Dido525
Evaluate the Intergal
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Dido525
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\[\int\limits\limits_{}^{}\frac{ \arcsin(x) }{ x^2 } dx\]
Dido525
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I have no idea... I am think about trig substitutions but I am not sure about this...
wio
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Not sure either, but what does \(x=\sin(u)\) get you?
Dido525
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Why would you do that?
AccessDenied
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I was thinking about trying integration by parts... have you tried that yet? :)
wio
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It would get rid of the arcsin.
wio
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It's just a guess
Dido525
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Well I was going to make a substitution and then use intergration by parts but I am not sure...
Dido525
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What if I said u=arcsin(x) ?
wio
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That's the same as x = sin(u)
Dido525
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Right. But if I said that I would get a squart root and I could use trigonometric substitution.
Dido525
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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
Dido525
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\[\frac{ -\arcsin(x) }{ x }-\int\limits_{}^{}\frac{ 1 }{ x \sqrt{1-x^2} }dx\]
Dido525
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I am kinda stuck at this point.
BAdhi
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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
Dido525
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Hmm let me see...
Dido525
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@BAdhi So I got to this point
\[\int\limits_{}^{}ucsc(u)\cot(u) du\]
Dido525
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I would use integration by parts here right?
BAdhi
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we know that $$\frac{d\csc(u)}{du}=-\csc(u)\cot(u)$$
then,
$$\begin{align*}\int u\csc(u)\cot(u)\,du&=\int u\frac{-d\csc(u)}{du}\,du\\
&=-u\csc(u)+\int \csc(u)\,du
\end{align*}$$
Dido525
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Yep I got that far.
Dido525
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Now I need to figure out the integral of csc(x).
Dido525
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Thanks I got this :) .
wio
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could use x = sin(u)
Dido525
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Thank you too wio :) .