What's the integral of int (tanx)^8 ?

1 Answer
Jun 7, 2016

tan^7 x / 7 - tan^5 x / 5 + tan^3 x / 3 - tan x + x + K

Explanation:

First, to prevent loss in presentation marks, we always tag on the dx on the end of the integral. That is, our goal is to find int (tan^8 x) dx.

Notice that tan^8 x = tan^6 x tan^2 x = tan^6 x (sec^2 x -1) = tan^6 x sec^2 x - tan^6 x

Then int (tan^8 x) dx = int(tan^6 x)(sec^2 x)dx - int(tan^6 x)dx

To solve the first integral, we let u = tan x which implies that du = sec^2 x dx, thus the integral is simplified to

int u^6 du = u^7/7 + K = tan^7 x / 7 + K.

Using similar methods to find int(tan^6 x)dx, int(tan^4 x)dx and int(tan^2 x)dx, we derive that

int(tan^8 x)dx = tan^7 x / 7 - tan^5 x / 5 + tan^3 x / 3 - tan x + x + K


For a slightly shorter method, we use the method of reduction formulae. Consider the general integral I_n=int(tan^n x)dx. Using the same trigonometric identity, we find that

int (tan^n x) dx = int(tan^(n-2) x)(sec^2 x)dx - int(tan^(n-2) x)dx

which stands to reason that I_n = tan^(n-1) x / (n-1) - I_(n-2).

All we now need to do is find I_8, which in turn requires us to find I_6, I_4, I_2 and I_0 respectively.

I_0 is fairly easy, since int(tan^0 x) dx = int 1 dx = x + K.

Using the formulae given (I_n = tan^(n-1) x / (n-1) - I_(n-2)), we can find everything else.

I_2 = tan x - I_0 = tan x - x - K

I_4 = tan^3 x / 3 - I_2 = tan^3 x / 3 - tan x + x + K

I_6 = tan^5 x / 5 - I_4 = tan^5 x / 5 - tan^3 x / 3 + tan x - x - K

I_8 = tan^7 x / 7 - I_6 = tan^7 x / 7 - tan^5 x / 5 + tan^3 x / 3 - tan x + x + K

Therefore, int(tan^8 x)dx = tan^7 x / 7 - tan^5 x / 5 + tan^3 x / 3 - tan x + x + K