integrate 1/((x^5)*sqrt(9*x^2-1))

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

integrate 1/((x^5)*sqrt(9*x^2-1))

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

I've gotten pretty far I think, set x= 1/3 sec(U)
then I got to the point where I had \[81 \int\limits_{}^{} (\cos(x))^4 dx\]
the I apply the reduction formula twice and sub back in values for x to get (81/4)*(1/(3*x))^(3)*((sqrt(9 x^2-1))/(3*x))+243/(8)*(1/(3*x))*((sqrt(9 x^2-1))/(3*x))+243

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

|dw:1327038086064:dw|
maple and wolfram tell me that there is supposed to be an arctan in there somehow. But I can't figure out where it would get added in.
I can show more steps in between, but what I have there matches maple except for the last term 243/8 is multiplied by -arctan(1/(sqrt(9 x^2-1))) for some reason.
It seems like if you have made a mistake when substituting back for x.
I did this |dw:1327038548902:dw|
I know I've done it correctly right up until the substitution
|dw:1327038727194:dw| then all I did was substitute the values for sin and cos I found from the triangle
oops that last term should be over 8
There's more than a way to substitute back, seems to me like wolfarm did this \[\tan{u}=\sqrt{9x^2-1} \implies u=\tan^{-1}{\sqrt{9x^2-1}}.\]
There should be a term in the form \(ku\) in your integral, k is a constant.
Back in the step where you had \[81\int\limits_{}^{}\cos ^{4}(u) du\] did you consider substituting \[\cos^{2}u = (1 + \cos2u) / 2\] and solving from there?
is the integral |dw:1327039058333:dw|
I think that's where my missing term might be
Yeah.
ok because I had it set to 1. Then I will do the tan substitution you mentioned before and it should work out
And this is the term where the arctan came from, because \(u=\tan^{-1}{\sqrt{9x^2-1}}\).
Exactly.
Yeah, thanks a lot for your help!
You're welcome.
If you guys are still online I ran into problems with how this was working out.
I somehow need u= -arctan(1/sqrt(9x^2−1))
Why would you need that?
|dw:1327040345712:dw|
that is what it's telling me I'm missing from the end
|dw:1327040463342:dw|
any ideas where this would come from. It's confusing since my first substitutions look right, but the u one isn't consistent with the same triangle

Not the answer you are looking for?

Search for more explanations.

Ask your own question