How do you find the integral of ∫cscn(x) if m or n is an integer?

2 Answers
Jim H · mason m
Oct 20, 2016

There is no m in the integral. To evaluate ∫cscn(x)dx for integer n use the reduction formula:

Explanation:

∫cscn(x)dx=−1n−1cotxcscn−2x+n−2n−1∫cscn−2xdx

Repeat until you get ∫csc2xdx which is −cotx or you get ∫cscxdx which is ln|cscx−cotx|

mason m
Oct 22, 2016

∫cscn(x)dx=n−2n−1∫cscn−2(x)dx−cot(x)cscn−2(x)n−1

Explanation:

If we want to derive the reduction formula:

I=∫cscn(x)dx=∫cscn−2(x)csc2(x)dx

Now, perform integration by parts on this, taking the form ∫udv=uv−∫vdu.

Let u=cscn−2(x). Differentiating this gives:

du=(n−2)cscn−3(x)⋅(−csc(x)cot(x))dx

du=−(n−2)cscn−2(x)cot(x)dx

And let dv=csc2(x). Integrating this gives v=−cot(x).

Plugging these in yields:

I=−cot(x)cscn−2(x)−(n−2)∫cscn−2(x)cot2(x)dx

Now, through the Pythagorean identity, let cot2(x)=csc2(x)−1.

I=−cot(x)cscn−2(x)−(n−2)∫cscn−2(x)(csc2(x)−1)dx

Distributing which gives:

I=−cot(x)cscn−2(x)−(n−2)∫cscn(x)dx+(n−2)∫cscn−2(x)dx

Since ∫cscn(x)dx=I:

I=−cot(x)cscn−2(x)−(n−2)I+(n−2)∫cscn−2(x)dx

Adding (n−2)I to both sides (this is weird to think about):

(n−1)I=−cot(x)cscn−2(x)+(n−2)∫cscn−2(x)dx

Solving for I=cscn(x)dx:

∫cscn(x)dx=−cot(x)cscn−2(x)n−1+n−2n−1∫cscn−2(x)dx

And then repeat this with ∫cscn−2(x)dx until it's either ∫csc2(x)dx=−cot(x) or ∫csc(x)dx=−ln|cot(x)+csc(x)|.

Related questions
See all questions in Integrals of Trigonometric Functions
Impact of this question
21349 views around the world
You can reuse this answer
Creative Commons License