Question

How to derive power reducing formula for `int(sec^nx)dx` and `int (tan^nx)dx` for integration?

Answer 1

`I_n = int \ sec^nx \ dx => I_n = 1/(n-1)tanx \ sec^(n-2)x + (n-2)/(n-1)I_(n-2)`

`J_n = int \ tan^nx \ dx => J_n = tan^(n-1)x/(n-1) - J_(n-2)`

Explanation:

We seek reduction formula for

A) `I_n = \ int \ sec^nx \ dx`
B) `J_n = \ int \ tan^nx \ dx`

Part (A):

If we assume that `n gt 2`, We can write:

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

We can then apply Integration By Parts. Let:

# { (u,=sec^(n-2)x, => (du)/dx,=(n-2)sec^(n-3)x \ secx \ tanx ), ((dv)/dx,=sec^2x , => v,=tanx ) :}#

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

We have:

`int \ (sec^(n-2)x)(sec^2x) \ dx = (sec^(n-2)x)(tanx) - int \ (tanx)((n-2)sec^(n-3)x \ secx \ tanx) \ dx`

`:. I_n = tanx \ sec^(n-2)x - (n-2) \ int \ tan^2x \ sec^(n-2)x \ dx`

Using `1+tan^2 theta -= sec^2 theta` we get:

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

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

`:. I_n = tanx \ sec^(n-2)x - (n-2) (I_n-I_(n-2))`

`:. I_n = tanx \ sec^(n-2)x - (n-2) I_n+(n-2)I_(n-2)`

`:. (n-1)I_n = tanx \ sec^(n-2)x + (n-2)I_(n-2)`

`:. I_n = 1/(n-1)tanx \ sec^(n-2)x + (n-2)/(n-1)I_(n-2)`

Part (B):

If we assume that `n gt 2`, We can write:

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

Using `1+tan^2 theta -= sec^2 theta` we get:

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

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

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

Now we can perform a substitution, Let:

`u=tanx => (du)/dx = sec^2x`

And substituting into the above result:

`J_n = int \ u^(n-2) \ du - J_(n-2)`

Which is trivial to integrate, so doing so give sus:

`J_n = u^(n-1)/(n-1) - J_(n-2)`

And restoring the substitution, we get:

`J_n = tan^(n-1)x/(n-1) - J_(n-2)`