How to derive power reducing formula for int(sec^nx)dx and int (tan^nx)dx for integration?
1 Answer
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
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
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
J_n = int \ tan^(n-2)x \ tan^2x \ dx
Using
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)