Use the sub x-1 = sqrt(5)tan(u) to evaluate 5∫3 1/(x^2-2x+6)^3/2
I don't know how to use the substitution
well, if tan u = (x-1)/sqrt5
then sin u= (x-1)/(x^2-2x+6)^1/2 (draw the triangle u out)
and sec u= (x^2-2x+6)^1/2 / sqrt5
and sec^2 u=(x^2-2x+6)/5
At this point you should be able now to substitute in the integral for x the u function.
x^2-2x+6 = (x-1)^2 + 5
if x-1 = √5 tanu then
(x-1)^2+5 = 5tan^2u+5 = 5sec^2u
and
dx = √5 sec^2u du
plug those into the integral and watch things simplify
Would it be possible to give me a more step by step? Im lost
x-1 = √5 tan(u)
x^2-2x+6 = 5sec^2(u)
dx = √5 sec^2(u) du
Now plug it into the integral
5∫3 1/(x^2-2x+6)^3/2 dx
5∫3 1/(5sec^2(u))^3/2 √5 sec^2(u) du
5∫3 (√5 sec^2(u))/(5√5 sec^3(u)) du
5∫3 1/(5 sec(u)) du
∫3 cos(u) du
3 sin(u) + C
now, if
x-1 = √5 tan(u),
tan(u) = (x-1)/√5
sin(u) = (x-1)/√(x^2-2x+6)
so the final answer is
3(x-1)/(x^2-2x+6) + C
Not sure just what the 5∫3 means, but if it's some kind of typo, I'm sure you can adjust things.
Oh, thank you. The 5∫3 just mean the upper and lower limts, I didn't know how else to write it
To use the substitution, we need to start by changing the variable in the integral. Let's solve the given equation for x in terms of u:
sub x - 1 = sqrt(5) * tan(u)
Adding 1 to both sides, we get:
x = sqrt(5) * tan(u) + 1
Now, let's find the derivative of x with respect to u to get dx in terms of du:
dx = sqrt(5) * sec^2(u) du
Next, substitute the value of x in the integral using the derived equation:
∫(1/(x^2 - 2x + 6)^3/2) dx
= ∫(1/((sqrt(5) * tan(u) + 1)^2 - 2(sqrt(5) * tan(u) + 1) + 6)^3/2) * (sqrt(5) * sec^2(u)) du
= ∫(1/((5tan^2(u) + 2sqrt(5)tan(u) + 1 + 6 - 2sqrt(5)tan(u) - 2)^3/2)) * (sqrt(5) * sec^2(u)) du
= ∫(1/((5tan^2(u) - 2sqrt(5)tan(u) + 5)^3/2)) * (sqrt(5) * sec^2(u)) du
Simplifying, we have:
= (1/5) ∫(1/(tan^2(u) - (2/sqrt(5)) tan(u) + 1)) * (sqrt(5) * sec^2(u)) du
Now, we can evaluate this integral using trigonometric identities or by applying partial fraction decomposition.