# [SOLVED] Using both u and trig substitution

• Sep 8th 2008, 10:43 AM
redman223
[SOLVED] Using both u and trig substitution
$
\int_0^\frac {\pi}{2} sin^5x~cos^6x~dx
$

I know I am supposed to strip one layer away, in this case from sin, to give me:
$
\int_0^\frac {\pi}{2} sin^4x~cos^6x~sinx~dx
$

But I don't know what to do next, only that it will probably involve u substitution.
• Sep 8th 2008, 10:48 AM
11rdc11
On the first $\int sinx(1-cos^2x)^2cos^6x~dx$ and then expand it out

On the second use half angle identity again
• Sep 8th 2008, 10:53 AM
Matt Westwood
You can solve this by parts, but it's tedious.

Pick either the sin or cos part to be u, and the other to be dv. You will end up with a bit you can evaluate and another bit you can integrate by parts again.

Then you integrate by parts again and you get another integration with reduced indices.

You will be able to deduce a rule which goes something like this (from my book of integrals):

$\int \sin^m x \cos^n dx = \frac{\sin^{m+1} x \cos^{n-1} x}{m+n}+ \frac {n-1}{m+n}\int \sin^m x \cos^{n-2} x dx$

or

$\int \sin^m x \cos^n dx = \frac{\sin^{m-1} x \cos^{n+1} x}{m+n}+ \frac {m-1}{m+n}\int \sin^{m-2} x \cos^n x dx$
• Sep 8th 2008, 11:02 AM
redman223
Quote:

Originally Posted by 11rdc11
On the first $\int sinx(1-cos^2x)^2cos^6x~dx$ and then expand it out

On the second use half angle identity again

I tried that and ended up with 3 separate integrals that don't seem any easier, did I do something wrong?
• Sep 8th 2008, 11:22 AM
TwistedOne151
Once to $\int(1-\cos^2x)^2\cos^6{x}\sin{x}\,dx$
Use u-substitution: $u=\cos{x}\;\to\;du=-\sin{x}\,dx$
so that
$\int(1-\cos^2x)^2\cos^6{x}\sin{x}\,dx=-int(1-u^2)^2u^6\,du$
Now expand, and pay attention to the change in the limits when doing the u-substitution

--Kevin C.
(Or if you knew about the beta and gamma functions, you could use the formula $\int_{0}^{\frac{\pi}{2}}\cos^m\theta\sin^n\theta\, d\theta=\frac{1}{2}\beta\left(\frac{m+1}{2},\frac{ n+1}{2}\right)$)
• Sep 8th 2008, 12:32 PM
redman223
$
-\int_0^\frac{\pi}{2} u^6-2u^8+u^{10}~du
$

This is what I came up with so far.

What do you mean about the change in limits?
• Sep 8th 2008, 12:39 PM
Chop Suey
Quote:

Originally Posted by redman223
What do you mean about the change in limits?

Since $u = \cos{x}$, the upper and lower limits change according to your substitution. So, with the sub you made, the limits will change to:
from 1 to .....
• Sep 8th 2008, 12:40 PM
Moo
Quote:

Originally Posted by redman223
$
-\int_0^\frac{\pi}{2} u^6-2u^8+u^{10}~du
$

This is what I came up with so far.

What do you mean about the change in limits?

It's x going from 0 to $\frac \pi 2$

If you change x ---> u=cos(x)

The boundaries of the integrals *have to* be the ones for u, not for x.

If x=0 ---> u=...
If x=pi/2 ---> u=...

These are the new boundaries/limits of the integral :)

So either you change it, either you substitute back u by cos x the integral you will get. But here, it'll be easier to change the boundaries/limits

NB : what you've got so far is correct
• Sep 8th 2008, 12:41 PM
redman223
Ok thanks, now I get it, I have to plug in the limits to cosx and re-evaluate the limits right? So it should be from 1 to 0 right?
• Sep 8th 2008, 12:44 PM
Moo
Quote:

Originally Posted by redman223
Ok thanks, now I get it, I have to plug in the limits to cosx and re-evaluate the limits right? So it should be from 1 to 0 right?

Yes !

Note that $\int_1^0 ~dx$ (it has to be in the same order as the old boundaries !) is $-\int_0^1 ~dx$, in case it's easier for you to deal with this order of boundaries :p
• Sep 8th 2008, 12:44 PM
Chop Suey
Quote:

Originally Posted by redman223
Ok thanks, now I get it, I have to plug in the limits to cosx and re-evaluate the limits right? So it should be from 1 to 0 right?

(Clapping)

And what Moo said... :p
• Sep 8th 2008, 01:07 PM
redman223
I can't seem to come up with the right answer, but this is what I have so far:

$
-(\frac {cos^70}{7} - \frac {cos^71}{7}) + (\frac {2cos^90}{9} - \frac {2cos^91}{9}) - (\frac {cos^{11}0}{11} - \frac {cos^{11}1}{11})
$
• Sep 8th 2008, 01:08 PM
Moo
Quote:

Originally Posted by redman223
I can't seem to come up with the right answer, but this is what I have so far:

$
-(\frac {cos^70}{7} - \frac {cos^71}{7}) + (\frac {2cos^90}{9} - \frac {2cos^91}{9}) - (\frac {cos^{11}0}{11} - \frac {cos^{11}1}{11})
$

Nooooo !

You have two possibilities :
- keep u in the answer and use boundaries 0 and 1
- substitute back u by cos and keep boundaries 0 and pi/2

!!!!! :p

If you prefer, values 0 and 1 are for u.
values 0 and pi/2 are for x.
Don't mix'em up
• Sep 8th 2008, 01:09 PM
Shyam
Quote:

Originally Posted by redman223
$
\int_0^\frac {\pi}{2} sin^5x~cos^6x~dx
$

I know I am supposed to strip one layer away, in this case from sin, to give me:
$
\int_0^\frac {\pi}{2} sin^4x~cos^6x~sinx~dx
$

But I don't know what to do next, only that it will probably involve u substitution.

$\int_0^\frac {\pi}{2} sin^5x~cos^6x~dx$

$=\int_0^\frac {\pi}{2} (1-\cos^2x)^2\cos^6{x}\sin{x}~dx$

$=\int_0^\frac {\pi}{2} (1-\cos^2x)^2\cos^6{x}\sin{x}~dx$

substitute, $u=\cos{x}\;\to\;du=-\sin{x}\,dx\; \Rightarrow -du=\sin{x}\,dx$

$as \; x\to \;0,\; u \to 1,\; and \; as \; x\to {\frac {\pi}{2}},\;\; u \to 0$

$=-\int_1^0 (1-u^2)^2 u^6~du$

$=-\int_1^0 (1-2u^2+u^4) u^6~du$

$=-\int_1^0 (v^6-2u^8+u^{10})~du$

$=-(0-0+0)+\left( \frac {1}{7} -\frac {2}{9} + \frac{1}{11} \right)$

$=\frac {8}{693}$
• Sep 8th 2008, 02:42 PM
Chris L T521
Quote:

Originally Posted by redman223
$
\int_0^\frac {\pi}{2} sin^5x~cos^6x~dx
$

I know I am supposed to strip one layer away, in this case from sin, to give me:
$
\int_0^\frac {\pi}{2} sin^4x~cos^6x~sinx~dx
$

But I don't know what to do next, only that it will probably involve u substitution.

See here to learn how to tackle integrals of this type.

--Chris