MHB Integrate by Parts: Solving Difficult Integrand

[cbarker1]

I am trying to integrate a difficult integrand.
\[1/2*\int \sin(\sqrt(3)/2x)*\sec(\sqrt(3)x)\, dx\]
I know that it requires to use integrate by parts.
Which function do I use to for the differential and integrable?

 

[alyafey22]

Well, I don't think this is solvable .

 

[Sudharaka]

I am trying to integrate a difficult integrand.
\[1/2*\int \sin(\sqrt(3)/2x)*\sec(\sqrt(3)x)\, dx\]
I know that it requires to use integrate by parts.
Which function do I use to for the differential and integrable?

Hi Cbarker1, :)

I think there is a little ambiguity in your integral due to the lack of parenthesis. Did you meant this,

\[\frac{1}{2}\int\sin\left(\frac{\sqrt{3}x}{2}\right)\sec(\sqrt{3}x)\,dx\]

or this,

\[\frac{1}{2}\int\sin\left(\frac{\sqrt{3}}{2x}\right)\sec(\sqrt{3}x)\,dx\]

Kind Regards,
Sudharaka.

 

[Opalg]

I am trying to integrate a difficult integrand.
\[1/2*\int \sin(\sqrt(3)/2x)*\sec(\sqrt(3)x)\, dx\]
I know that it requires to use integrate by parts.
Which function do I use to for the differential and integrable?

I am assuming that the first of Sudharaka's readings is the one that is intended: $\frac12{\displaystyle\int} \sin\bigl(\frac{\sqrt3}2x\bigr)\sec(\sqrt3x)\,dx$. If you write $$\sec(\sqrt3x) = \frac1{\cos(\sqrt3x)} = \frac1{2\cos^2 \bigl(\frac{\sqrt3}2x\bigr) -1}$$ and then make the substitution $u = \cos\bigl(\frac{\sqrt3}2x\bigr)$, the integral becomes $\displaystyle-\frac{\sqrt3}4 \int\frac{du}{2u^2-1}$, which you can integrate using partial fractions.