Definite Intergral (subtuition)

[Weave]

Homework Statement

Evaluate:
$$\displaystyle \int_{\pi/6}^{\pi/2} \frac{\cos(z)}{\sin^{9}(z)}\, dz$$

Homework Equations

$$u=sin(z)\longrightarrow\space\,du=cos(z)dz$$

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.

 

[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.

 

[Weave]

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

 

[Dick]

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

 

[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?

 

[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!

 

[Weave]

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!

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!

 

[Dick]

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

 

[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$$\displaystyle \int_{9}^{16} 2u\sqrt{u}\,du$$ ??

 

[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)?

 

[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?

 

[Dick]

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