# Thread: Stuck on Trigonometric Integral

1. ## 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.

2. Just a hint:

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

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

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

Now what? Or did we actually get anywhere?

3. 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.

4. Use the general reduction formula:

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

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

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

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

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

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

Now, add $\int sec^{3}(x)dx$ to both sides and finish.