Question

integrate ∫dx/(64−x2) from 0 to 8. Below is my work so far. txs

Answer 1

you want du/ ( 1 - u^2) , so first do some work

Answer 2

substitute
x=8sin(t)
dx=8cos(t)dt

Answer 3

\[\int\limits_{0}^{8} dx/(64-x^{2}) \]

I have \[x=8\tan \theta \]
\[dx=8\sec \theta \ d \theta \]
\[=\int\limits_{0}^{8} 8\sec^2\theta d \theta/(64(1-\tan^2 \theta)) \]
Is this the correct course to take?
I was told to solve it with a trig substitution. Thanks for your time.

Answer 4

Thanks for the quick reply sami-21

Answer 5

\[\int\limits_{0}^{8} dx/(64-x^{2}) = \lim_{a \rightarrow 8} \int\limits_{0}^{a} dx/(64-x^{2}) 0 \]
\[x=8\sin \theta \] \[dx=8\cos \theta\] ......

\[\lim_{a \rightarrow 8} 1/8\int\limits_{0}^{a} \sec \theta d \theta = \lim_{a \rightarrow 8} 1/8\ \ln \left| (1/8)( x+\sqrt{64-x^{2}}) \right| from\ 0\ to \ a\]
Have a made a mistake here?

Answer 6

\[\lim_{a \rightarrow 8}1/8 \ln \left| (x+8)/\sqrt{64-x^{2}} \right|\] from 0 to a
is this right?

Answer 7

\[\tan \theta = x/\sqrt{64-x^2}\]
\[\sec \theta = 8/ \sqrt{64-x^{2}}\]
I've never used a trig substitution in this case before. I am just making sure that this doesn't converge. I think it does.