# Integration problem ∫1/(√3 sinx+ cosx) dx

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## Homework Statement

∫1/(√3 sinx+ cosx) dx is
Ans. as per textbook 1/2(log(tan((x/2)+(pi/12))
2. The attempt at a solution
Attempt 1-I changed it to half angles (x/2) and then multiplied and divided sec^2(x/2) to the numerator and denominator, then putting tan(x/2)=t I got 2∫1/(√3t+1-t^2) I wrote the quadratic expression as (√7/2)^2-(t-(√3/2)^2) [completing the square], then applying the standard integral form of 1/(a^2-x^2) I substituted the values into 1/2a(ln(a+x)-ln(a-x)) but I don't know how to match it with the answer
Attempt 2- I took 1/2 common from denominator to convert it into sin(pi/3)sinx+cos(pi/3)cosx=cos(x-(pi/3))
then I integrated using ∫secx=log(secx+tanx) and ended up with 1/2(log(sec(x-(pi/3))+tan(x-(pi/3)) Again, I cant match with the answer. Would appreciate some help

## Answers and Replies

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fresh_42
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vela
Staff Emeritus
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Attempt 2- I took 1/2 common from denominator to convert it into sin(pi/3)sinx+cos(pi/3)cosx=cos(x-(pi/3))
then I integrated using ∫secx=log(secx+tanx) and ended up with 1/2(log(sec(x-(pi/3))+tan(x-(pi/3)) Again, I cant match with the answer. Would appreciate some help
I don't know if this is the most straightforward way. Let $u = (x-\pi/3)/2$ and rewrite the argument of the log in terms of sines and cosines. You should be able to show that
$$\sec\left(x-\frac \pi 3\right) + \tan\left(x-\frac \pi 3\right) = \frac{1+\sin 2u}{\cos 2u}.$$ Use the double-angle identities and eventually you can simplify it to
$$\frac{\cos u + \sin u}{\cos u - \sin u},$$ which you can show equals $\tan(u+\pi/4)$. Finish up by rewriting in terms of $x$.

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Please have a look on our LaTeX help page:
https://www.physicsforums.com/help/latexhelp/

I read the page, I'm still finding it very unnatural to use...is there a proper tutorial I can see somewhere,
I don't know if this is the most straightforward way. Let $u = (x-\pi/3)/2$ and rewrite the argument of the log in terms of sines and cosines. You should be able to show that
$$\sec\left(x-\frac \pi 3\right) + \tan\left(x-\frac \pi 3\right) = \frac{1+\sin 2u}{\cos 2u}.$$ Use the double-angle identities and eventually you can simplify it to
$$\frac{\cos u + \sin u}{\cos u - \sin u},$$ which you can show equals $\tan(u+\pi/4)$. Finish up by rewriting in terms of $x$.
I got my answer this way, isn't there a shorter method though- this'd be quite hard to intuitively guess in an exam

vela
Staff Emeritus
Homework Helper
Here's a slightly faster way:
$$\frac{1+\sin\left(x-\frac \pi 3\right)}{\cos\left(x-\frac \pi 3\right)} = \frac{\sin \frac\pi 2+\sin\left(x-\frac \pi 3\right)}{\cos \frac \pi 2 + \cos\left(x-\frac \pi 3\right)}$$ Then use the sum-to-product identities.

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Here's a slightly faster way:
$$\frac{1+\sin\left(x-\frac \pi 3\right)}{\cos\left(x-\frac \pi 3\right)} = \frac{\sin \frac\pi 2+\sin\left(x-\frac \pi 3\right)}{\cos \frac \pi 2 + \cos\left(x-\frac \pi 3\right)}$$ Then use the sum-to-product identities.
Got it! thank you

ehild
Homework Helper
I read the page, I'm still finding it very unnatural to use...is there a proper tutorial I can see somewhere,

I got my answer this way, isn't there a shorter method though- this'd be quite hard to intuitively guess in an exam
You do not need to guess intuitively, you certaninly were thought the methods to be used. First, if you see a formula asin(x)+bcos(x), write it in the form A sin(x+θ) where tan(θ)=b/a, $A= \sqrt{a^2+b^2}$ and a=A cos(θ), b=A sin(θ). In this case, A=2 and θ=pi/6. With the substitution u=x+π/6, the integrand becomes
$\int{\frac{1}{2\sin(u)}du}$.
It is also a regular method to integrate rational functions of sine, cosine, tangent, that we write the functions in terms of tan(u/2), and use the substitution t=tan(u/2) with $du = \frac{2}{1+t^2}dt$
$\sin(x)=\frac{2\tan(x/2)}{1+\tan^2(x/2)}$, $\cos(x)=\frac{1-\tan^2(x/2}{1+\tan^2(x/2)}$. These formulas are easy to remember or derive.

fresh_42
Mentor
I read the page, I'm still finding it very unnatural to use...is there a proper tutorial I can see somewhere
There are presumably many, but not easier to read. I used to click on the reply button just to see how others write it. You just have to remember to delete it again. After a short time the software with its automatic save will remember it next time you come back to the thread, so make sure there are no remains from your lookup if you'll answer again.