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Integrating Trignometric stuff

I have to integrate 1/cos^2x sin^2x

I get to the point with 2(integralsign)1/1-cos(u)du where u= 4x

I don't know what to do next. :frown::confused:
Original post by Nirm
I have to integrate 1/cos^2x sin^2x

I get to the point with 2(integralsign)1/1-cos(u)du where u= 4x

I don't know what to do next. :frown::confused:


I'd go back a step to where you had (sin(2x))^2 in the denominator, which becomes (cosec(2x))^2 when you flip it up, and it's then a standard integral.
(edited 9 years ago)
Reply 2
Avatar for TeeEm
TeeEm
19
Original post by ghostwalker
I'd go back a step to where you had (sin(2x))^2 in the denominator, which becomes (cosec(2x)^2 when you flip it up, and it's then a standard integral.


two ways of doing this but I am not sure of your notation.

EDIT: I am gone...:colondollar:
Reply 3
Avatar for Nirm
Nirm
OP
13
Original post by ghostwalker
I'd go back a step to where you had (sin(2x))^2 in the denominator, which becomes (cosec(2x))^2 when you flip it up, and it's then a standard integral.

I never had a (sin(2x))^2 for some reason :eek:. I just made cos^2x sin^2x= 1-cos(4x)/8 (all over 8).
I am probably not seeing a simple step right now :colondollar: not sure how to get (sin(2x))^2
(edited 9 years ago)
Original post by Nirm
I never had a (sin(2x))^2 for some reason :eek:. I just made cos^2x sin^2x= 1-cos(4x)/8 (all over 8).
I am probably not seeing a simple step right now :colondollar: not sure how to get (sin(2x))^2


Sin(2x)=2sin(x)cos(x)
Squaring both sides gives a big hint in the right direction. You will soon have your integral in the form of a nice one.


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Reply 5
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davros
16
Original post by Nirm
I have to integrate 1/cos^2x sin^2x

I get to the point with 2(integralsign)1/1-cos(u)du where u= 4x

I don't know what to do next. :frown::confused:


Is this supposed to be

sin⁡2xcos⁡2x\dfrac{\sin^2x}{\cos^2x}

or

1cos⁡2xsin⁡2x\dfrac{1}{\cos^2x \sin^2x}

If you don't know Latex, at least help us out with a few extra brackets :smile:
1sin⁡2xcos⁡2x=sin⁡2x+cos⁡2xsin⁡2xcos⁡2x=1cos⁡2x+1sin⁡2x\displaystyle \frac{1}{\sin^2{x}\cos^2{x}} = \frac{\sin^2{x}+\cos^2{x}}{\sin^2{x}\cos^2{x}} = \frac{1}{\cos^2{x}}+\frac{1}{ \sin^2{x}}
Reply 7
Avatar for Nirm
Nirm
OP
13
Original post by davros
Is this supposed to be

sin⁡2xcos⁡2x\dfrac{\sin^2x}{\cos^2x}

or

1cos⁡2xsin⁡2x\dfrac{1}{\cos^2x \sin^2x}

If you don't know Latex, at least help us out with a few extra brackets :smile:

Yeah i dont know latex, need to learn it. Makes everything so clear. Also it is the bottom one.
Reply 8
Avatar for Nirm
Nirm
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13
Original post by Skαteboαrder
1sin⁡2xcos⁡2x=sin⁡2x+cos⁡2xsin⁡2xcos⁡2x=1cos⁡2x+1sin⁡2x\displaystyle \frac{1}{\sin^2{x}\cos^2{x}} = \frac{\sin^2{x}+\cos^2{x}}{\sin^2{x}\cos^2{x}} = \frac{1}{\cos^2{x}}+\frac{1}{ \sin^2{x}}

Ah okay i will try that. Thanks :smile:
Reply 9
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TeeEm
19
Original post by Nirm
Yeah i dont know latex, need to learn it. Makes everything so clear. Also it is the bottom one.


look at Q317 in the link

http://www.madasmaths.com/archive/maths_booklets/further_topics/integration/integration_indefinite_mixed_up.pdf

sorry but the down load time is over 5 min
Reply 10
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Nirm
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13
Original post by TeeEm

Its okay. Having a look at it/loading the page right now.

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