Question

Using arcsecu+c identity integrate the following

Answer 1

\[\int\limits_{}^{} \frac{ 5}{ x \sqrt{5x^{2}-9} } dx\]

Answer 2

i understand that \[\int\limits_{}^{} \frac{ 1}{ u \sqrt{u^{2}-1} } dx = arcsecu+c\]

Answer 3

But... i'm not sure how to take out the 5 and 9 :S

Answer 4

more time for englis version

Answer 5

take the numerator 5 in denominator and then into sq roots n try some thing i think this is only thing we can do for now meanwhile lemme try and reach you again if noone answers before me

Answer 6

..i think not @Meinme

Answer 7

\[\int\frac{5}{x\sqrt{5x^2-9}}dx\]
I think the following sub should work:
\[x=\frac{3}{\sqrt5}\sec u\\
dx=\frac{3}{\sqrt5}\sec u\tan u\;du\\
\int\frac{5}{\frac{3}{\sqrt5}\sec u\sqrt{5\left(\frac{3}{\sqrt5}\sec u\right)^2-9}}\;\left(\frac{3}{\sqrt5}\sec u\tan u\;du\right)\\
\frac{5}{3} \int\frac{\sec u\tan u}{\sec u\sqrt{\sec^2 u-1}}du\]
(I skipped some steps, so I'd advise you to work out the algebra for yourself in case I messed up somewhere.)
Now make another substitution:
\[t=\sec u\\
dt=\sec u\tan u\;du\]
Now the integral becomes
\[\frac{5}{3} \int\frac{1}{t\sqrt{t^2-1}}dt,\]
which is what you can match up with the identity. Of course, you'll still have to sub back to get the answer in terms of x.

Answer 8

ah! perfect thanks.