Integral help please

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

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

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.

Actually, looking at it; it's probably $\int \frac{3}{15+9 \cos x} \mathrm{d}x$

Yeah that is exactly the one

Yeah that is exactly the one

The t-substitution quoted above is the standard way to tackle integrals like this.

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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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"?

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

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.

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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No need for partial fractions. Integrates to arctan directly after the sub.

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)