Question

integral of 1/t^2*sqrt(t^2-25)

Answer 1

i already started with trig sub

Answer 2

good, that's what I was thinking at a glance

Answer 3

Im working through it now because after i posted the question i realized i left something out

Answer 4

ok nvm i got it haha thanks for responding though

Answer 5

this is precisely a trig sub. particular sin(x) sub.

Answer 6

\[I=\int\limits \frac{ 1 }{ t^2\sqrt{t^2-25} }dt\]
put\(t=1/y\)
\[dt=\frac{ -1 }{ y^2 }dy\]
\[I=\int\limits \frac{ \frac{ -1 }{ y^2 }dy }{ \frac{ 1 }{ y^2 } \sqrt{\frac{ 1 }{ y^2 }-25}}=\int\limits \frac{ -ydy }{ \sqrt{1-25y^2}}\]
\[=-\int\limits \left( 1-25y^2 \right)^{\frac{ -1 }{ 2 }}\left( y~dy \right)=?\]