You can see the final answer here.
This is going to end up using the following major pieces of knowledge:
- Completing the Square
- Trigonometric Substitution
- The trick to intsecthetad theta∫secθdθ (involves u-substitution)
- Properties of Logarithms
int 1/sqrt(t^2 + t - 2)dt∫1√t2+t−2dt
t^2 + t - 2 = (t - 1)(t + 2)t2+t−2=(t−1)(t+2)
Although you CAN factor this, I'm going to complete the square instead. I want to get rid of the square root.
t^2 + t - 2 = t^2 + t + 1/4 - 1/4 - 2 = (t + 1/2)^2 - 9/4t2+t−2=t2+t+14−14−2=(t+12)2−94
=> int 1/sqrt((t + 1/2)^2 - 9/4)dt⇒∫1√(t+12)2−94dt
Surprisingly, now we have the form:
sqrt(x^2 - a^2)√x2−a2
Now things are looking up for Trigonometric Substitution. I was taught to use x = a"trig"thetax=atrigθ, so with x = t + 1/2x=t+12 and a = 3/2a=32, let:
t + 1/2 = 3/2secthetat+12=32secθ
(t + 1/2)^2 = 9/4sec^2theta(t+12)2=94sec2θ
dt = 3/2secthetatanthetad thetadt=32secθtanθdθ
sqrt((t+1/2)^2 - 9/4) = sqrt(9/4sec^2theta - 9/4) = 3/2sqrt(sec^2theta - 1) = 3/2sqrt(tan^2theta) = 3/2tantheta√(t+12)2−94=√94sec2θ−94=32√sec2θ−1=32√tan2θ=32tanθ
Now, we have:
=> int 1/(cancel(3/2)cancel(tantheta))cancel(3/2)secthetacancel(tantheta)d theta
= int secthetad theta
Not so bad, I guess... but you have to know this integral by now. Not because I expect that you should know more, but you'll see it eventually if you haven't already (it's pretty popular...). At least, I've been taught to do this integral. If you haven't learned it yet, here is how to do intsecthetad theta:
With malice aforethought... multiply by (sectheta+tantheta)/(sectheta+tantheta).
= int (sectheta(sectheta+tantheta))/(sectheta+tantheta)d theta
= int (sec^2theta+secthetatantheta)/(sectheta+tantheta)d theta
Now, let:
u = sectheta+tantheta
du = secthetatantheta + sec^2thetad theta
=> int 1/udu
WOAH! Much easier now!
= ln|u|
= ln|sectheta + tantheta| + C
Going back to our substitutions:
sectheta = 2/3(t+1/2)
tantheta = 2/3sqrt((t+1/2)^2 - 9/4) = 2/3sqrt(t^2 + t - 2)
So the final result is:
= ln|2/3(t+1/2) + 2/3sqrt(t^2 + t - 2)| + C
= ln|1/3(2(t+1/2) + 2sqrt(t^2 + t - 2))| + C
= ln|1/3(2t + 1 + 2sqrt(t^2 + t - 2))| + C
Now, with the properties of logarithms, separate ln out.
= ln|2sqrt(t^2 + t - 2) + 2t + 1| + ln(1/3) + C
but ln(1/3) = -ln 3, both of which are just constants. So now, we get a new constant D which we can embed into C and call it C again... oh well. It's all still a constant, and the point of it is to say that there are infinite vertical shifts of the antiderivative that can be differentiated to give 1/sqrt(t^2 + t - 2). In other words, the value of the constant is not relevant. Finally, we get:
= color(blue)(ln|2sqrt(t^2 + t - 2) + 2t + 1| + C)
