Question

fed up

find the integral sqrt(81-x^2)/x^2

I know it's trig sub, but I get stuck at the end.

I've been on this question for the past hour...

Answer 1

what did you substitute ?

"stuck at the end" ....can you show the steps what you did ? maybe i could help you spot the error :)

Answer 2

x = 9sin theta
dx = 9cos theta

\[\sqrt(81 - 81\sin^2 \theta)/(81\sin^2\theta)9\cos \theta d \theta\]

\[(1/3) (\cos^2 \theta)/(\sin^2 \theta) d \theta\]

\[\cos^2 \theta = (1 + \cos2 \theta) / 2\]

Answer 3

\[\sin^2 \theta = (1 - \cos2 \theta) / 2\]

\[1+\cos(2\theta) / 2 * 2/1-\cos(2\theta)\]

Answer 4

firstly, how did 1/3 come up ??
numerator and denominator, both have 81 ?

secondly, you got cos^2 t/ sin^2 t
which is actually cot^2 t
and there's a really easy way to get integral of cot^2 t , know it ?

Answer 5

The numerator sqrt(81-81sin^2) = sqrt(81(1-sin^2)) = 9sqrt(1-sin^2)

and no, I don't.

Answer 6

and dx = 9 cos t, so this 9 will also go in numerator right ?

Answer 7

Yeah, I see now

Answer 8

hint : \(\Large 1+\cot^2x = \csc^2 x\)

Answer 9

so whats cot^2 t from there ??

Answer 10

got it. thanks. =)

Answer 11

welcome ^_^
and since you're new here,

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