Question

Integral of sqrt(81-x^2)dx using the substitution of x=9sint

Answer 1

dx=9cost dt\[\int\limits_{}^{}\sqrt{81-(9sint)^2} *9cost dt=\int\limits{}^{}\sqrt{81-81\sin^2(t)}*9cost dt\]
\[=\int\limits_{}^{}\sqrt{81}\sqrt{1-\sin^2{t}}*9cost dt=\sqrt{81}*9\int\limits_{}^{}\cos(t)*\cos(t)dt=81\int\limits_{}^{}\cos^2(t) dt\]

Answer 2

yeah ditto

Answer 3

use a trig identity
cos^2(t)=1/2*(1+cos(2x))

Answer 4

any questions?

Answer 5

I'm using a web-base homework system and I plugged that in but it was incorrect.. Could you go a little further so that I may see where I went wrong?

Answer 6

\[\frac{1}{2}\int\limits_{}^{}(1+\cos(2x))dx\]

Answer 7

what did you put in?
remember to put it in terms of x at the end
t= sin^-1(x/9)

Answer 8

oops that should be in terms of t above k?

Answer 9

Thanks!! Got it :)

Answer 10

\[\frac{1}{2}\int\limits_{}^{}1 dt +\frac{1}{2}\int\limits_{}^{}\cos(2t)dt\]