Question

Calculus 2 Help please!!!

Answer 1

\[\int\limits\frac{ \sin^5(x) }{1+\cos^2(x) }dx\]

Answer 2

I got this far:
\[\int\limits \frac{ (1-\cos^2)^2 }{ 1+\cos^2 }dx\]
u=cos(x) du=-sin(x)dx dx=-1/sin(x)du
\[\int\limits \frac{ (1-u^2)^2 }{1+u^2 }du\]

I don't know what to do after this...

Answer 3

That isn't quite right. \[\int\limits \frac {\sin^5 (x)}{1 + \cos^2 (x)}dx = \int\limits \frac {(1- \cos^2 (x))^2 \sin (x)}{1+\cos^2(x)}dx\]The rest of your problem looks good.

Answer 4

XD thats what I meant to type

Answer 5

Also, make sure you specify that \[dx = -\frac {du}{sin(x)}\]

Answer 6

From here, try and use integration by parts to solve this equation.

Answer 7

Also, need to add a negative sign to your equation.

Answer 8

so would it just turn the whole integration negative?

Answer 9

Yes. This is because \[dx = - \frac {du}{sin(x)}\]the negative sign on this substitution.

Answer 10

So..
\[-\int\limits\limits \frac{ 1-2u^2+u^4 }{ 1+u }du\]
=\[-\int\limits(\frac{ 1 }{ 1+u^2 })du +\int\limits(\frac{ 2u^2 }{ 1+u^2})du-\int\limits \frac{ u^4 }{ 1+u^2 }du\]

Answer 11

first one is integration of tan inverse x...I am stuck with the rest 2...

Answer 12

With \[\int\limits \frac{ 2u^2 }{ 1+u^2 }du\]
would we split it like\[\int\limits \frac{ 2 }{ 1+u^2 }\frac{ u^2 }{ 1+u^2 } du\]

Answer 13

no can't do that...

Answer 14

That isn't right. That is the equivalent of \[\int \frac {2u^2}{(1+u^2)^2}\]What you need to do is use integration by parts: \[\int u(x)v'(x) = u(x)v(x) - \int u'(x)v(x)dx\]

Answer 15

how do you guys use the integration symbols in here teach me quick...

Answer 16

what would we use as u and v

Answer 17

lol got it..

Answer 18

You can either use the Equation button beneath the chat box to input equations, or use LaTeX notation to manually type it out.

Answer 19

\[u(x) = 2u^2\] or the numerator and \[v(x) = \frac {1}{1+u^2}\]

Answer 20

\[\left(\frac{u^2-\left(1+u^2\right)}{1+u^2}+1\right)\]

Answer 21

\[\left(2\int\limits \frac{u^2-\left(1+u^2\right)}{1+u^2}du+\int\limits \:1du\right)\]

Answer 22

wow its so hard to type these stuff

Answer 23

It's nice once you get used to it. I'm off for the night. Good luck. I know the solution, but I didn't do the work to get it, if you want me to post it.

Answer 24

form that you get tan inverse u + x

Answer 25

please give me the solution

Answer 26

\[\int\limits2u^2/(1+u2)\]

Answer 27

thats how u break that

Answer 28

\[\int \frac {sin^5(x)}{1+cos^2(x)} = \frac {cos^3(x)}{3} -3cos(x) +4tan^{-1}(cos(x)) + C\]Courtesy of Wolfram Alpha

Answer 29

Thank you for helping, Good night

Answer 30

Another solution by plugging the original equation into Wolfram Alpha is \[\int \frac {sin^5(x)}{1+cos^2(x)}dx = \frac {11cos(x)}{4} - \frac {1}{12}cos(3x) - 4tan^{-1}(cos(x)) +C\]