Integral help please

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shady2.0
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#1
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#1
Evaluate the integral of 3dx/15+9cosx
this question is giving me a hard time,I tried to let the cosx=2cos^2(x/2)-1,But cant get the answer so need help
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Andy98
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(Original post by shady2.0)
Evaluate the integral of 3dx/15+9cosx
this question is giving me a hard time,I tried to let the cosx=2cos^2(x/2)-1,But cant get the answer so need help
Do you mean 3 \int 15 + 9 \cos x \mathrm{d}x?
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Slowbro93
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(Original post by shady2.0)
Evaluate the integral of 3dx/15+9cosx
this question is giving me a hard time,I tried to let the cosx=2cos^2(x/2)-1,But cant get the answer so need help
Can you take a photo of the question?
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atsruser
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(Original post by shady2.0)
Evaluate the integral of 3dx/15+9cosx
this question is giving me a hard time,I tried to let the cosx=2cos^2(x/2)-1,But cant get the answer so need help
If it's as written, look up the Weierstrass substitution: t=\tan \frac{\theta}{2} - Google is your friend.
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Andy98
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(Original post by atsruser)
If it's as written, look up the Weierstrass substitution: t=\tan \frac{\theta}{2} - Google is your friend.
You don't need that as such.
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Andy98
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Actually, looking at it; it's probably \int \frac{3}{15+9 \cos x} \mathrm{d}x
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shady2.0
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(Original post by Andy98)
Actually, looking at it; it's probably \int \frac{3}{15+9 \cos x} \mathrm{d}x
Yeah that is exactly the one
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davros
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(Original post by shady2.0)
Yeah that is exactly the one
The t-substitution quoted above is the standard way to tackle integrals like this.
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Andy98
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#9
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(Original post by shady2.0)
Yeah that is exactly the one
Ahh well I'd cancel the 3 out first

(Original post by davros)
The t-substitution quoted above is the standard way to tackle integrals like this.
Is it? Can you not split the fraction?
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username1162972
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(Original post by Andy98)
Ahh well I'd cancel the 3 out first



Is it? Can you not split the fraction?
Hmm, maybe split it but i dont see that getting it anywhere.


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Andy98
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(Original post by physicsmaths)
Hmm, maybe split it but i dont see that getting it anywhere.


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Oh

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davros
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(Original post by Andy98)
Ahh well I'd cancel the 3 out first



Is it? Can you not split the fraction?
How exactly would you propose "splitting the fraction"?
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shady2.0
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#13
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(Original post by davros)
How exactly would you propose "splitting the fraction"?
The three will cancel out,but tried to use the the double angle formula for cos and still dont get the answer
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Andy98
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#14
(Original post by davros)
How exactly would you propose "splitting the fraction"?
Ahhhh *******s, just realised I was thinking of it as if was the other way round. Meaning I don't have a clue, which is worrying because I've literally just finished FP2

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atsruser
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(Original post by shady2.0)
The three will cancel out,but tried to use the the double angle formula for cos and still dont get the answer
As I mentioned above, use t=\tan\frac{\theta}{2} - this is a standard substitution whose details you can find via Google. Look for "Weierstrass substitution".

You will end up with a polynomial fraction in t for which you will probably need partial fractions.
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Andy98
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(Original post by atsruser)
As I mentioned above, use t=\tan\frac{\theta}{2} - this is a standard substitution whose details you can find via Google. Look for "Weierstrass substitution".

You will end up with a polynomial fraction in t for which you will probably need partial fractions.
Is there not an easier way?

That looks horrible

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Zacken
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#17
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(Original post by atsruser)

You will end up with a polynomial fraction in t for which you will probably need partial fractions.
(Original post by Andy98)
Is there not an easier way?

That looks horrible

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No need for partial fractions. Integrates to arctan directly after the sub.
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Andy98
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#18
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#18
(Original post by Zacken)
No need for partial fractions. Integrates to arctan directly after the sub.
Arctan of tan? Yikes

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Zacken
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#19
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#19
(Original post by Andy98)
Arctan of tan? Yikes

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(Not in this case)

But usually, arctan of tan is nice and definitely not yucky.

Arctan(tan x) = x (as long as x lies in a suitable domain)
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Andy98
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(Original post by Zacken)
(Not in this case)

But usually, arctan of tan is nice and definitely not yucky.

Arctan(tan x) = x (as long as x lies in a suitable domain)
Ahhhh....

Dunno if you noticed but I have nightmares about trig - it's terrifying

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