Question

Trigonometric substitution problem





\[\int{\frac{{{x}^{2}}}{\sqrt{16-{{x}^{2}}}}}\]

Answer 1

try either \(x=2\sin(\theta)\) or \(u=2\cos(\theta)\) either should work

Answer 2

ok that was silly, i meant
\[u=4\sin(\theta)\] because then \(\sqrt{16-(4\sin(\theta))^2}=\sqrt{16-16\sin^2(\theta)}=4\cos(\theta)\)

Answer 3

why a 2 upfront whats the logic ?

Answer 4

a mistake on my part, not logical at all

Answer 5

I tried 4sin(theta) but nothing at all

Answer 6

i wrote it out above

Answer 7

4*INT(sinu/cosu )

Answer 8

thats basically where I stuck forever

Answer 9

you should have
\[16\int\sin^2(\theta)d\theta\]

Answer 10

true

Answer 11

don't forget if \(u=4\sin(\theta)\) then \(du=4\cos(\theta)d\theta\) and the cosines cancel top and bottom

Answer 12

thanks

Answer 13

yw

Answer 14

sorry for bothering you again @satellite73 but can you tell me the next 1-2 steps ? :D