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How do you integrate trig functions with fractions??? eg. cosecx or secx
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Hi 1a2b3c2212;
Here is one way, but it requires really good tables.
For trig function compositions I use this table here:
http://staff.jccc.net/swilson/trig/compositions.htm
Say
Also from a table.
Look at the composition of trig functions on the above page. They have:
So:
Now use the sub from above.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Is it possible to integrate from here?
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Hi;
If you are familiar with the derivatives of your trig functions then you know that:
So the integral is the reverse operation.
Since csc(x) = 1 / sin(x) and csc(x)^2 = 1 / (sin(x))^2. So going backwards.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Here another way to it
∫▒〖1/sin^2x dx〗
let t=tan〖x/2〗
dt/dx=1/2 sec^2〖x/2〗
therefore dx=2dtcos^2 x/2
now tan x/2=t
thefore cos x/2=1/√(1+t^2 )
sin〖x/2〗=t/√(1+t^2 )
now sinx=2sin x/2 cos x/2
hence sinx=2t/(1+t^2 ) therefore sin^2〖x=4t^2/(1+t^2 )^2 〗
now substituting x by t we get
∫▒1/sin^2x dx= ∫▒(1+t^2 )^2/(4t^2 )×2/((1+t^2)) dt
=1/2 ∫▒(1+t^2)/t^2 dt
=1/2 ∫▒1/t^2 +1 dt
=1/2 [(-1 )/t+t]+c
=-[(1-t^2)/2t]+c
=-[(1-tan^2〖x/2〗)/(2tan x/2)]+c
now tanx=(2tan x/2)/(1-tan^2〖x/2〗 ) hence cotx=(1-tan^2〖x/2〗)/(2tan x/2)
therefore ∫▒〖1/〖sin〗^2x dx〗=-cotx+c
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substitution is the only way?
i was taught to increase the power by 1 and divide by the power+ 1.
can this apply to trig functions too?
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Hi 1a2b3c2212;
No substitution is not the only way. Post #4 is the easiest way, I think.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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what if i am not familiar with d cotx /dx = -csc(x)² ?
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Hi 1a2b3c2212;
Post #2 is another way and post #5 is yet another. I haven't been able to think of anything else except integration by parts which I didn't have much luck with.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
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Hi 1a2b3c2212;
You have been shown a number of ways to do this one, You are expected to know the integrals and derivatives of the the trig functions. At least you would be allowed access to a table, perhaps even on a test. In the real world you would definitely be able to use a table.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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this makes sense to me but i cannot get the negative sign. is there something wrong?
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Hi;
The fourth step is not right.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hi1 a2b3c2212
bobbym Moderator is right
riad zaidan
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why is it wrong to take out cot²x ?
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Hi;
You cannot pull the variable x passed the integral sign. Only constants! In this case it is a miraculous that your final answer is only off by a negative sign.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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oh. i cant take out unknown... how coicidental. thanks for the help
Last edited by 1a2b3c2212 (2010-07-09 00:21:53)
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Hi;
It looks like you just floated the cot(x)^2 right passed the integral sign. Is that what you did? Whatever you algebraically do to the integrand for the variable that you are integrating with respect too, stays on the right side.
This is fine:
This is not:
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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how to solve modulus ineqality
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Hi mukesh;
What is the example you want worked on?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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HI 1a2b3c2212;
There is another way to do your integral it uses a reduction formula.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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