can anyone take me through this answer plz?
use substitution x = sin& to integrate :
1 / (1  x²) ^ 3/2 dx.
Hard Integration Question!!!

 Follow
 1
 05042008 15:15

 Follow
 2
 05042008 15:17
Well you've been given the substitution so what have you done so far?

 Follow
 3
 05042008 15:17
(Original post by Eager PPe ist)
can anyone take me through this answer plz?
use substitution x = sin& to integrate :
1 / (1  x²) ^ 3/2 dx.
then you must differentiate x = sin& to get dx/d&=cos&
know what to do next? 
 Follow
 4
 05042008 15:24
The way you have set out the question is not very clear.
If you mean:
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.\int{\frac{1}{(1x^2)^\frac{3}{2}}dx
Then using a substitution of think trig identities.Last edited by AEsp; 05042008 at 15:58. 
 Follow
 5
 05042008 15:27
that's what i mean AEsp.
i have used tirg identities and have integrated and have arrived here:
I = 0.5sin2x ( 1 + cos2x) ^ 1/2 
 Follow
 6
 05042008 15:49
Ok then i believe your answer is wrong.
Then using substitution x = sin u
and trig identity
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.\frac{dx}{du} = cos u
\displaystyle \int {(cos^2u)^{\frac{3}{2}}cos u\, du
\displaystyle = \int \frac{cos u}{\sqrt {(cos^2u)^3}}\, du
\displaystyle = \int \frac{cosu}{cos^3u}\, du
\displaystyle = \int \frac{1}{cos^2u}\, du
\displaystyle = \int sec^2u\, du
Im sure you recognise the integral ofLast edited by AEsp; 05042008 at 16:19. 
 Follow
 7
 08092011 12:19
http://www.wolframalpha.com/input/?i=integrate+%281%2F%281++x%C2%B2%29%29^%283%2F2%29+dx.+
integral 1/(1/(1x^2))^(3/2) dx
Expanding the integrand 1/(1/(1x^2))^(3/2) gives 2 sqrt(1/(1x^2)) x^2+sqrt(1/(1x^2))+sqrt(1/(1x^2)) x^4:
= integral (2 sqrt(1/(1x^2)) x^2+sqrt(1/(1x^2))+sqrt(1/(1x^2)) x^4) dx
Integrate the sum term by term and factor out constants:
= integral sqrt(1/(1x^2)) dx2 integral x^2 sqrt(1/(1x^2)) dx+ integral x^4 sqrt(1/(1x^2)) dx
For the integrand x^2 sqrt(1/(1x^2)), simplify powers:
= integral sqrt(1/(1x^2)) dx2 integral x^2/sqrt(1x^2) dx+ integral x^4 sqrt(1/(1x^2)) dx
For the integrand, x^2/sqrt(1x^2) substitute x = sin(u) and dx = cos(u) du. Then sqrt(1x^2) = sqrt(1sin^2(u)) = cos(u) and u = sin^(1)(x):
= 2 integral sin^2(u) du+ integral sqrt(1/(1x^2)) dx+ integral x^4 sqrt(1/(1x^2)) dx
For the integrand x^4 sqrt(1/(1x^2)), simplify powers:
= 2 integral sin^2(u) du+ integral sqrt(1/(1x^2)) dx+ integral x^4/sqrt(1x^2) dx
For the integrand, x^4/sqrt(1x^2) substitute x = sin(s) and dx = cos(s) ds. Then sqrt(1x^2) = sqrt(1sin^2(s)) = cos(s) and s = sin^(1)(x):
= integral sin^4(s) ds2 integral sin^2(u) du+ integral sqrt(1/(1x^2)) dx
For the integrand sqrt(1/(1x^2)), simplify powers:
= integral sin^4(s) ds2 integral sin^2(u) du+ integral 1/sqrt(1x^2) dx
The integral of 1/sqrt(1x^2) is sin^(1)(x):
= integral sin^4(s) ds2 integral sin^2(u) du+sin^(1)(x)
Write sin^2(u) as 1/21/2 cos(2 u):
= integral sin^4(s) ds2 integral (1/21/2 cos(2 u)) du+sin^(1)(x)
Integrate the sum term by term and factor out constants:
= integral sin^4(s) ds2 integral 1/2 du+ integral cos(2 u) du+sin^(1)(x)
The integral of 1/2 is u/2:
= integral sin^4(s) dsu+ integral cos(2 u) du+sin^(1)(x)
