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Dido525

  • 2 years ago

Evaluate the Intergal

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  1. Dido525
    • 2 years ago
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    \[\int\limits\limits_{}^{}\frac{ \arcsin(x) }{ x^2 } dx\]

  2. Dido525
    • 2 years ago
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    I have no idea... I am think about trig substitutions but I am not sure about this...

  3. wio
    • 2 years ago
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    Not sure either, but what does \(x=\sin(u)\) get you?

  4. Dido525
    • 2 years ago
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    Why would you do that?

  5. AccessDenied
    • 2 years ago
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    I was thinking about trying integration by parts... have you tried that yet? :)

  6. wio
    • 2 years ago
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    It would get rid of the arcsin.

  7. wio
    • 2 years ago
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    It's just a guess

  8. Dido525
    • 2 years ago
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    Well I was going to make a substitution and then use intergration by parts but I am not sure...

  9. Dido525
    • 2 years ago
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    What if I said u=arcsin(x) ?

  10. wio
    • 2 years ago
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    That's the same as x = sin(u)

  11. Dido525
    • 2 years ago
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    Right. But if I said that I would get a squart root and I could use trigonometric substitution.

  12. Dido525
    • 2 years ago
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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

  13. Dido525
    • 2 years ago
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    \[\frac{ -\arcsin(x) }{ x }-\int\limits_{}^{}\frac{ 1 }{ x \sqrt{1-x^2} }dx\]

  14. Dido525
    • 2 years ago
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    I am kinda stuck at this point.

  15. BAdhi
    • 2 years ago
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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

  16. Dido525
    • 2 years ago
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    Hmm let me see...

  17. Dido525
    • 2 years ago
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    @BAdhi So I got to this point \[\int\limits_{}^{}ucsc(u)\cot(u) du\]

  18. Dido525
    • 2 years ago
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    I would use integration by parts here right?

  19. BAdhi
    • 2 years ago
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    we know that $$\frac{d\csc(u)}{du}=-\csc(u)\cot(u)$$ then, $$\begin{align*}\int u\csc(u)\cot(u)\,du&=\int u\frac{-d\csc(u)}{du}\,du\\ &=-u\csc(u)+\int \csc(u)\,du \end{align*}$$

  20. Dido525
    • 2 years ago
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    Yep I got that far.

  21. Dido525
    • 2 years ago
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    Now I need to figure out the integral of csc(x).

  22. Dido525
    • 2 years ago
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    Thanks I got this :) .

  23. wio
    • 2 years ago
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    could use x = sin(u)

  24. Dido525
    • 2 years ago
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    Thank you too wio :) .

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