## Astrophysics one year ago Elementary integral problems

1. Astrophysics

|dw:1442388808400:dw|

2. Astrophysics

http://puu.sh/kd3Dz/a5f1d4e81e.png if it's not clear enough

3. Astrophysics

|dw:1442388847137:dw| so far...maybe these can be a little tricky

4. Empty

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

Maybe I'll post solutions once I'm done all of them, just for fun

6. imqwerty

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

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

|dw:1442389097067:dw| part 2 http://puu.sh/kd3Qd/129ef4383c.png

9. Astrophysics

Yeah try what ever one you like out lol

10. Jhannybean

$\sf \#12 \\ \\ \int \frac{x}{x^4+x^2+1}dx$

11. Astrophysics

Want to know how?

12. Jhannybean

No, not yet

13. nincompoop

font looks familiar

14. Jhannybean

You have an x at the top and you have different "variations" of x at the bottom

15. Astrophysics

Stewart calc nin lol

16. Jhannybean

Wouldnt a u-sub work?/

17. nincompoop

I know ... LOL

18. Jhannybean

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

Hint: $\sf \#12 \\ \\ \int\limits \frac{x}{(x^2)^2+x^2+1}dx$

20. Jhannybean

OKAY!

21. nincompoop

omg why did you give a hint

22. Jhannybean

$u=x^2~,~ du = 2xdx$$2\int\frac{du}{(u)^2+u+1}$

23. Astrophysics

$\checkmark$

24. nincompoop

LOLLLLL go jhanny go jhanny aren't you supposed to be doing chemistry?

25. Jhannybean

.......... yeah

26. nincompoop

I am a sober now, so I can help

27. zzr0ck3r

good old Stewart

28. IrishBoy123

#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

$\int\limits_{-1}^{1} \frac{ e ^{arc \tan y} }{ 1+y^2 }dy$ $put ~\tan^{-1} y=t,\frac{ dy }{ 1+y^2 }=dt$ 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

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

31.put $x=\cos \theta,dx=-\sin \theta~d \theta$

32. anonymous

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

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

48$I=\int\limits \frac{ dx }{ x \sqrt{4x+1} }$ put$x=\frac{ 1 }{ t },dx=-\frac{ 1 }{ t^2 }dt$ $I=\int\limits \frac{ \frac{ -1 }{ t^2 }dt }{ \frac{ 1 }{ t }\sqrt{\frac{ 4 }{ t }+1} }=-\int\limits \frac{ dt }{\sqrt{4+t}=? }$