secxtanx=1?

**dbakeg00** · November 16th 2010, 01:14 PM

Can I use identities to make sec(x)tan(x)=1?

Moderator edit: This question is not relevant to the Calculus subforum. However, the relevance of this thread to the Calculus subforum only becomes clear with post #3.

**Plato** · November 16th 2010, 01:20 PM

Originally Posted by dbakeg00

Can I use identities to make sec(x)tan(x)=1?

What do you mean by 'make'?

Do you mean solve $\sec(x)\tan(x)=1$ for $x~?$

**dbakeg00** · November 16th 2010, 01:33 PM

This is the wrong forum for this but my original problem is $\displaystyle\int\frac{secxtanx}{9+4sec^2x}dx$

So I used the integral identity $\displaystyle\int\frac{1}{a^2+x^2}dx=\frac{1}{a}ta n^{-1}\frac{x}{a}+c$

Then I rewrote the problem $\displaystyle\int\ secxtanx*\frac{1}{3^3+2^2sec^2x}dx$

$\displaystyle\int secxtanx*\frac{1}{\frac{1}{4}[(\frac{3}{2})^2+sec^2x]}dx$

So, if i could find a way to transform that secxtanx, then I would be home free...is that clear?

**Plato** · November 16th 2010, 01:37 PM

You do know that $\dfrac{{d\sec (x)}}{{dx}} = \sec (x)\tan (x)~?$

**dbakeg00** · November 16th 2010, 01:44 PM

Originally Posted by Plato

You do know that $\dfrac{{d\sec (x)}}{{dx}} = \sec (x)\tan (x)~?$

I do..and maybe I'm missing something because I'm still new to integrals but the answer given in the book is:

$\displaystyle\frac{1}{6}tan^{-1}(\frac{2secx}{3})+c$

which is what I come up with when I integrate this part

$\displaystyle\frac{1}{\frac{1}{4}[(\frac{3}{2})^2+sec^2x]}dx$

so I had the secxtanx left over and didnt know what to do with it

**Plato** · November 16th 2010, 01:52 PM

$\displaystyle \int {\frac{{\sec (x)\tan (x)}} {{9 + 4\sec ^2 (x)}}dx = \frac{1} {4}\int {\frac{{\sec (x)\tan (x)}} {{\left( {\frac{3} {2}} \right)^2 + \sec ^2 (x)}}dx = \frac{1} {4}\left( {\frac{2} {3}} \right)\arctan \left( {\frac{{2x}} {3}} \right)} } $

**dbakeg00** · November 16th 2010, 01:57 PM

So what happened to the sec(x)tan(x)? shouldn't it be sec(x)?

**Plato** · November 16th 2010, 02:09 PM

Originally Posted by dbakeg00

So what happened to the sec(x)tan(x)? shouldn't it be sec(x)?

Let $u=\sec(x)$ then $\displaystyle \[\frac{1} {4}\int {\frac{{\sec (x)\tan (x)dx}}{{\left( {\frac{3} {2}} \right)^2 + \sec ^2 (x)}}} = \frac{1}{4}\int {\frac{{du}} {{\left( {\frac{3}{2}} \right)^2 + u^2 }}}$

**dbakeg00** · November 16th 2010, 02:15 PM

there goes the light bulb! Thanks, I really appreciate it!

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