos@
dx=-sin@d@

x=0, @=pi/2
x=1, @=0

[maths]\int_{0}^{1}\frac{x^7dx}{\sqrt{1-x^2}}[/maths]

[maths]\int_{\pi /2}^{0}\frac{\cos^7\Theta (-\sin\Theta d\Theta )}{sin\Theta }[/maths]

[maths]\int_{0}^{\pi /2}\cos^7\Theta d\Theta[/maths]

then use the reduction formula from the previous part.

cutemouse · Apr 13, 2009

Hi, I got one more question. Could someone please help me on this one? Thanks

Drongoski · Apr 13, 2009

jm01 said:
Hi, I got one more question. Could someone please help me on this one? Thanks

That's easy!

jet · Apr 13, 2009

Continuing on for azureus
$\int^{\frac{\pi}{2}}_0 cos^7 \theta d\theta \\ = \left [ \frac{1}{7}cos^6\theta sin\theta \right ]^{\frac{\pi}{2}}_0 + \frac{6}{7}\int^{\frac{\pi}{2}}_0 cos^{5} \theta d\theta \\ = \frac{6}{7}\left ( \left [ \frac{1}{5}cos^4 \theta sin \theta \right ]^{\frac{\pi}{2}}_0 + \frac{4}{5}\int^{\frac{\pi}{2}}_0 cos^{3}\theta d\theta \right ) \\ = \frac{6}{7} \left ( \frac{4}{5} \left (\left[\frac{1}{3}cos^2 \theta sin \theta \right ]^{\frac{\pi}{2}}_0 + \frac{2}{3} \int^{\frac{\pi}{2}}_0 cos\theta d\theta \right ) \right ) \\ = \frac{6}{7}\left (\frac{4}{5} \left ( \frac{2}{3} \left (\left [ sin\theta \right ]^{\frac{\pi}{2}}_0 \right ) \right ) \right ) \\ = \frac{6}{7}\left ( \frac{4}{5} \left ( \frac{2}{3} \left [ 1 \right ] \right ) \right ) \\ = \frac{6 \cdot 4 \cdot 2 }{7 \cdot 5 \cdot 3} \\ = \frac{48}{105} = \frac{16}{35}$

jet · Apr 13, 2009

Heres the other question:
$sec^n x = sec^{n-2}xsec^2x \\ \text{but } sec^2 x = \frac{d}{dx} (tanx) \\ \therefore \int sec^n x \, dx = \int sec^{n-2}\frac{d}{dx}(tanx) \, dx \\ \\ \text{let } u = sec^{n-2}x, du = (n-2) sec^{n-3}x \cdot secxtanx = (n-2)sec^{n-2}tanx \\ dv = \frac{d}{dx} (tanx) dx, v = tanx \\ \therefore \int sec^{n}x \, dx = \int sec^{n-2}\frac{d}{dx}(tanx) \, dx = sec^{n - 2}x \, tanx - (n - 2)\int sec^{n - 2} x \, tan^2x \, dx \\ = sec^{n - 2}x \, tanx - (n - 2)\left [ \int sec^{n - 2} x \, (sec^2x - 1) \, dx \right ] \\ \int sec^n x \, dx = sec^{n - 2}x \, tanx - (n - 2) \int sec^n{x} \, dx + (n - 2) \int sec^{n-2} x \, dx \\ (n - 1) \int sec^n x \, dx = sec^{n - 2} x tanx + (n-2) \int sec^{n-2} x \, dx \\ \int sec^{n} x \, dx = \frac{sec^{n - 2} x tanx}{n - 1} + \frac{n - 2}{n - 1} \int sec^{n-2} x \, dx$

cutemouse · Apr 13, 2009

Hey Jetblack, thanks for your help!

Just one question though, how do I know when to apply integration by parts or to use the other 'miscellaneous' method?

Thanks

cutemouse · Apr 13, 2009

Hi again,

I have another question, lol.. It's a part of the one I previously posted.

Hence find ∫sec³x dx <-- I can do it until I have to find ∫secx dx, how would I do that?

Thanks for your help so far ppl!

azureus88 · Apr 13, 2009

[maths]\int \sec\Theta d\Theta [/maths]

[maths]=\int \frac{\sec\Theta (\sec\Theta +\tan\Theta )}{\sec\Theta +\tan\Theta }d\Theta [/maths]

[maths]=\ln(\sec\Theta +\tan\Theta )[/maths] as the derivative of demoninator is the numberator.

Alternatively, you could just use the t-formula.

Drongoski · Apr 13, 2009

$\int sec\ x\ dx\ \ =\ \ \int \frac{sec\ x(sec\ x+tan\ x)}{sec\ x \ +\ tan\ x}\ dx\\ \\ \ =\ \int \frac{sec^{2}\ x\ +\ secx\ tanx}{sec\ x\ +\ tan\ x}\ dx\ \ =\ \ ln\left |sec\ x\ +\ tan\ x \right |\ +\ C$

Edit: Above learnt from the text. I wonder how many of us could have figured this technique out ourselves. So it is one of those you 'memorise'.

cutemouse · Apr 13, 2009

azureus88 said:
[maths]\int \sec\Theta d\Theta [/maths]

[maths]=\int \frac{\sec\Theta (\sec\Theta +\tan\Theta )}{\sec\Theta +\tan\Theta }d\Theta [/maths]

[maths]=\ln(\sec\Theta +\tan\Theta )[/maths] as the derivative of demoninator is the numberator.

Alternatively, you could just use the t-formula.

Hi, thanks for your help... Just wondering though, how am I supposed to know this? lol... is there a 'list' of integrals like these that I should know and be able to manipulate?

