How do I derive the integration reduction formula for tangent?
Integral of (tan x)^n dx = ...
I can do the derivations for sin/cosine, but I'm getting stuck on tan.
Thanks!
tan^n(x) =
sin^2(x)/cos^2(x)tan^(n-2)(x)
sin^2(x)/cos^2(x) =
[1-cos^2(x)]/cos^2(x) =
1/cos^2(x) - 1
So:
tan^n(x) = tan^(n-2)(x)/cos^2(x) -
tan^(n-2)(x)
Integral of tan^(n-2)(x)/cos^2(x)dx =
tan^(n-1)(x)/(n-1)
makes perfect sense. Thanks Count!
To derive the integration reduction formula for tangent, you will need to use integration by parts and the original integration formula for tangent. Here's how you can approach it step by step:
1. Start with the original integration formula for tangent:
∫ (tan x)^n dx
2. Rewrite tangent as the ratio of sine and cosine:
∫ ((sin x)/(cos x))^n dx
3. Simplify the expression by multiplying both the numerator and denominator by (cos x)^n:
∫ (sin x)^n / (cos x)^(n-1) dx
4. Now, let's perform integration by parts. The formula for integration by parts is:
∫ u * dv = u * v - ∫ v * du
Choose u = (sin x)^(n-1)
Choose dv = (sin x) dx
Differentiate u to find du:
du = (n-1) * (sin x)^(n-2) * cos x dx
Integrate dv to find v:
v = -cos x
5. Apply the integration by parts formula using the choices made above:
∫ (sin x)^n / (cos x)^(n-1) dx = - (sin x)^(n-1) * cos x + ∫ (n-1) * (sin x)^(n-2) * cos^2 x dx
6. Now, simplify the expression by using the trigonometric identity cos^2 x = 1 - sin^2 x:
∫ (sin x)^n / (cos x)^(n-1) dx = - (sin x)^(n-1) * cos x + ∫ (n-1) * (sin x)^(n-2) * (1 - sin^2 x) dx
7. Expand the expression and distribute:
∫ (sin x)^n / (cos x)^(n-1) dx = - (sin x)^(n-1) * cos x + ∫ (n-1) * (sin x)^(n-2) dx - ∫ (n-1) * (sin x)^n dx
8. Simplify the second integral by integrating:
∫ (sin x)^n / (cos x)^(n-1) dx = - (sin x)^(n-1) * cos x + (n-1) * ∫ (sin x)^(n-2) dx - ∫ (n-1) * (sin x)^n dx
9. Finally, rewrite the integration formula in terms of the original variable x:
∫ (tan x)^n dx = - (sin x)^(n-1) * cos x + (n-1) * ∫ (tan x)^(n-2) dx - ∫ (n-1) * (tan x)^n dx
Rearrange the equation to solve for the integral of (tan x)^n:
∫ (tan x)^n dx = (sin x)^(n-1) * cos x + (n-1) * ∫ (tan x)^(n-2) dx / (n - 1)
Therefore, the integration reduction formula for tangent is:
∫ (tan x)^n dx = (sin x)^(n-1) * cos x + (n-1) * ∫ (tan x)^(n-2) dx / (n - 1)
I hope this explanation helps you understand how to derive the integration reduction formula for tangent!