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Integrals: Trig Substitution, or so I thought...
- Thread starter Overcore
- Start date
Hello, Overcore!
Under the radical, complete the square.
\(\displaystyle x^2 - 8x \color{red}{+ 16} \color{blue}{- 16} \;=\;(x^2-8x + 16) - 16 \;=\;(x-4)^2-16\)
The integral becomes: .\(\displaystyle \displaystyle 4\int\frac{dx}{\sqrt{(x-4)^2 - 16}}\)
Trig substitution: .let \(\displaystyle x-4 \:=\:4\sec\theta\)
Got it?
\(\displaystyle \displaystyle\int \frac{4}{\sqrt{x^2 - 8x}} dx\)
Under the radical, complete the square.
\(\displaystyle x^2 - 8x \color{red}{+ 16} \color{blue}{- 16} \;=\;(x^2-8x + 16) - 16 \;=\;(x-4)^2-16\)
The integral becomes: .\(\displaystyle \displaystyle 4\int\frac{dx}{\sqrt{(x-4)^2 - 16}}\)
Trig substitution: .let \(\displaystyle x-4 \:=\:4\sec\theta\)
Got it?
D
Deleted member 4993
Guest
almost. I'm left with the integral of secant, which is
ln(abs(sec(u) + tan(u)))
Now I have to go back into values of X, but how would I go about doing that?
Replacing the secant is easy since I know x-4 = 4sec(u)
but what about the tan?
tan2(Θ) = sec2(Θ) - 1
D
Deleted member 4993
Guest
Does "completing the square" work in all cases of trig substitution?
That's a tall order - nothing that I know of works in all the cases. Then again I do not know it all....
That's a tall order - nothing that I know of works in all the cases. Then again I do not know it all....
Figured out the Trig substituton stuff. Got the basic jist of it. So don't worry about it.
Basically, to sum it up. You look at the linear equation with a and x. You match it to a trig formula. You do the problem and usually it ends up as some different trig name. For instance, it might start out with sin and then the final answer will have sec. So then you match up the sec with it's corresponding sec intergal formula.