Thanks

azureus88 · Apr 13, 2009

I guess you just know from experience. You should also learn the integral of cosec@ which uses the same method. If you're unsure though, you could still use the t formula as last resort

cutemouse · Apr 13, 2009

How would I use t results in this case? I tried and miserably failed

azureus88 · Apr 13, 2009

let t=tan@/2
dt/d@=(1/2)sec^2(@/2)=(1/2)(1+t^2)
d@=2dt/(1+t^2)

[maths]\int \sec\Theta d\Theta [/maths]

[maths]\int (\frac{1+t^2}{1-t^2})(\frac{2dt}{1+t^2})[/maths]

[maths]\int \frac{2dt}{(1-t)(1+t)}[/maths]

Then procede with partial fractions.

cutemouse · Apr 13, 2009

Yeah I got there, and got two log expressions, but I don't know how to make them into the other form (secx+tanx)

Sorry if this is really obvious, I'm pretty slow at this stuff

lolokay · Apr 13, 2009

use log law and the t identities

also, the integral of secx can be made into a standard integral if you let u=secx

azureus88 · Apr 13, 2009

You dont really need it in that form but if you want, you can do this:

[maths]\int (\frac{1}{1-t}+\frac{1}{1+t})dt [/maths]

[maths]=\ln(\frac{1+t}{1-t})[/maths]

[maths]=\ln[\frac{(1+t)^2}{1-t^2}][/maths]

[maths]=\ln(\frac{1+t^2}{1-t^2}+\frac{2t}{1-t^2})[/maths]

[maths]=\ln(\sec\Theta +\tan\Theta )[/maths]

However, you cant always expect the result to turn out neatly like this case, since in integration, the result may differ by a constant.

By the way, nice method lolokay.

cutemouse · Apr 14, 2009

Hi, Sorry but I have more questions...

1. If I_{m, n}=∫sin^mx cosⁿx dx, prove that
I_{m, n}=∫sin^m-1x (d/dx)[(-cosⁿ⁺¹x)/(n+1)] dx = [-sin^m-1x cosⁿ⁺¹x]/(n+1) + (m-1)/(n+1)[I_m-2,n - I_m,n]

Hence show that I_{m, n} = [-sin^m-1x cosⁿ⁺¹x]/(n+1) + (m-1)/(n+1)*I_m-2,n

2. If U_n=∫₀^pi/2 θsinⁿθdθ, n>1 prove that:

U_n=(n-1)/n * U_n-2 + 1/n²

3. If U_n=∫(cos2nx)/sinx dx, show that U_n=[2cos(2n-1)x/(2n-1)]+U_n-1

Thanks alot for your help!

Trebla · Apr 14, 2009

jm01 said:
3. If U_n=∫(cos2nx)/sinx dx, show that U_n=[2cos(2n-1)x/(2n-1)]+U_n-1

Thanks alot for your help!

$$Consider $ U_n - U_{n-1} = \displaystyle\int\dfrac{\cos 2nx}{\sin x}\,dx - \displaystyle\int\dfrac{\cos 2(n-1)x}{\sin x}\,dx\\= \displaystyle\int\dfrac{1}{\sin x}(\cos 2nx - \cos (2nx - 2x))\,dx\\= \displaystyle\int\dfrac{1}{\sin x}(\cos [(2nx-x)+x] - \cos [(2nx-x)-x])\,dx\\= \displaystyle\int\dfrac{1}{\sin x}(\cos (2nx-x)\cos x-\sin(2nx-x)\sin x -\cos (2nx-x)\cos x-\sin(2nx-x)\sin x)\,dx\\= -2\displaystyle\int\dfrac{1}{\sin x}(\sin(2nx-x)\sin x)\,dx\\= -2\displaystyle\int\sin((2n-1)x)\,dx\\= \dfrac{2\cos((2n-1)x)}{2n-1}\\\therefore U_n = \dfrac{2\cos((2n-1)x)}{2n-1} + U_{n-1}\\$Note the constant of integration is ignored because they are incorporated in the integrals $ U_n$ and $U_{n-1}$

azureus88 · Apr 14, 2009

jm01 said:
2. If U_n=∫₀^pi/2 θsinⁿθdθ, n>1 prove that:

U_n=(n-1)/n * U_n-2 + 1/n²

I dont like working with U_n and @ so i changed it to I_n and x. Anyway, here it is:

[maths]I_n=\int_{0}^{\pi/2}x\sin^nxdx\\=\int_{0}^{\pi/2}\sin^{n-1}x(x\sin x)dx\\=\int_{0}^{\pi/2}\sin^{n-1}x\frac{\mathrm{d} }{\mathrm{d} x}(-x\cos x+\sin x)dx\\=[-x\cos x\sin^{n-1}x+\sin^nx]-(n-1)\int_{0}^{\pi/2}(-x\cos x+\sin x)(\sin^{n-2}x\cos x)dx\\=1+(n-1)\int_{0}^{\pi/2}(x\sin^{n-2}x(1-\sin^2x)-\sin^{m-1}x\cos x)dx\\=1+(n-1)[I_{n-2}-I_n]-(n-1)\int_{0}^{\pi/2}\sin^{n-1}x\cos xdx\\=1+(n-1)[I_{n-2}-I_n]-\frac{n-1}{n}[\sin^nx]^{\pi/2}_0[/maths]

[maths]I_n(n-1+1)=1+(n-1)I_{n-2}-\frac{n-1}{n}\\I_n=\frac{1}{n}+\frac{n-1}{n}I_{n-2}-\frac{n-1}{n^2}\\=\frac{n-1}{n}I_{n-2}+\frac{1}{n^2}[/maths]

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