Stuck on Trigonometric Integral

**study777** · June 27th 2008, 11:31 AM

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.

**TKHunny** · June 27th 2008, 11:39 AM

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?

**study777** · June 27th 2008, 11:43 AM

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.

**galactus** · June 27th 2008, 11:49 AM

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.

**Mathstud28** · June 27th 2008, 01:04 PM

http://www.mathhelpforum.com/math-he...-integral.html

Thread: Stuck on Trigonometric Integral

LinkBack

Thread Tools

Display

Stuck on Trigonometric Integral

Similar Math Help Forum Discussions

Stuck in this trigonometric equation...

Stuck on a problem in Homework (Trigonometric Identities and Formulas)

Stuck on trying to find integral of a mixture of trigonometric functions.

Stuck with integration, trigonometric function

Proof by induction stuck on trigonometric sum :help:

Search Tags