trig substitution for integration (1 Viewer)

[shsshs]

hi, my thinking has got me a little confused.

say you need to integrate

root(4 - x^2)

then i let x = 2sin[FONT='*Ｓ 明朝']θ , dx = 2cos[FONT='*Ｓ 明朝']θ d[FONT='*Ｓ 明朝']θ[/FONT][/FONT][/FONT]
[FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][/FONT][/FONT][/FONT]
[FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝']but when i simplify 2 x root(1 - sin^2 [FONT='*Ｓ 明朝']θ) , how do i know whether to put cos[FONT='*Ｓ 明朝']θ or -cos[FONT='*Ｓ 明朝']θ ???[/FONT][/FONT][/FONT][/FONT][/FONT][/FONT]
[FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT]
[FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝']i cant figure out the domain restrictions or whatever you need to determine it. I know the answer is +cos[FONT='*Ｓ 明朝']θ i juss dont know why[/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT]
[FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT]
[FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][FONT='*Ｓ 明朝'][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT]

 

[ianc]

no, you solve this simply using the trig identity sin²θ + cos²θ = 1, hence:
sqrt(1-sin²θ) = sqrt(cos²θ) = cosθ

edit: I understand your question now. The original integration question was sqrt(4 - x^2), not ±sqrt(4-x^2), hence it simplifies down to a positive cosθ

 

[pLuvia]

Int.sqrt{4-x²}dx
x=2sinθ dx=2cosθdθ
I=int.sqrt{4-4sin²θ}(2cosθdθ)
=int.2sqrt{1-sin²θ}(2cosθdθ)
=int.2sqrt{cos²θ}(2cosθdθ)
=int.2cosθ(2cosθdθ)
=int.4cos²θdθ
=4int.1/2(1+cos2θ)dθ
=2[θ+1/2sin2θ]+C
=2θ+sin2θ+C
=2sin^-1x/2+1/2(xsqrt{4-x²}+C

 

[A l]

hi, my thinking has got me a little confused.

say you need to integrate

root(4 - x^2)

then i let x = 2sin[FONT='ＪＳ 明朝']θ , dx = 2cos[FONT='ＪＳ 明朝']θ d[FONT='ＪＳ 明朝']θ[/FONT][/FONT][/FONT]
[FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][/FONT][/FONT][/FONT]
[FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝']but when i simplify 2 x root(1 - sin^2 [FONT='ＪＳ 明朝']θ) , how do i know whether to put cos[FONT='ＪＳ 明朝']θ or -cos[FONT='ＪＳ 明朝']θ ???[/FONT][/FONT][/FONT][/FONT][/FONT][/FONT]
[FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT]
[FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝']i cant figure out the domain restrictions or whatever you need to determine it. I know the answer is +cos[FONT='ＪＳ 明朝']θ i juss dont know why[/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT]
[FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT]
[FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][FONT='ＪＳ 明朝'][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT]

Think about the value of cos²θ and sin²θ. Since sin θ and cos θ can only be between -1 and 1 inclusive, cos²θ and sin²θ can only be between 0 and 1 inclusive. Therefore when you have 1 - sin²θ the answer is always between 0 and 1, thus it is always positive.

 

[_ShiFTy_]

Could you can say that 'θ' determines the sign of cosθ, so a ± is not necessary?

 

[pLuvia]

If you have this situation
sqrt{cos²θ}
=|cosθ|

Can this prove that it can only be positive cosθ?

 

[shsshs]

but going from root(cos^2 θ) to cosθ is what troubles me

for example since you are saying root(cos^2 θ) = cosθ

if you let θ = 180

then RHS = 1
LHS = -1

thats what bugs me which i still cant explain.

 

[zeek]

well if you look at the equation:

root(cos^2@)

This is the modulus of cos @ or the right side of the equation (cos @). So in other words, you can re-write the equation to be come...

|cos@|=cos@

This is why you get LHS = 1 and RHS = -1

 

[shsshs]

thankyou

so theres no case when you use the negative squareroot right?