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Astrophysics
 one year ago
Elementary integral problems
Astrophysics
 one year ago
Elementary integral problems

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442388808400:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0http://puu.sh/kd3Dz/a5f1d4e81e.png if it's not clear enough

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442388847137:dw so far...maybe these can be a little tricky

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Oooh 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.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Maybe I'll post solutions once I'm done all of them, just for fun

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0sry astro i i havent yet given the answer to that quadrilateral question :) i will do that after some time 2weeks ig :) hehe :)

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Ah I just figured out 12 I think, but 21... Hmmm... I'll have to try IBP maybe but I'm kinda scared to lol

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442389097067:dw part 2 http://puu.sh/kd3Qd/129ef4383c.png

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Yeah try what ever one you like out lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sf \#12 \\ \\ \int \frac{x}{x^4+x^2+1}dx\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Want to know how?

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1font looks familiar

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You have an x at the top and you have different "variations" of x at the bottom

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Stewart calc nin lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wouldnt a usub work?/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Like whats the easiest way of making the top a derivative of something? now I know you're either going to usub \(\sf x^2\) and \(\sf x^4\)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Hint: \[\sf \#12 \\ \\ \int\limits \frac{x}{(x^2)^2+x^2+1}dx\]

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1omg why did you give a hint

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ u=x^2~,~ du = 2xdx\]\[2\int\frac{du}{(u)^2+u+1}\]

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1LOLLLLL go jhanny go jhanny aren't you supposed to be doing chemistry?

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1I am a sober now, so I can help

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0#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 doable 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{1}^{1} \frac{ e ^{arc \tan y} }{ 1+y^2 }dy\] \[put ~\tan^{1} y=t,\frac{ dy }{ 1+y^2 }=dt\] when y=1,\[\tan^{1} \left( 1 \right)=t,t= \frac{ \pi }{ 4 }\] when y=1,\[t=\frac{ \pi }{ 4 }\] \[I=\int\limits_{\frac{ \pi }{ 4 }}^{\frac{ \pi }{ 4 }}e^tdt=e^t\frac{ \pi }{ 4 }\rightarrow \frac{ \pi }{ 4 }\] \[=e ^{\frac{ \pi }{ 4 }}e ^{\frac{ \pi }{ 4 }}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.017.\[I=\int\limits_{0}^{\pi}t \cos ^2t~dt=\int\limits_{0}^{\pi}\left( \pit \right)\cos ^2\left( \pit \right)dt\] \[=\int\limits_{0}^{\pi}\pi \cos ^2t~ dt\int\limits_{0}^{\pi}t~\cos ^2t~dt=\pi \int\limits_{0}^{\pi}\left( \frac{ 1+\cos 2t }{ 2 } \right)dtI\] \[=\pi \frac{ 1 }{ 2 }\int\limits_{0}^{\pi}\left( 1+\cos 2t \right)dt\] \[I=\frac{ \pi }{ 4 }?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.031.put \[x=\cos \theta,dx=\sin \theta~d \theta \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.041.\[I=\int\limits \theta \tan ^2\theta d \theta=\int\limits \theta \left( \sec ^2\theta1 \right)d \theta =\int\limits \theta \sec ^2\theta d \theta \frac{ \theta^2 }{ 2 }+c\] integrate by parts

anonymous
 one year ago
Best ResponseYou've already chosen the best response.045\[\int\limits x^5e ^{x^3}dx=\int\limits x^3e ^{x^3}x^2dx\] put\[x^3=t,3x^2dx=dt\] ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.048\[I=\int\limits \frac{ dx }{ x \sqrt{4x+1} }\] 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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