# Homework Help: Definite Intergral (subtuition)

1. Apr 30, 2007

### Weave

1. The problem statement, all variables and given/known data
Evaluate:
$$\displaystyle \int_{\pi/6}^{\pi/2} \frac{\cos(z)}{\sin^{9}(z)}\, dz$$

2. Relevant equations
$$u=sin(z)\longrightarrow\space\,du=cos(z)dz$$

3. The attempt at a solution
After making the substitution and simplifying I got the$$\displaystyle \int_a^b g(u)\,du$$ where:
g(u) =$$\frac{1}{u^9}$$
a =1
b = .5
Then I do $$\displaystyle\int_{.5}^{1}\frac{1}{u^9}\,du\longrightarrow\frac{-1}{8sin(z)^8}$$
I evaluated that on the interval [.5,1] but I got wrong answer, I know I went wrong somewhere.

Last edited: Apr 30, 2007
2. Apr 30, 2007

### Dick

Did you evaluate z from [.5,1] or u? You should evaluate u over that interval or if you want to change back to z, then use the z limits.

3. Apr 30, 2007

### Weave

Ah,
I think I evaluated:
$$\frac{-1}{8sin(z)^8}$$
z from .5 to 1.
so I evaluate u first?

4. Apr 30, 2007

### Dick

You already changed the z limits to u limits. You may as well just evaluate -1/(8*u^8) between them.

5. Apr 30, 2007

### Weave

Ah thanks,
One more question for you
How would I go about evaluating:
$$\displaystyle \int_0^1 x^2\sqrt{7 x + 9}\,dx$$
what sub would help me out?

6. Apr 30, 2007

### Dick

It would help you a lot. You would get sqrt(u) times a polynomial in u. And then everything would turn into fractional powers. Try it! You'll like it!

7. Apr 30, 2007

### Weave

LOL.

$$\displaystyle \int_0^1 x^2\sqrt{7 x + 9}\,dx$$
$$u=7x+9\longrightarrow\,du=7dx$$
So
$$\displaystyle \int_{9}^{16} x^2\sqrt{u}\,du$$
What do I do to rid of the x^2!

Last edited: Apr 30, 2007
8. Apr 30, 2007

### Dick

Solve for x in terms of u. BTW what happened to the 1/7 as in dx=du/7.?

9. Apr 30, 2007

### Weave

$$\frac{1}{7}\displaystyle \int_{9}^{16} x^2\sqrt{u}\,du$$
Oh you mean that one?

So do you mean:
$$\frac{d}{du}=2u$$?
then$$\displaystyle \int_{9}^{16} 2u\sqrt{u}\,du$$ ??

10. Apr 30, 2007

### Dick

Yes, I mean that one. But on the other question I mean why not put x=(u-9)/7? So x^2=(u-9)^2/49. Then expand and multiply by u^(1/2)?

11. Apr 30, 2007

### Weave

Ahhh I see what you mean:
$$\frac{1}{7}\displaystyle \int_{9}^{16} x^2\sqrt{u}\,du$$
$$x=(u-9)/7$$So:
$$\frac{1}{7}\displaystyle \int_{9}^{16} (\frac{u-9}{7})^2\sqrt{u}\,du$$
$$\frac{1}{7}\displaystyle \int_{9}^{16} (\frac{(u-9)^2}{49})\sqrt{u}\,du$$
I then can expand and evaluate from there, correct?

12. Apr 30, 2007

### Dick

Sure then you'll just get stuff like u^(5/2) etc. Easy to handle for one of your obvious abilities! Sorry, gotta go. zzzzzzz.