Integration by Parts Homework Help

[Jbreezy]

Homework Statement

Hi, I attached the question.Just integral trouble.

Homework Equations

The Attempt at a Solution

 

[SteamKing]

I don't understand the expression you have written after "This is what I want". You have an integral in the denominator of an expression. I don't see how this is supposed to result in a recursion formula. Have you checked you attachment to make sure all expressions are correctly written?

 

[Jbreezy]

There isn't it is (n-1)/n (times) the integral It isn't supposed to be in the denominator.

∫〖sin〗^n x dx= (-〖sin〗^(n-1) xcosx)/n)+ ((n-1)/n)∫〖sin〗^(n-2) x dx

thx

 

[haruspex]

There isn't it is (n-1)/n (times) the integral It isn't supposed to be in the denominator.

∫〖sin〗^n x dx= (-〖sin〗^(n-1) xcosx)/n)+ ((n-1)/n)∫〖sin〗^(n-2) x dx

thx

I find that hard to read. Do you mean $\int \sin^n(x).dx = -\frac1n \sin^{n-1}(x)\cos(x) + \frac{n-1}n \int \sin^{n-2}(x).dx$? Looks right.

 

[Jbreezy]

Yeah that is what I mean.

See I'm trying to figure out how to get to that. On my attachment there is a problem before that where you get

\int \sin^n(x).dx = -\frac1n \sin^{n-1}(x)\cos(x) + \frac{n-1}n \int \sin^{n-2}(x)cos^{2}.dx
They tell you to replace cosx^2 in the second integral and get to
\int \sin^n(x).dx = -\frac1n \sin^{n-1}(x)\cos(x) + \frac{n-1}n \int \sin^{n-2}(x).dx

I'm having trouble with that.

 

[Jbreezy]

How come it doesn't work then I copy and pasted your code for the equation? Anyways my first equation in the attachment I need to get to the one you have written above. Just look at my attachment it is all there/

 

[haruspex]

For some reason, I can never read the .docx files you attach. I see the plain text, but equations and diagrams are blank.
In the latex, you left out the double hash (##) at start and end.

$\int \sin^n(x).dx = -\frac1n \sin^{n-1}(x)\cos(x) + \frac{n-1}n \int \sin^{n-2}(x)cos^{2}.dx$
They tell you to replace cosx^2 in the second integral and get to
$\int \sin^n(x).dx = -\frac1n \sin^{n-1}(x)\cos(x) + \frac{n-1}n \int \sin^{n-2}(x).dx$

I'm having trouble with that.

I don't understand how you got the first of those two equations. You should have
$\int \sin^n(x).dx = [-\sin^{n-1}(x)\cos(x)] + \int (n-1)\sin^{n-2}(x)\cos^2(x).dx$
Replacing the cos² in that gives:
$\int \sin^n(x).dx = [-\sin^{n-1}(x)\cos(x)] + (n-1)\int \sin^{n-2}(x)(1-\sin^2(x)).dx$
$\int \sin^n(x).dx = [-\sin^{n-1}(x)\cos(x)] + (n-1)\int \sin^{n-2}(x).dx - (n-1)\int \sin^{n}(x).dx$
$n\int \sin^n(x).dx = [-\sin^{n-1}(x)\cos(x)] + (n-1)\int \sin^{n-2}(x).dx$
etc.