goformit100
  • goformit100
Integrate :-
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
hartnn
  • hartnn
my favourite topic :D
goformit100
  • goformit100
\[\int\limits \frac{\cos ^3x + \cos ^5x }{ \sin ^2x + \sin ^4x } dx\]
AkashdeepDeb
  • AkashdeepDeb
@hartnn Not anymore? :D

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hartnn
  • hartnn
this is a bummer, need to think a bit.... and ofcourse it still is :)
.Sam.
  • .Sam.
Wolfram crashed >.>
hartnn
  • hartnn
:O
.Sam.
  • .Sam.
But its doable, \[\int\limits \frac{\cos ^3 x+\cos ^5 x}{\sin ^2 x+\sin ^4 x} \, dx=-\frac{1}{2} (\csc x) (cos(2x)+12sin(x)tan^{-1}(sin(x))+3)+c\]
hartnn
  • hartnn
i would try t= sin x
hartnn
  • hartnn
cos^3 + cos^5 dx = (cos^2 + cos^4) cos x dx t= sin x dt = -cos x dx
hartnn
  • hartnn
cos^2 + cos^4 is easy to convert into sin^2 terms can you @goformit100 ?
goformit100
  • goformit100
Can I get to know the Process ?? @.Sam.
hartnn
  • hartnn
would you not like to try on your own first ?
goformit100
  • goformit100
@hartnn I am Weak At Integrals :(
hartnn
  • hartnn
at trigonometry ? can you convert cos^2 x+ cos^4 x into only sin terms ?
goformit100
  • goformit100
yes
hartnn
  • hartnn
ok, what you got for cos^2 x+ cos^4 x ?
hartnn
  • hartnn
there ?
goformit100
  • goformit100
I am unable.
hartnn
  • hartnn
you know sin^2 x + cos^2 x=1 ?
goformit100
  • goformit100
yes I know
hartnn
  • hartnn
so, whats cos^2 x from there ? and whats cos^4 x ?
goformit100
  • goformit100
Not getting Please Explain
hartnn
  • hartnn
sin^2 x + cos^2 x = 1 so, cos^2 x =.... ?
goformit100
  • goformit100
1 - sin^2
hartnn
  • hartnn
good, so cos^2 x = 1-sin^2 x what about cos^4 x ?
goformit100
  • goformit100
Not getting really.
hartnn
  • hartnn
cos^4x =(cos^2)^2
anonymous
  • anonymous
I can help, lets start with algebra. What ways do you see to simplify this fraction? it's always better to make the thing you're integrating simpler before you try to integrate it.
anonymous
  • anonymous
@goformit100
goformit100
  • goformit100
Ya. But I know I have to use the substitution. But How, don't know .
anonymous
  • anonymous
Well, there are several algebraic ways you can simplify the fraction (the thing between the integral sign and dx) For instance, you could try factoring something out. Do you see anything that you can factor out of the numerator? Let's start by seeing how this can be factored. Why don't you go ahead and factor what you can out of the numerator and denominator?
goformit100
  • goformit100
Ok.
anonymous
  • anonymous
Ok please do it on the board, and then post it.
anonymous
  • anonymous
One thing I notice is that since all the powers in the numerator are odd, if I' factor out a cos (x), I'll have a spare term of c0s (x) dx
goformit100
  • goformit100
Ya, okay. After then ?
anonymous
  • anonymous
can I see how it looks like?
abb0t
  • abb0t
lies and slander. You're not even in Calculus, goformit!
goformit100
  • goformit100
@abb0t Mind your LANG^
UsukiDoll
  • UsukiDoll
oh my that is one long integration problem
anonymous
  • anonymous
\(\color{blue}{\text{Originally Posted by}}\) @goformit100 \[\int\limits \frac{\cos ^3x + \cos ^5x }{ \sin ^2x + \sin ^4x } dx\] \(\color{blue}{\text{End of Quote}}\) \[ \int \frac{\cos x(\cos^2x+\cos^4 t)}{\sin^2x+\sin^4x }dx = \int \frac{\cos x((1-\sin^2x)+(1-\sin^2 t)^2)}{\sin^2x+\sin^4x }dx \] \(\color{blue}{\text{Originally Posted by}}\) @hartnn i would try t= sin x \(\color{blue}{\text{End of Quote}}\) \(dt = \cos x dx\) \[ \int \frac{(1-\sin^2x)+(1-\sin^2 t)^2}{\sin^2x+\sin^4x} \cos xdx = \int \frac{(1-t^2)+(1-t^2)^2}{t^2+t^4}dt \]
hartnn
  • hartnn
long, but not so difficult
UsukiDoll
  • UsukiDoll
alright going to bed guys. good luck night
anonymous
  • anonymous
Good night @UsukiDoll!
UsukiDoll
  • UsukiDoll
thanks for helping me earlier @wio :3

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