1. Mar 20, 2008

### BuBbLeS01

1. The problem statement, all variables and given/known data

Find the integral involving sin, cos, sec, and tan:
A. sin^3x * sqrt(cosx)dx
B. sec^3(2x) * tan(2x)dx

2. Relevant equations

3. The attempt at a solution
A. sin^3x * sqrt(cosx) dx
sin^2*x * sqrt(cosx) * sinx dx
(1-cos^2*x) * sqrt(cosx) * sinx dx
sinx * sqrt(cosx) - sinx * cosx dx

but not I don't know what to do??

2. Mar 20, 2008

### Gib Z

You did well, but just stop on the second last line, make a substitution u= sin x, then you have a simple integral in powers of u no?

3. Mar 20, 2008

### BuBbLeS01

So I stop here....

(1-cos^2*x) * sqrt(cosx) * sinx dx

???

I am not understanding how to use u-substitution with this? u = sinx, du = cosx but thats not it?

4. Mar 20, 2008

### Gib Z

SOrry sorry! I meant u=cos x, du = -sin x dx.

5. Mar 20, 2008

### BuBbLeS01

Umm but I am still not understanding lol im sorry...

(1-cos^2*x) * sqrt(cosx) * sinx dx

u=cos x, du = -sin x dx

So if I do u^1/2 the integral is 2/3 * u^3/2

but that only gets me....... sqrt(cosx) * -sinx dx

right??

6. Mar 20, 2008

### Gib Z

When you make the substitution,

$$\int (1- \cos^2 x) \sqrt{\cos x} \sin x dx = - \int (1-u^2) \sqrt{u} du = -\int ( u^{1/2} - u^{5/2}) du$$

You should be able to finish it off with the power rule, and replace back in u=cos x at the end.

Last edited: Mar 20, 2008
7. Mar 20, 2008

### BuBbLeS01

so if I leave it at...

1 - u^2 * sqrt(u) du

I could write it as

x - cosx^2 * sqrt(cosx)

8. Mar 20, 2008

### Gib Z

Did you follow my last post :( ? Do you know how to integrate those terms I had in the last post?

9. Mar 20, 2008

### BuBbLeS01

woops that wasn't done right...

(x - 1/3*cosx^3) * 2/3*sqrt(cosx)^3/2

can I write it like this...I know its not completely reduced yet but I will do that.

10. Mar 20, 2008

### Gib Z

Well when I evaluate the integral I posted, I get $$2\cos^{3/2} x \left( \frac{1}{3} - \frac{\cos^2 x}{7}} \right)$$

11. Mar 20, 2008

### BuBbLeS01

Why am I not understanding this....

(1-u^2) * 2/3 u^3/2

so I plug u=cosx back in and integrate...

(x - 1/3*cosx^3) * 2/3*(cosx)^3/2

12. Mar 20, 2008

### Gib Z

Do you follow post 6? I get my answer from post 6 by directly using the power rule.

13. Mar 20, 2008

### BuBbLeS01

I thought it would be....

u^3/2 - u^5/2

14. Mar 20, 2008

### rocomath

$$\int\sin^3 x\sqrt{\cos x}dx$$

$$\int(1-\cos^2 x)\sqrt{\cos x}\sin xdx$$

$$u=\cos x$$
$$du=-\sin xdx$$

$$-\int(1-u^2)\sqrt udu$$

$$-\int(u^{\frac 1 2}-u^{\frac 5 2})du$$

15. Mar 20, 2008

### Gib Z

Check the power rule again.

16. Mar 21, 2008

### BuBbLeS01

Okay I don't know why I am not seeing what you are doing but when you have...
(1-u^2) * sqrt u

don't you multiply u^1/2 by (1-u^2) ???

and get u^1/2 - u^3/2

so I don't understand why the second term is ^5/2?

17. Mar 21, 2008

### rocomath

BuBbLeS ... what is 1.5 and 2.5 in fractions?

18. Mar 21, 2008

### Snazzy

$$x^a\cdot x^b = x^a^+^b$$