For the integrand cos(2 u), substitute p = 2 u and dp = 2 du:
= 1/2 integral cos(p) dp+ integral sin^4(s) dsu+sin^(1)(x)
The integral of cos(p) is sin(p):
= (sin(p))/2+ integral sin^4(s) dsu+sin^(1)(x)
Use the reduction formula, integral sin^m(s) ds = (cos(s) sin^(m1)(s))/m + (m1)/m integral sin^(2+m)(s) ds, where m = 4:
= (sin(p))/21/4 sin^3(s) cos(s)+3/4 integral sin^2(s) dsu+sin^(1)(x)
Write sin^2(s) as 1/21/2 cos(2 s):
= (sin(p))/21/4 sin^3(s) cos(s)+3/4 integral (1/21/2 cos(2 s)) dsu+sin^(1)(x)
Integrate the sum term by term and factor out constants:
= (sin(p))/21/4 sin^3(s) cos(s)+3/4 integral 1/2 ds3/8 integral cos(2 s) dsu+sin^(1)(x)
For the integrand cos(2 s), substitute w = 2 s and dw = 2 ds:
= (sin(p))/21/4 sin^3(s) cos(s)+3/4 integral 1/2 dsu3/16 integral cos(w) dw+sin^(1)(x)
The integral of 1/2 is s/2:
= (sin(p))/2+(3 s)/81/4 sin^3(s) cos(s)u3/16 integral cos(w) dw+sin^(1)(x)
The integral of cos(w) is sin(w):
= (sin(p))/2+(3 s)/81/4 sin^3(s) cos(s)u(3 sin(w))/16+sin^(1)(x)+constant
Substitute back for w = 2 s:
= (sin(p))/2+(3 s)/81/4 sin^3(s) cos(s)3/8 sin(s) cos(s)u+sin^(1)(x)+constant
Substitute back for p = 2 u:
= (3 s)/81/4 sin^3(s) cos(s)3/8 sin(s) cos(s)u+sin(u) cos(u)+sin^(1)(x)+constant
Substitute back for s = sin^(1)(x):
= u+sin(u) cos(u)3/8 sqrt(1x^2) x1/4 sqrt(1x^2) x^3+11/8 sin^(1)(x)+constant
Substitute back for u = sin^(1)(x):
= 5/8 sqrt(1x^2) x1/4 sqrt(1x^2) x^3+3/8 sin^(1)(x)+constant
Factor the answer a different way:
= 1/8 (x sqrt(1x^2) (52 x^2)+3 sin^(1)(x))+constant
Which is equivalent for restricted x values to:
= 1/8 sqrt(1/(1x^2)) (2 x^57 x^3+3 sqrt(1x^2) sin^(1)(x)+5 x)+constant 
 Follow
 8
 08092011 13:06
(Original post by g118)
..
See also the Guide to Posting and what it has to say about posting full solutions.
If you read the thread, you'd also see that we don't think that's what the OP was actually trying to solve. (Although it is what he said he was trying to solve). 
 Follow
 9
 08092011 13:10
(Original post by g118)
http://www.wolframalpha.com/input/?i=integrate+%281%2F%281++x%C2%B2%29%29^%283%2F2%29+dx.+
integral 1/(1/(1x^2))^(3/2) dx
Expanding the integrand 1/(1/(1x^2))^(3/2) gives 2 sqrt(1/(1x^2)) x^2+sqrt(1/(1x^2))+sqrt(1/(1x^2)) x^4:
= integral (2 sqrt(1/(1x^2)) x^2+sqrt(1/(1x^2))+sqrt(1/(1x^2)) x^4) dx
Integrate the sum term by term and factor out constants:
= integral sqrt(1/(1x^2)) dx2 integral x^2 sqrt(1/(1x^2)) dx+ integral x^4 sqrt(1/(1x^2)) dx
For the integrand x^2 sqrt(1/(1x^2)), simplify powers:
= integral sqrt(1/(1x^2)) dx2 integral x^2/sqrt(1x^2) dx+ integral x^4 sqrt(1/(1x^2)) dx
For the integrand, x^2/sqrt(1x^2) substitute x = sin(u) and dx = cos(u) du. Then sqrt(1x^2) = sqrt(1sin^2(u)) = cos(u) and u = sin^(1)(x):
= 2 integral sin^2(u) du+ integral sqrt(1/(1x^2)) dx+ integral x^4 sqrt(1/(1x^2)) dx
For the integrand x^4 sqrt(1/(1x^2)), simplify powers:
= 2 integral sin^2(u) du+ integral sqrt(1/(1x^2)) dx+ integral x^4/sqrt(1x^2) dx
For the integrand, x^4/sqrt(1x^2) substitute x = sin(s) and dx = cos(s) ds. Then sqrt(1x^2) = sqrt(1sin^2(s)) = cos(s) and s = sin^(1)(x):
= integral sin^4(s) ds2 integral sin^2(u) du+ integral sqrt(1/(1x^2)) dx
For the integrand sqrt(1/(1x^2)), simplify powers:
