Question

Evaluate the Intergal

Answer 1

\[\int\limits\limits_{}^{}\frac{ \arcsin(x) }{ x^2 } dx\]

Answer 2

I have no idea... I am think about trig substitutions but I am not sure about this...

Answer 3

Not sure either, but what does \(x=\sin(u)\) get you?

Answer 4

Why would you do that?

Answer 5

I was thinking about trying integration by parts... have you tried that yet? :)

Answer 6

It would get rid of the arcsin.

Answer 7

It's just a guess

Answer 8

Well I was going to make a substitution and then use intergration by parts but I am not sure...

Answer 9

What if I said u=arcsin(x) ?

Answer 10

That's the same as x = sin(u)

Answer 11

Right. But if I said that I would get a squart root and I could use trigonometric substitution.

Answer 12

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

Answer 13

\[\frac{ -\arcsin(x) }{ x }-\int\limits_{}^{}\frac{ 1 }{ x \sqrt{1-x^2} }dx\]

Answer 14

I am kinda stuck at this point.

Answer 15

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

Answer 16

Hmm let me see...

Answer 17

@BAdhi So I got to this point

\[\int\limits_{}^{}ucsc(u)\cot(u) du\]

Answer 18

I would use integration by parts here right?

Answer 19

we know that $$\frac{d\csc(u)}{du}=-\csc(u)\cot(u)$$
then,
$$\begin{align*}\int u\csc(u)\cot(u)\,du&=\int u\frac{-d\csc(u)}{du}\,du\\
&=-u\csc(u)+\int \csc(u)\,du
\end{align*}$$

Answer 20

Yep I got that far.

Answer 21

Now I need to figure out the integral of csc(x).

Answer 22

Thanks I got this :) .

Answer 23

could use x = sin(u)

Answer 24

Thank you too wio :) .