Question

Integrate :-

Answer 1

my favourite topic :D

Answer 2

\[\int\limits \frac{\cos ^3x + \cos ^5x }{ \sin ^2x + \sin ^4x } dx\]

Answer 3

@hartnn
Not anymore? :D

Answer 4

this is a bummer, need to think a bit....
and ofcourse it still is :)

Answer 5

Wolfram crashed >.>

Answer 6

:O

Answer 7

But its doable,
\[\int\limits \frac{\cos ^3 x+\cos ^5 x}{\sin ^2 x+\sin ^4 x} \, dx=-\frac{1}{2} (\csc x) (cos(2x)+12sin(x)tan^{-1}(sin(x))+3)+c\]

Answer 8

i would try t= sin x

Answer 9

cos^3 + cos^5 dx = (cos^2 + cos^4) cos x dx

t= sin x
dt = -cos x dx

Answer 10

cos^2 + cos^4 is easy to convert into sin^2 terms

can you @goformit100 ?

Answer 11

Can I get to know the Process ?? @.Sam.

Answer 12

would you not like to try on your own first ?

Answer 13

@hartnn I am Weak At Integrals :(

Answer 14

at trigonometry ?

can you convert cos^2 x+ cos^4 x into only sin terms ?

Answer 15

yes

Answer 16

ok, what you got for
cos^2 x+ cos^4 x ?

Answer 17

there ?

Answer 18

I am unable.

Answer 19

you know sin^2 x + cos^2 x=1 ?

Answer 20

yes I know

Answer 21

so, whats cos^2 x from there ?
and whats cos^4 x ?

Answer 22

Not getting Please Explain

Answer 23

sin^2 x + cos^2 x = 1

so, cos^2 x =.... ?

Answer 24

1 - sin^2

Answer 25

good,
so cos^2 x = 1-sin^2 x
what about cos^4 x ?

Answer 26

Not getting really.

Answer 27

cos^4x =(cos^2)^2

Answer 28

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.

Answer 29

@goformit100

Answer 30

Ya. But I know I have to use the substitution. But How, don't know .

Answer 31

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?

Answer 32

Ok.

Answer 33

Ok please do it on the board, and then post it.

Answer 34

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

Answer 35

Ya, okay. After then ?

Answer 36

can I see how it looks like?

Answer 37

lies and slander. You're not even in Calculus, goformit!

Answer 38

@abb0t Mind your LANG^

Answer 39

oh my that is one long integration problem

Answer 40

\(\color{blue}{\text{Originally Posted by}}\) @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
\(\color{blue}{\text{End of Quote}}\)
\(dt = \cos x dx\)
\[
\int \frac{(1-\sin^2x)+(1-\sin^2 t)^2}{\sin^2x+\sin^4x} \cos xdx = \int \frac{(1-t^2)+(1-t^2)^2}{t^2+t^4}dt
\]

Answer 41

long, but not so difficult

Answer 42

alright going to bed guys. good luck night

Answer 43

Good night @UsukiDoll!

Answer 44

thanks for helping me earlier @wio :3