Question

How do you find the integral of dt/(sqrt(5-3t^2))? I've gotten to the point where I factor the 1/sqrt(3) out, but I don't know the right substitution to put in.

Answer 1

\[\int\limits \frac{ dt }{ \sqrt{5-3t^2} }\]

Answer 2

\[\int\limits \frac{ dx }{ \sqrt{a^2-x^2} }=\sin^{-1} \frac{ x }{ a }\]

Answer 3

put
\[\sqrt{3}t=\sqrt{5}\sin \theta \]
\[\sqrt{3}dt=\sqrt{5}\cos \theta d \theta \]

Answer 4

\[dt=\frac{ \sqrt{5} }{ \sqrt{3} }\cos \theta d \theta \]

Answer 5

you are correct proceed further.

Answer 6

according to you
\[t=\sqrt{\frac{ 5 }{ 3 }}\sin \theta \]

Answer 7

\[dt=\sqrt{\frac{ 5 }{ 3 }}\cos \theta d \theta \]

Answer 8

where does the \[\sqrt{3}d=\sqrt{5} \sin \theta\] come from though?

Answer 9

differential of t is dt
and similarly otherside.

Answer 10

oops sorry, i meant \[\sqrt{3}t\] and I meant where do you get it from in the equation? I don't get how you got sin when it isn't in the starting equation. I know the equation will eventually be arcsin, I just don't see how the sin fits in to all of this.

Answer 11

\[I=\int\limits \frac{ dt }{ \sqrt{5-3t^2} }=\int\limits \frac{ \sqrt{\frac{ 5 }{ 3 }} \cos \theta d \theta }{ \sqrt{5-5 \sin ^2 \theta } }\]
\[=\sqrt{\frac{ 5 }{ 3 }}*\frac{ 1 }{ \sqrt{5} }\int\limits \frac{ \cos \theta d \theta }{ \sqrt{1-\sin ^2\theta } }\]
\[=\frac{ 1 }{ \sqrt{3} }\int\limits \frac{ \cos \theta d \theta }{ \cos \theta }=\frac{ 1 }{ \sqrt{3} }\int\limits d \theta \]
\[=\frac{ 1 }{ \sqrt{3} }\theta =?\]