Posted by **john** on Thursday, June 3, 2010 at 11:22pm.

find the indefinite integral:

integration of tan^3(7x) dx

- calculus -
**Count Iblis**, Friday, June 4, 2010 at 11:55am
Yo can find a reduction formula for the integral of tan^n(x) as follows. We have that

tan^n(x) = sin^n(x)/cos^n(x) =

sin^(n-2)(x)/cos^n(x) sin^2(x) =

sin^(n-2)(x)/cos^n(x) [1-cos^2(x)] =

sin^(n-2)(x)/cos^n(x) -

sin^(n-2)(x)/cos^(n-2)(x) =

sin^(n-2)(x)/cos^(n-2)(x) 1/cos^2(x) -

sin^(n-2)(x)/cos^(n-2)(x) =

tan^(n-2)(x) 1/cos^2(x) - tan^(n-2)(x)

Now, 1/cos^2(x) is the derivate of

tan(x), so you can immediately integrate the first term:

Integral of tan^(n-2)(x) 1/cos^2(x)dx =

1/(n-1) tan^(n-1)(x)

The integral of the second term is, of course, a similar problem as the original problem, but with a lower value for n, so you can interate the formula until you end up at n = 1 or n = 0.

So, denoting the integral of tan^n(x)dx by I_n, we have:

I_n = 1/(n-1) tan^(n-1)(x) - I_{n-2}

For n = 3, we get:

I_3 = 1/2 tan^2(x) - I_1

and I_1 is the integral of tan(x)dx, which is -Log|cos(x)|.

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