Question

How do you find the integral of `int tan^n(x)` if n is an integer?

Answer 1

See the explanation section below.

Explanation:

For `n >= 3`, use Integration by substitution to decrease the exponent on `tanx`

`int tan^nx dx = int tan^(n-2) tan^2x dx`

`= int tan^(n-2) (sec^2x - 1) dx`

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

The first integral is `int u^(n-2) du`, so we get

`int tan^nx dx = 1/(n+1) tan^(n-1) x - int tan^(n-2) x dx`

The second integral is of the same form as the first. So, repaet the process until you arrive at

(for `n` even): `int tan^2 x dx = int (sec^2x -1) dx = tanx - x +C`

or

(for `n` odd): `int tan x dx = int (sinx/cosx) dx = ln abs(secx) +C`.

Answer 2

The general formula is too difficult for elementary calculus classes.

Click for "Answer"

You have to determine these the hard way, sorry. I will do the first three.

`int tanxdx`

Multiply by `secx/secx`:

`= int (secxtanx)/secxdx`

Let:
`u = secx`
`du = secxtanxdx`

`=> int 1/udu`

`= ln|u|`

`= color(blue)(ln|secx| + C)`

Ironically, the second degree `tan` is easier to differentiate.

`int tan^2xdx`

Use trig identity and follow up:

`= int sec^2x - 1dx`

`= color(blue)(tanx - x + C)`

And the third:

`int tan^3xdx`

Split into familiar first and second-degree `tan` terms:

`= int tan^2x tanx dx`

Use trig identity:

`= int tanx(sec^2x - 1)dx`

`= inttanxsec^2x - tanxdx`

Split into familiar `u` and `du` pairs:

`= int secxtanx*secx - tanxdx`

Let:
`u = secx`
`du = secxtanxdx`

`=> int udu - inttanxdx`

Recall the first-degree `tan` solution and re-use it.

`= color(blue)(sec^2x/2 - ln|secx| + C)`

Hopefully that gave you the general techniques you can use to approach higher degree problems.

Answer 3

See the explanation.

Explanation:

Maybe this helps:

Let `n=2k+1`, then:

`int tan^nxdx= int tan^(2k+1)xdx=int (tan^2x)^k tanxdx=I`

`tan^2x=t => 2tanxsec^2xdx=dt => tanxdx=dt/(2(t+1))`

since: `sec^2x=tan^2x+1`

`I=1/2 int t^k/(t+1)dt`

Dividing polynomial `t^k` with `t+1` gives us following result:

`t^k:(t+1)=t^(k-1)-t^(k-2)+t^(k-3)-t^(k-4)+...`

and reminder

`1/(t+1)`

which sign depends on `k` (if `k` is even it's positive and vice versa)

Hence,

`t^k/(t+1)=sum_(i=1)^k (-1)^(i+1) t^(k-i) + (-1)^k 1/(t+1)`

`I=1/2 int (sum_(i=1)^k (-1)^(i+1) t^(k-i) + (-1)^k 1/(t+1)) dt`

`I=1/2 (sum_(i=1)^k (-1)^(i+1) int t^(k-i)dt+(-1)^k int dt/(t+1))`

`I=1/2 (sum_(i=1)^k (-1)^(i+1) t^(k-i+1)/(k-i+1) + (-1)^k ln(t+1))`

`I=1/2 (sum_(i=1)^k (-1)^(i+1) (tan^2x)^(k-i+1)/(k-i+1) + (-1)^k ln(tan^2x+1))`

`I=1/2 (sum_(i=1)^k (-1)^(i+1) (tan^2x)^(k-i+1)/(k-i+1) + (-1)^k lnsec^2x)`

When `n=2k`:

`int tan^nxdx= int tan^(2k)xdx = I`

`tanx=t => sec^2xdx=dt => dx=dt/(t^2+1)`

And hence:

`I=int t^(2k)/(t^2+1)dt`

`t^(2k)/(t^2+1)=sum_(i=1)^k (-1)^(i+1) t^(2(k-i)) + (-1)^k 1/(t^2+1)`

Hence,

`I=int ( sum_(i=1)^k (-1)^(i+1) t^(2(k-i)) + (-1)^k 1/(t^2+1) ) dt`

`I= sum_(i=1)^k (-1)^(i+1) int t^(2(k-i)) dt + (-1)^k int 1/(t^2+1) dt`

`I= sum_(i=1)^k (-1)^(i+1) t^(2(k-i)+1)/(2(k-i)+1) + (-1)^k arctant`

`I= sum_(i=1)^k (-1)^(i+1) (tanx)^(2(k-i)+1)/(2(k-i)+1) + (-1)^k arctan(tanx)`

`I= sum_(i=1)^k (-1)^(i+1) (tanx)^(2(k-i)+1)/(2(k-i)+1) + (-1)^k x`