# Thread: [SOLVED] Using both u and trig substitution

1. ## [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.

2. On the first $\int sinx(1-cos^2x)^2cos^6x~dx$ and then expand it out

On the second use half angle identity again

3. 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

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

4. 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?

5. 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 $\int_{0}^{\frac{\pi}{2}}\cos^m\theta\sin^n\theta\, d\theta=\frac{1}{2}\beta\left(\frac{m+1}{2},\frac{ n+1}{2}\right)$)

6. $
-\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?

7. 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 .....

8. 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

9. 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?

10. 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

11. 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?

And what Moo said...

12. 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})
$

13. 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

!!!!!

If you prefer, values 0 and 1 are for u.
values 0 and pi/2 are for x.
Don't mix'em up

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

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

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

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

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

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

$=\frac {8}{693}$

15. 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