Question

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

Answer 1

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

Explanation:

`int csc^n(x) dx = (-1)/(n-1)cotx csc^(n-2)x+(n-2)/(n-1)intcsc^(n-2)x dx`

Repeat until you get `int csc^2 x dx` which is `-cotx` or you get `intcscx dx` which is `ln abs(cscx-cotx)`

Answer 2

`intcsc^n(x)dx=(n-2)/(n-1)intcsc^(n-2)(x)dx-(cot(x)csc^(n-2)(x))/(n-1)`

Explanation:

If we want to derive the reduction formula:

`I=intcsc^n(x)dx=intcsc^(n-2)(x)csc^2(x)dx`

Now, perform integration by parts on this, taking the form `intudv=uv-intvdu`.

Let `u=csc^(n-2)(x)`. Differentiating this gives:

`du=(n-2)csc^(n-3)(x)*(-csc(x)cot(x))dx`

`color(white)(du)=-(n-2)csc^(n-2)(x)cot(x)dx`

And let `dv=csc^2(x)`. Integrating this gives `v=-cot(x)`.

Plugging these in yields:

`I=-cot(x)csc^(n-2)(x)-(n-2)intcsc^(n-2)(x)cot^2(x)dx`

Now, through the Pythagorean identity, let `cot^2(x)=csc^2(x)-1`.

`I=-cot(x)csc^(n-2)(x)-(n-2)intcsc^(n-2)(x)(csc^2(x)-1)dx`

Distributing which gives:

`I=-cot(x)csc^(n-2)(x)-(n-2)intcsc^n(x)dx+(n-2)intcsc^(n-2)(x)dx`

Since `intcsc^n(x)dx=I`:

`I=-cot(x)csc^(n-2)(x)-(n-2)I+(n-2)intcsc^(n-2)(x)dx`

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

`(n-1)I=-cot(x)csc^(n-2)(x)+(n-2)intcsc^(n-2)(x)dx`

Solving for `I=csc^n(x)dx`:

`intcsc^n(x)dx=(-cot(x)csc^(n-2)(x))/(n-1)+(n-2)/(n-1)intcsc^(n-2)(x)dx`

And then repeat this with `intcsc^(n-2)(x)dx` until it's either `intcsc^2(x)dx=-cot(x)` or `intcsc(x)dx=-lnabs(cot(x)+csc(x))`.