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Date Posted: 7/24/2008 1:18:57 AM  Status: Live
find antiderivative using trigonometric substitution
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i know that that the above is arcsin(x) but im confused because my teacher said the next step is the integral of sin(u)^2 and im confused where the sin^(x) came from. any help would be appreciated.
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Guru
Karma Points: 1,730
(University of Kentucky)
Date Posted: 7/24/2008 1:33:43 AM  Status: Live
Asker's Rating: Helpful   
Response:
First let   then   and substituting back in we have....
 
 
Remembering the indentity    we have that 
 
∴ 
 
 
Now we can solve this integral by first noting the following identity....
 
 
∴ Substituting back in we have....
 
 
We break this up into two integrals remembering to carry the negative sign....
 
 
 
Now since we originally set    we see that  so substituting back in we have our final answer of .......
 
 
    ANSWER
 
Hope this helps!
 
Jonathan
 
Jonathan Ross
University of Kentucky
B.A. Mathematical Economics

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Sage
Karma Points: 5,455
(Melbourne University)
Date Posted: 7/24/2008 1:59:35 AM  Status: Live
Asker's Rating: Helpful   
Response:
The ANSWER ABOVE IS DEAD WRONG
I WOULD SERIOUSLY ADVISE THE PERSON ABOVE IF U CANT DO THE INTEGRATION RIGHT THEN DON'T DO IT. IT'S THE SECOND TIME IN ONE DAY I HAVE FOUND HIM DOING WRONG


:
Let  ..........................................>
 
∴
 
Using the indentity  ..................................................>
 
∴ 
 
 
Using double angle formula sin2x=1/2(1-cos2x).
 
∴ Substituting back in we have....
 
 

 
.....................................>convert sin2θ using double angle formula sin2θ=2sinθcosθ.........................................>∴
 
  ........................................................>
∴
 
 
 

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Oracle
Karma Points: 8,949
Date Posted: 7/24/2008 5:54:19 AM  Status: Live
Asker's Rating: Lifesaver   
Response:
The point of trig substitution is to get rid of the square root.
Relate the square root to the pythagorean theoreom.
a2+b2 = c2
To get a minus sign under a square root, the square root must be one of the legs of a triangle.
a = √(c2 - b2)
so: a = √(12 - x2) and a = leg, 1 = hypotenuse, x = leg
For trig substitution, I always set up the triangle for the square root right away because you need to substitute everything out in terms of θ anyway to be able to integrate. 

Generally, you need to write three things from this triangle.
1) substitution for the square root
2) substitution for x
3) substitution for dx

1) = 1/cosθ
2)
3) [from equation 2] dx = cosθdθ

Substituting in θ expressions in place of x expressions:

Simplify:

power-reducing identity:
= =
= (1/2)θ -
u-sub on second integral:
u = 2θ
du = 2dθ => du/2 = dθ
= (1/2)θ -
= (1/2)θ - (1/4)sin(u) = (1/2)θ - (1/4)sin(2θ) + C
Simplify [double angle identity: sin(2θ) = 2sinθcosθ]: = (1/2)θ - (1/2)sinθcosθ + C

Resubstitute in x-expressions by using trig identities from the triangle:
You get θ from your x = expression: x = sinθ; sin-1(x) = θ
= (1/2)sin-1(x) - (1/2)(x)(√(1-x2)) + C
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