Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

Evaluate the Intergal

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

\[\int\limits\limits_{}^{}\frac{ \arcsin(x) }{ x^2 } dx\]
I have no idea... I am think about trig substitutions but I am not sure about this...
Not sure either, but what does \(x=\sin(u)\) get you?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Why would you do that?
I was thinking about trying integration by parts... have you tried that yet? :)
It would get rid of the arcsin.
It's just a guess
Well I was going to make a substitution and then use intergration by parts but I am not sure...
What if I said u=arcsin(x) ?
That's the same as x = sin(u)
Right. But if I said that I would get a squart root and I could use trigonometric substitution.
Allright so I so far got... Use intragtion by parts: u=arcsinx) \[du=\frac{ 1 }{ \sqrt{1-x^2} }\] dv=x^-2 dx v=-x^-1
\[\frac{ -\arcsin(x) }{ x }-\int\limits_{}^{}\frac{ 1 }{ x \sqrt{1-x^2} }dx\]
I am kinda stuck at this point.
First do the substitution $$x=\sin(u)\implies dx=\cos(u)du$$ then $$\int \frac{\arcsin{x}}{x^2}\;dx=\int \frac{u\cos(u)}{\sin^2(u)}\;du=\int u\csc(u)\cot(u)\;du$$ $$\int u\frac{d(-\csc(u))}{du}du$$ do the integration by parts
Hmm let me see...
@BAdhi So I got to this point \[\int\limits_{}^{}ucsc(u)\cot(u) du\]
I would use integration by parts here right?
we know that $$\frac{d\csc(u)}{du}=-\csc(u)\cot(u)$$ then, $$\begin{align*}\int u\csc(u)\cot(u)\,du&=\int u\frac{-d\csc(u)}{du}\,du\\ &=-u\csc(u)+\int \csc(u)\,du \end{align*}$$
Yep I got that far.
Now I need to figure out the integral of csc(x).
Thanks I got this :) .
could use x = sin(u)
Thank you too wio :) .

Not the answer you are looking for?

Search for more explanations.

Ask your own question