- #1
Nano-Passion
- 1,291
- 0
Hey. I was reviewing some trig substitution and I know how to solve the problem but I was curious if you can also solve it by using two different substitutions/change of varaibles? If so, how would you relate the functions? Consider for example:
∫ (1-x^2)^1/2 dx, -1<= x <= 1
x = sin Θ
dx = cos Θ dΘ
-pi/2 <= Θ <= pi/2
∫ (1-x^2)^1/2 dx = ∫ 1 - sin^2Θ dΘ
u = 2Θ
du = 2 dΘ
-pi <= u <= pi
Then the integral is equal to:
Θ - 1/2Θ - 1/2 sin u(Θ(x)) + C
where Θ(x) = sin^-1x.
Would u(Θ(x)) = 2arcsinx be true?
Which would give us:
arcsin x - 1/2arcsin x - 1/2 sin (2arcsin x) + C
However ugly it is, would this be correct with the adjusted interval? Does it make sense?
∫ (1-x^2)^1/2 dx, -1<= x <= 1
x = sin Θ
dx = cos Θ dΘ
-pi/2 <= Θ <= pi/2
∫ (1-x^2)^1/2 dx = ∫ 1 - sin^2Θ dΘ
u = 2Θ
du = 2 dΘ
-pi <= u <= pi
Then the integral is equal to:
Θ - 1/2Θ - 1/2 sin u(Θ(x)) + C
where Θ(x) = sin^-1x.
Would u(Θ(x)) = 2arcsinx be true?
Which would give us:
arcsin x - 1/2arcsin x - 1/2 sin (2arcsin x) + C
However ugly it is, would this be correct with the adjusted interval? Does it make sense?