= integral sin^4(s) ds2 integral sin^2(u) du+ integral 1/sqrt(1x^2) dx
The integral of 1/sqrt(1x^2) is sin^(1)(x):
= integral sin^4(s) ds2 integral sin^2(u) du+sin^(1)(x)
Write sin^2(u) as 1/21/2 cos(2 u):
= integral sin^4(s) ds2 integral (1/21/2 cos(2 u)) du+sin^(1)(x)
Integrate the sum term by term and factor out constants:
= integral sin^4(s) ds2 integral 1/2 du+ integral cos(2 u) du+sin^(1)(x)
The integral of 1/2 is u/2:
= integral sin^4(s) dsu+ integral cos(2 u) du+sin^(1)(x)
For the integrand cos(2 u), substitute p = 2 u and dp = 2 du:
= 1/2 integral cos(p) dp+ integral sin^4(s) dsu+sin^(1)(x)
The integral of cos(p) is sin(p):
= (sin(p))/2+ integral sin^4(s) dsu+sin^(1)(x)
Use the reduction formula, integral sin^m(s) ds = (cos(s) sin^(m1)(s))/m + (m1)/m integral sin^(2+m)(s) ds, where m = 4:
= (sin(p))/21/4 sin^3(s) cos(s)+3/4 integral sin^2(s) dsu+sin^(1)(x)
Write sin^2(s) as 1/21/2 cos(2 s):
= (sin(p))/21/4 sin^3(s) cos(s)+3/4 integral (1/21/2 cos(2 s)) dsu+sin^(1)(x)
Integrate the sum term by term and factor out constants:
= (sin(p))/21/4 sin^3(s) cos(s)+3/4 integral 1/2 ds3/8 integral cos(2 s) dsu+sin^(1)(x)
For the integrand cos(2 s), substitute w = 2 s and dw = 2 ds:
= (sin(p))/21/4 sin^3(s) cos(s)+3/4 integral 1/2 dsu3/16 integral cos(w) dw+sin^(1)(x)
The integral of 1/2 is s/2:
= (sin(p))/2+(3 s)/81/4 sin^3(s) cos(s)u3/16 integral cos(w) dw+sin^(1)(x)
The integral of cos(w) is sin(w):
= (sin(p))/2+(3 s)/81/4 sin^3(s) cos(s)u(3 sin(w))/16+sin^(1)(x)+constant
Substitute back for w = 2 s:
= (sin(p))/2+(3 s)/81/4 sin^3(s) cos(s)3/8 sin(s) cos(s)u+sin^(1)(x)+constant
Substitute back for p = 2 u:
= (3 s)/81/4 sin^3(s) cos(s)3/8 sin(s) cos(s)u+sin(u) cos(u)+sin^(1)(x)+constant
Substitute back for s = sin^(1)(x):
= u+sin(u) cos(u)3/8 sqrt(1x^2) x1/4 sqrt(1x^2) x^3+11/8 sin^(1)(x)+constant
Substitute back for u = sin^(1)(x):
= 5/8 sqrt(1x^2) x1/4 sqrt(1x^2) x^3+3/8 sin^(1)(x)+constant
Factor the answer a different way:
= 1/8 (x sqrt(1x^2) (52 x^2)+3 sin^(1)(x))+constant
Which is equivalent for restricted x values to:
= 1/8 sqrt(1/(1x^2)) (2 x^57 x^3+3 sqrt(1x^2) sin^(1)(x)+5 x)+constant 
 Follow
 10
 08092011 13:10
(Original post by DFranklin)
There's no need to post more than the link.
See also the Guide to Posting and what it has to say about posting full solutions.
If you read the thread, you'd also see that we don't think that's what the OP was actually trying to solve. (Although it is what he said he was trying to solve).Post rating:1 
 Follow
 11
 08092011 13:25
(Original post by arob752)
3 years on, I think they may have worked it outPost rating:1
Write a reply…
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following:
Sign in
Not got an account? Sign up now
Updated: September 8, 2011
Share this discussion:
Tweet
Related discussions:
 Hard integration by substitution question?
 Hard integration by substitution question
 Edexcel 06/06/2014 FP2 Paper Discussion
 Very hard integration: BEWARE
 integration question help please
 really easy? Or hard?
 Integration by parts question
 Integrationsubstitution hard
 C4 integration by substitution question help
 Hard dynamics question!
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by: