# Stuck on Trigonometric Integral

• Jun 27th 2008, 11:31 AM
study777
Stuck on Trigonometric Integral
I can get to a point in the problem where I have the integral of sine squared over cosine cubed dx, But I'm not even sure if getting to that point is correct. Any help would be much appreciated and if someone could please show the steps that would just be AWESOME! Thank you.

http://img262.imageshack.us/img262/6524/problemmu4.png
• Jun 27th 2008, 11:39 AM
TKHunny
Just a hint:

$\displaystyle \int\;\sec^{3}(x)\;dx\;=\;$

$\displaystyle \int\;\sec(x)*\sec^{2}(x)\;dx\;=\;$

$\displaystyle \int\;\sec(x)\;d(\tan(x))\;=\;\sec(x)*\tan(x)\;-\;\int\;\sec(x)*\tan^{2}(x)\;dx$

Now what? Or did we actually get anywhere?
• Jun 27th 2008, 11:43 AM
study777
This is as far as I have gotten on my own. I tried break up the tan squared and that just seemed to make a bigger mess.
• Jun 27th 2008, 11:49 AM
galactus
Use the general reduction formula:

$\displaystyle \int sec^{n}(x)dx=\frac{sec^{n-2}(x)tan(x)}{n-1}+\frac{n-2}{n-1}\int sec^{n-2}(x)dx$

If you must reinvent the wheel, try using parts. That is how the above formula is derived.

$\displaystyle u=sec(x), \;\ dv=sec^{2}(x)dx, \;\ v=\int sec^{2}(x)dx=tan(x), \;\ du=sec(x)tan(x)dx$

$\displaystyle \int sec^{3}(x)dx = sec(x)tan(x)-\int sec(x)tan^{2}(x)dx$

$\displaystyle =sec(x)tan(x)-\int sec(x)(sec^{2}(x)-1)dx$

$\displaystyle =sec(x)tan(x)-\int sec^{3}(x)dx+\int sec(x)dx$

Now, add $\displaystyle \int sec^{3}(x)dx$ to both sides and finish.
• Jun 27th 2008, 01:04 PM
Mathstud28