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.
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.
This is the wrong forum for this but my original problem is $\displaystyle \displaystyle\int\frac{secxtanx}{9+4sec^2x}dx$
So I used the integral identity $\displaystyle \displaystyle\int\frac{1}{a^2+x^2}dx=\frac{1}{a}ta n^{-1}\frac{x}{a}+c$
Then I rewrote the problem $\displaystyle \displaystyle\int\ secxtanx*\frac{1}{3^3+2^2sec^2x}dx$
$\displaystyle \displaystyle\int secxtanx*\frac{1}{\frac{1}{4}[(\frac{3}{2})^2+sec^2x]}dx$
So, if i could find a way to transform that secxtanx, then I would be home free...is that clear?
I do..and maybe I'm missing something because I'm still new to integrals but the answer given in the book is:
$\displaystyle \displaystyle\frac{1}{6}tan^{-1}(\frac{2secx}{3})+c$
which is what I come up with when I integrate this part
$\displaystyle \displaystyle\frac{1}{\frac{1}{4}[(\frac{3}{2})^2+sec^2x]}dx$
so I had the secxtanx left over and didnt know what to do with it
$\displaystyle \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)} }
$