Simple Integral, Still Having Trouble

[Blues_MTA]

Homework Statement

I am trying to take the integral of 1/((-4t²+4t+3)^1/2)

I know that i need to complete the square and it should come out to an inverse sin function but i don't understand how the completed square in the denominator is equal to (4-(2t-1)²)^1/2

Homework Equations

The Attempt at a Solution

 

[issacnewton]

just expand the square and you see they are equal

 

[Blues_MTA]

But how do you come to that? I understand its equal, i just don't know how they calculated it, am I missing something completely obvious or...?

 

[issacnewton]

have you studied how to complete square ? if you have a quadratic equation

ax^2+bx+c = 0 then you add and subtract middle term square divided by

4 times the first term

ax^2+bx+c+\frac{b^2x^2}{4ax^2}-\frac{b^2x^2}{4ax^2}+c=0

a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2})=\frac{b^2}{4a}-c

a(x+\frac{b}{2a})^2=\frac{b^2}{4a}-c

so use this procedure. in your case there is just a quadratic term, no equation. but
this approach can still be used

 

[Blues_MTA]

Yeah, I have, and for some reason this problem completely stumps me, If you use those formulas , you end up with 4(t+1/2)^2 = -2

I don't see how that is equivalent to 4-(2t-1)^2

 

[issacnewton]

-4t^2+4t+3

-4t^2+4t +\frac{(4t)^2}{4(-4t^2)}-\frac{(4t)^2}{4(-4t^2)}+3

-4t^2+4t-1+1+3

-(4t^2-4t+1)+4

4-(4t^2-4t+1)

4-(2t-1)^2

\smile

 

[Blues_MTA]

Ok, never mind i see that they are equivalent but how do you know to rearrange them in that manner to take the integral? is there any sort of method or is it just through practice

 

[Blues_MTA]

oooook..Im sorry I've been so troublesome, thank you so much!

 

[issacnewton]

you have to arrange the integrand in the form

\frac{1}{\sqrt{a^2-x^2}}

since there is standard integral in that form

http://www.sosmath.com/tables/integral/integ13/integ13.html

 

[issacnewton]

mta, i think you should REALLY know how to complete squares BEFORE you take on calculus. have you had pre-calculus ?