integral help please?

**silverbells213** · September 14th 2009, 07:09 PM

can someone help me find the integral of sec cubed/tan?
I would really appreciate it!

**seld** · September 14th 2009, 07:22 PM

$\int\frac{sec^3\theta}{tan \theta}d\theta$

You can convert secant and tangent into sines and cosines.

$\int\frac{\frac{1}{cos^3\theta}}{\frac{sin\theta}{ cos\theta}}d\theta$

$=\int\frac{cos\theta}{sin\theta * cos^3\theta}d\theta$

$=\int\frac{1}{sin\theta*cos^2\theta}d\theta$

$=\int sec^2\theta*csc\theta d\theta$

can you take it from there, you do it by parts.

**redsoxfan325** · September 14th 2009, 07:57 PM

Use $u=\csc\theta$ and $dv=\sec^2\theta\,d\theta$

**silverbells213** · September 15th 2009, 03:46 AM

thank you so much!

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