Trig Integration

**WhoCares357** · October 13th, 2009, 11:29 AM

$\int sec(x)^5dx$

http://www.wolframalpha.com/input/?i=integrate+sec%28x%29^5

I know that there is a "reduction formula" but I want to know how to do it without that formula.

**Aryth** · October 13th, 2009, 01:26 PM

We know that we have to set this:

$I_n = \int sec^n(x) ~dx$

We know that:

$\int sec^2 ~dx = tan(x) + C$

So:

$I_n = \int sec^{n-2}(x)sec^2(x) ~dx$

$I_n = \int sec^{n-2}(x)d(tan(x)) ~dx$

$I_n = sec^{n-2}(x)tan(x) - \int tan(x)d(sec^{n-2}(x))$

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

$I_n = sec^{n-2}(x) tan(x) - (n-2)\int sec^{n-2}(x)tan^2(x) ~dx$

We know that:

$tan^2(x) = sec^2(x) - 1$

So that:

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

$I_n = sec^{n-2}(x)tan(x) - (n-2)[\int sec^{n}(x) ~dx - \int sec^{n-2}(x) ~dx]$

$I_n = sec^{n-2}(x) tan(x) - (n-2)[I_n - I_{n-2}]$

$(n-1)I_n = sec^{n-2}(x)tan(x) + (n-2)I_{n-2}$

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

I believe that is the reduction for Secant.

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