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Astrophysics

  • one year ago

Elementary integral problems

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  1. Astrophysics
    • one year ago
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    |dw:1442388808400:dw|

  2. Astrophysics
    • one year ago
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    http://puu.sh/kd3Dz/a5f1d4e81e.png if it's not clear enough

  3. Astrophysics
    • one year ago
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    |dw:1442388847137:dw| so far...maybe these can be a little tricky

  4. Empty
    • one year ago
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    Oooh damn nice. I think 21 is kinda famous, or at least it looks hard. I don't know about the other one if that one's gonna be easy or hard. I'm gonna guess you complete the square for that one or something.

  5. Astrophysics
    • one year ago
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    Maybe I'll post solutions once I'm done all of them, just for fun

  6. imqwerty
    • one year ago
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    sry astro i i havent yet given the answer to that quadrilateral question :) i will do that after some time 2weeks ig :) hehe :)

  7. Empty
    • one year ago
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    Ah I just figured out 12 I think, but 21... Hmmm... I'll have to try IBP maybe but I'm kinda scared to lol

  8. Astrophysics
    • one year ago
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    |dw:1442389097067:dw| part 2 http://puu.sh/kd3Qd/129ef4383c.png

  9. Astrophysics
    • one year ago
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    Yeah try what ever one you like out lol

  10. Jhannybean
    • one year ago
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    \[\sf \#12 \\ \\ \int \frac{x}{x^4+x^2+1}dx\]

  11. Astrophysics
    • one year ago
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    Want to know how?

  12. Jhannybean
    • one year ago
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    No, not yet

  13. nincompoop
    • one year ago
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    font looks familiar

  14. Jhannybean
    • one year ago
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    You have an x at the top and you have different "variations" of x at the bottom

  15. Astrophysics
    • one year ago
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    Stewart calc nin lol

  16. Jhannybean
    • one year ago
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    Wouldnt a u-sub work?/

  17. nincompoop
    • one year ago
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    I know ... LOL

  18. Jhannybean
    • one year ago
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    Like whats the easiest way of making the top a derivative of something? now I know you're either going to u-sub \(\sf x^2\) and \(\sf x^4\)

  19. Astrophysics
    • one year ago
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    Hint: \[\sf \#12 \\ \\ \int\limits \frac{x}{(x^2)^2+x^2+1}dx\]

  20. Jhannybean
    • one year ago
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    OKAY!

  21. nincompoop
    • one year ago
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    omg why did you give a hint

  22. Jhannybean
    • one year ago
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    \[ u=x^2~,~ du = 2xdx\]\[2\int\frac{du}{(u)^2+u+1}\]

  23. Astrophysics
    • one year ago
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    \[\checkmark \]

  24. nincompoop
    • one year ago
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    LOLLLLL go jhanny go jhanny aren't you supposed to be doing chemistry?

  25. Jhannybean
    • one year ago
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    .......... yeah

  26. nincompoop
    • one year ago
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    I am a sober now, so I can help

  27. zzr0ck3r
    • one year ago
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    good old Stewart

  28. IrishBoy123
    • one year ago
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    #21 \(\int arctan \sqrt{x} \,\, dx\) |dw:1442396993523:dw| \(\frac{1}{2}x^{-1/2}dx = 2 \tan \theta \sec^2 \theta \, d\theta \) \(dx = 2 \tan \theta \ \sec^2 \theta \, d\theta \) [looking do-able already] \(\implies \int \theta . 2 \tan \theta \ \sec^2 \theta \, d\theta\,\, dx\) [parts] \(u = \theta, u' = 1, v'= 2 \tan \theta \ \sec^2, v = \tan^2 \theta \) and that's most of it

  29. anonymous
    • one year ago
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    \[\int\limits_{-1}^{1} \frac{ e ^{arc \tan y} }{ 1+y^2 }dy\] \[put ~\tan^{-1} y=t,\frac{ dy }{ 1+y^2 }=dt\] when y=-1,\[\tan^{-1} \left( -1 \right)=t,t= -\frac{ \pi }{ 4 }\] when y=1,\[t=\frac{ \pi }{ 4 }\] \[I=\int\limits_{-\frac{ \pi }{ 4 }}^{\frac{ \pi }{ 4 }}e^tdt=e^t|-\frac{ \pi }{ 4 }\rightarrow \frac{ \pi }{ 4 }\] \[=e ^{\frac{ \pi }{ 4 }}-e ^{-\frac{ \pi }{ 4 }}\]

  30. anonymous
    • one year ago
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    17.\[I=\int\limits_{0}^{\pi}t \cos ^2t~dt=\int\limits_{0}^{\pi}\left( \pi-t \right)\cos ^2\left( \pi-t \right)dt\] \[=\int\limits_{0}^{\pi}\pi \cos ^2t~ dt-\int\limits_{0}^{\pi}t~\cos ^2t~dt=\pi \int\limits_{0}^{\pi}\left( \frac{ 1+\cos 2t }{ 2 } \right)dt-I\] \[=\pi \frac{ 1 }{ 2 }\int\limits_{0}^{\pi}\left( 1+\cos 2t \right)dt\] \[I=\frac{ \pi }{ 4 }?\]

  31. anonymous
    • one year ago
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    31.put \[x=\cos \theta,dx=-\sin \theta~d \theta \]

  32. anonymous
    • one year ago
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    41.\[I=\int\limits \theta \tan ^2\theta d \theta=\int\limits \theta \left( \sec ^2\theta-1 \right)d \theta =\int\limits \theta \sec ^2\theta d \theta -\frac{ \theta^2 }{ 2 }+c\] integrate by parts

  33. anonymous
    • one year ago
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    45\[\int\limits x^5e ^{-x^3}dx=\int\limits x^3e ^{-x^3}x^2dx\] put\[x^3=t,3x^2dx=dt\] ?

  34. anonymous
    • one year ago
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    48\[I=\int\limits \frac{ dx }{ x \sqrt{4x+1} }\] put\[x=\frac{ 1 }{ t },dx=-\frac{ 1 }{ t^2 }dt\] \[I=\int\limits \frac{ \frac{ -1 }{ t^2 }dt }{ \frac{ 1 }{ t }\sqrt{\frac{ 4 }{ t }+1} }=-\int\limits \frac{ dt }{\sqrt{4+t}=? }\]

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