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- #1

- Thread starter Cbarker1
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- Thread starter
- #1

- Jan 17, 2013

- 1,667

Well, I don't think this is solvable .

- Feb 5, 2012

- 1,621

Hi 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?

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.

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- #4

- Feb 7, 2012

- 2,702

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.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?