Question

Help in integration?

Answer 1

\[\int\limits\limits_{}^{}\frac{ dy }{ \sqrt{y} (9-y) }\]

Answer 2

if I let
\[t= \sqrt{y} \]
then
\[t ^{2}= y\]
and
\[dy= 2tdt\]
Substituting these to the original integral, I will have
\[\int\limits\limits_{}^{}\frac{ 2tdt }{ t (9-t ^{2}) }\]
How can I integrate this?

Answer 3

Wait, from \(t=\sqrt{y}\),
\[dt=\frac{1}{2\sqrt y }dy \\ \\ 2dt=\frac{1}{\sqrt y } dy\]
You get
\[=2 \int\limits \frac{1}{9-t^2} \, dt\]

Answer 4

Looks like you need trig substitution for that

Answer 5

I see, I can now use the inverse one. And yes, I should have computed for the dt not the dy! Thanks!

Answer 6

alright, yw

Answer 7

Wait. But there is no square root below. Can I use the inverse formula?

Answer 8

if u= y^(1/2)

Answer 9

You can do it like this,
\[\frac{2}{9} \int\limits \frac{1}{1-\frac{t^2}{9}} \, dt \\ \\ \]
Let \[h=\frac{t}{3} \\ \\ dh=\frac{1}{3}dt\]
\[\frac{2}{3} \int\limits \frac{1}{1-h^2} \, dh\]
Integrate that you get hyperbolic function
\[\frac{2}{3}\tanh^{-1}(h)+c\]