Question

TRIG SUB Q:

Answer 1

look it up in khan academy if they have itt

Answer 2

Here is the problem:
\[\int\limits_{}^{}\sqrt{16-x ^{2}}\]

Answer 3

And here is ALL my work: http://imgur.com/K6tn1

Answer 4

I think I pretty much have it solved, I just am unsure about that last conversion

Answer 5

the minus suggests .... something

sin^2 + cos^2 = 1

sin^2 = 1-cos^2 ; cos^2 = 1-sin^2
--------------------------------

tan^2 + 1 = sec; sec^2 = tan^2 - 1

some such conflagerations

Answer 6

\[x=4 \sin (\theta )\]\[{dx}=4 \text{d$\theta $} \cos (\theta )\]
\[\sqrt{16-(4 \sin (\theta ))^2}\]
\[\sqrt{16-16 \sin ^2(\theta )}\]
\[4 \sqrt{1-\sin ^2(\theta )}\]
\[4 \sqrt{\cos ^2(\theta )}\]
\[\int 4 \text{Cos}[\theta ] 4\text{Cos}[\theta ]\text{d$\theta $}\]
\[16\int \text{Cos}[\theta ]^2\text{d$\theta $}\]