[br]\frac{dx}{du} = cos u[br][br]\displaystyle \int {(cos^2u)^{-\frac{3}{2}}cos u\, du[br][br]\displaystyle = \int \frac{cos u}{\sqrt {(cos^2u)^3}}\, du[br][br]\displaystyle = \int \frac{cosu}{cos^3u}\, du[br][br]\displaystyle = \int \frac{1}{cos^2u}\, du[br][br]\displaystyle = \int sec^2u\, du[br]

Im sure you recognise the integral of

sec^2u

Reply 6

g118

http://www.wolframalpha.com/input/?i=integrate+%281%2F%281+-+x%C2%B2%29%29^-%283%2F2%29+dx.+

integral 1/(1/(1-x^2))^(3/2) dx
Expanding the integrand 1/(1/(1-x^2))^(3/2) gives -2 sqrt(1/(1-x^2)) x^2+sqrt(1/(1-x^2))+sqrt(1/(1-x^2)) x^4:
= integral (-2 sqrt(1/(1-x^2)) x^2+sqrt(1/(1-x^2))+sqrt(1/(1-x^2)) x^4) dx
Integrate the sum term by term and factor out constants:
= integral sqrt(1/(1-x^2)) dx-2 integral x^2 sqrt(1/(1-x^2)) dx+ integral x^4 sqrt(1/(1-x^2)) dx
For the integrand x^2 sqrt(1/(1-x^2)), simplify powers:
= integral sqrt(1/(1-x^2)) dx-2 integral x^2/sqrt(1-x^2) dx+ integral x^4 sqrt(1/(1-x^2)) dx
For the integrand, x^2/sqrt(1-x^2) substitute x = sin(u) and dx = cos(u) du. Then sqrt(1-x^2) = sqrt(1-sin^2(u)) = cos(u) and u = sin^(-1)(x):
= -2 integral sin^2(u) du+ integral sqrt(1/(1-x^2)) dx+ integral x^4 sqrt(1/(1-x^2)) dx
For the integrand x^4 sqrt(1/(1-x^2)), simplify powers:
= -2 integral sin^2(u) du+ integral sqrt(1/(1-x^2)) dx+ integral x^4/sqrt(1-x^2) dx
For the integrand, x^4/sqrt(1-x^2) substitute x = sin(s) and dx = cos(s) ds. Then sqrt(1-x^2) = sqrt(1-sin^2(s)) = cos(s) and s = sin^(-1)(x):
= integral sin^4(s) ds-2 integral sin^2(u) du+ integral sqrt(1/(1-x^2)) dx
For the integrand sqrt(1/(1-x^2)), simplify powers:
= integral sin^4(s) ds-2 integral sin^2(u) du+ integral 1/sqrt(1-x^2) dx
The integral of 1/sqrt(1-x^2) is sin^(-1)(x):
= integral sin^4(s) ds-2 integral sin^2(u) du+sin^(-1)(x)
Write sin^2(u) as 1/2-1/2 cos(2 u):
= integral sin^4(s) ds-2 integral (1/2-1/2 cos(2 u)) du+sin^(-1)(x)
Integrate the sum term by term and factor out constants:
= integral sin^4(s) ds-2 integral 1/2 du+ integral cos(2 u) du+sin^(-1)(x)
The integral of 1/2 is u/2:
= integral sin^4(s) ds-u+ 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) ds-u+sin^(-1)(x)
The integral of cos(p) is sin(p):
= (sin(p))/2+ integral sin^4(s) ds-u+sin^(-1)(x)
Use the reduction formula, integral sin^m(s) ds = -(cos(s) sin^(m-1)(s))/m + (m-1)/m integral sin^(-2+m)(s) ds, where m = 4:
= (sin(p))/2-1/4 sin^3(s) cos(s)+3/4 integral sin^2(s) ds-u+sin^(-1)(x)
Write sin^2(s) as 1/2-1/2 cos(2 s):
= (sin(p))/2-1/4 sin^3(s) cos(s)+3/4 integral (1/2-1/2 cos(2 s)) ds-u+sin^(-1)(x)
Integrate the sum term by term and factor out constants:
= (sin(p))/2-1/4 sin^3(s) cos(s)+3/4 integral 1/2 ds-3/8 integral cos(2 s) ds-u+sin^(-1)(x)
For the integrand cos(2 s), substitute w = 2 s and dw = 2 ds:
= (sin(p))/2-1/4 sin^3(s) cos(s)+3/4 integral 1/2 ds-u-3/16 integral cos(w) dw+sin^(-1)(x)
The integral of 1/2 is s/2:
= (sin(p))/2+(3 s)/8-1/4 sin^3(s) cos(s)-u-3/16 integral cos(w) dw+sin^(-1)(x)
The integral of cos(w) is sin(w):
= (sin(p))/2+(3 s)/8-1/4 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)/8-1/4 sin^3(s) cos(s)-3/8 sin(s) cos(s)-u+sin^(-1)(x)+constant
Substitute back for p = 2 u:
= (3 s)/8-1/4 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(1-x^2) x-1/4 sqrt(1-x^2) x^3+11/8 sin^(-1)(x)+constant
Substitute back for u = sin^(-1)(x):
= 5/8 sqrt(1-x^2) x-1/4 sqrt(1-x^2) x^3+3/8 sin^(-1)(x)+constant
Factor the answer a different way:
= 1/8 (x sqrt(1-x^2) (5-2 x^2)+3 sin^(-1)(x))+constant
Which is equivalent for restricted x values to:
= 1/8 sqrt(1/(1-x^2)) (2 x^5-7 x^3+3 sqrt(1-x^2) sin^(-1)(x)+5 x)+constant

Reply 7

DFranklin

g118

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).