1. ## u-sub integration help

guys first timer here. cant quite figure this one out.

ʃ dt/ t√(5tē-3)

it looks to me to be a sec^-1u type of question but i cant seem to find the correct u value. any help from you smart peoples?

2. Originally Posted by lj212
guys first timer here. cant quite figure this one out.

ʃ dt/ t√(5tē-3)

it looks to me to be a sec^-1u type of question but i cant seem to find the correct u value. any help from you smart peoples?
$
5t^2 - 3 = 3 \left(\frac{5}{3}t^2 - 1 \right) = 3 \left( \left(\frac{\sqrt{5}}{\sqrt{3}} t\right)^2 - 1\right)
$
and $u = \frac{\sqrt{5}}{\sqrt{3}} t$

3. Originally Posted by danny arrigo
$
5t^2 - 3 = 3 \left(\frac{5}{3}t^2 - 1 \right) = 3 \left( \left(\frac{\sqrt{5}}{\sqrt{3}} t\right)^2 - 1\right)
$
and $u = \frac{\sqrt{5}}{\sqrt{3}} t$
Since $t=\frac{\sqrt{3}}{\sqrt{5}}u$ and $dt = \frac{\sqrt{3}}{\sqrt{5}}\,du$, you have: $\int\frac{dt}{t\sqrt{5t^2-3}} = \int\frac{\frac{\sqrt{3}}{\sqrt{5}}\,du}{\frac{\sq rt{3}}{\sqrt{5}}u\sqrt{3(u^2-1)}}$ $= \frac{1}{\sqrt{3}}\int\frac{du}{u\sqrt{u^2-1}}$

Spoiler:
$=\frac{1}{\sqrt{3}}\sec^{-1}(u) = \boxed{\frac{1}{\sqrt{3}}\sec^{-1}\left(\sqrt{\frac{5}{3}}\cdot t\right)}$