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# integration any1? watch

1. need to use the substitution x=sin theta
to integrate 1/(1-x^2)^(3/2)

ive worked out dx = dtheta cos theta
and subbed values in but i cant seem to get the right answer which is x/(1-x^2)^(1/2)

ive got 1/cos^2 theta dtheta so far...
2. and 1/cos^2 theta = sec^2 theta , when u integrate that u get tan theta, no?
3. [QUOTE=cj_134;30583041]and 1/cos^2 theta = sec^2 theta , when u integrate that u get tan theta, no?[/QU
yh .. but thats not the answer ..
4. yes, then replace ur theta with your x, not directly, notice tan theta = sin theta / cos theta , sin theta = x and cos theta = (1-x^2)^1/2 , which gives the answer you want in terms of x, and dont forget your constant
5. (Original post by cj_134)
yes, then replace ur theta with your x, not directly, notice tan theta = sin theta / cos theta , sin theta = x and cos theta = (1-x^2)^1/2 , which gives the answer you want in terms of x, and dont forget your constant
why is cos theta = (1-x^2)^1/2?
6. (Original post by Thisisj)
why is cos theta = (1-x^2)^1/2?
7. (Original post by Thisisj)
why is cos theta = (1-x^2)^1/2?
ok, you know sin^2 theta + cos^2 theta = 1
so cos^2 theta = 1-sin^2 theta
cos theta = (1 - sin^2 theta) ^1/2
you know sin theta = x
hence cos theta = (1-x^2)^1/2
8. (Original post by EEngWillow)

(Original post by cj_134)
ok, you know sin^2 theta + cos^2 theta = 1
so cos^2 theta = 1-sin^2 theta
cos theta = (1 - sin^2 theta) ^1/2
you know sin theta = x
hence cos theta = (1-x^2)^1/2
oooooo right...
thanks guys
9. (Original post by cj_134)
ok, you know sin^2 theta + cos^2 theta = 1
so cos^2 theta = 1-sin^2 theta
cos theta = (1 - sin^2 theta) ^1/2
you know sin theta = x
hence cos theta = (1-x^2)^1/2
do you know any good websites to help me or show a step by step method because im stuck on a few of these types of questions???

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