# a couple integrals

• Apr 17th 2008, 09:49 AM
rsanford
a couple integrals
1.
$\displaystyle \int \frac{sinx}{cos^3x} dx$

2. $\displaystyle \int x \sqrt {1-x} dx$

3. if F(x)= $\displaystyle \int_{0}^{\sqrt x} \sqrt {t^2+20} dx$ find F'(16)

Hi, any help would be wonderful. I really need the explanation more than the answer, though.
• Apr 17th 2008, 09:57 AM
PaulRS
1. Subsitute $\displaystyle u=\tan(x)$ remember that $\displaystyle u'=\frac{1}{cos^2(x)}$

2. Subsitute $\displaystyle u=\sqrt[]{1-x}$

3. Define $\displaystyle G(x)=\int_0^{x}{t^2+20}dt$ so that $\displaystyle G(\sqrt[]{x})=F(x)$(1)
Now differentiate in (1) using the chain rule and remember that $\displaystyle G'(x)=x^2+20$
• Apr 17th 2008, 09:59 AM
TKHunny

1. A substitution looks easy enough. u = ???

2. Begging for Integration by Parts if you want to learn something, but a simple substitution of everything under the radical might be a lot easier.

3. A derivative of an integral? Don't forget the Chain Rule. This one likely is ill-formed, Do you mean "dt"?

Let's see what you get.
• Apr 18th 2008, 11:21 AM
Moo
Hello,

For the first one, a substitution u=cos(x) is quite a valid way to solve it too ^^
• Apr 18th 2008, 12:20 PM
colby2152
Quote:

Originally Posted by Moo
Hello,

For the first one, a substitution u=cos(x) is quite a valid way to solve it too ^^