# Homework Help: : Integrate exactly

1. Jul 21, 2007

URGENT: Integrate exactly

1. The problem statement, all variables and given/known data
Using the substitution $$x = \sec ^2 u$$, evaluate $$\int_2^4 {\frac{{dx}}{{x^2 \sqrt {x - 1} }}}$$ exactly

2. The attempt at a solution
I have used the substitution and ended up with (after some simplification)
$$\int_2^4 { - 2\cos ^2 u.\sin u.\sqrt {\sec ^2 u - 1} }$$. What do i do now? How do I integrate that, I am very confused on how to go about this now. There are others which I need to solve, but are harder than this one, so I will post them in the future once i understand this one and if i still need help with them. Many thanks to those who help

2. Jul 21, 2007

### chaoseverlasting

$$sec^2u-1=tan^2u$$. That gives you $$-2cosu sin^2u$$. Substitute t=sinu, dt=cosu du[/tex]. You should be able to do the rest.

3. Jul 21, 2007

### VietDao29

Well, when you perform a u-substitution, you should also change the lower limit, and upper limit for the integral.

x = sec2u, where u is restricted to be on the interval $$\left[ 0 ;\ \frac{\pi}{2} \right[$$, so that tan(u) can be positive.

x = sec2u ~~~> dx = (2 sinu)/(cos3u) du
x = 2 ~~~> 1/cos2u = 2 ~~~> cos2(u) = 1/2 ~~~> cos(u) = 1/sqrt(2) ~~~> u = pi/4

x = 4 ~~~> 1/cos2u = 4 ~~~> cos2(u) = 1/4 ~~~> cos(u) = 1/2 ~~~> u = pi/3

So, your integral will become:

$$\int_{\frac{\pi}{4}} ^ \frac{\pi}{3} \ \cos ^ 4 u \frac{2 \sin u}{\cos ^ 3 u} \frac{du}{\sqrt{\sec ^ 2 u - 1}} = 2 \int_{\frac{\pi}{4}} ^ \frac{\pi}{3} \ \cos u \sin u \frac{du}{\sqrt{\tan ^ 2 u}}$$

$$= 2 \int_{\frac{\pi}{4}} ^ \frac{\pi}{3} \ \cos u \sin u \frac{du}{|\tan u|}$$

Since $$u \in \left[ 0 ;\ \frac{\pi}{2} \right[$$, tan(u) will be non-negative, so, we have:

$$= 2 \int_{\frac{\pi}{4}} ^ \frac{\pi}{3} \ \cos u \sin u \frac{du}{\tan u} = ...$$

Can you go from here? :)

4. Jul 21, 2007