Question

What is `int csc^4(x) cot^6(x) dx`?

Answer 1

`-cot^9(x)/9-cot^7(x)/7+C`

Explanation:

Notice that when we have `cot(x)` and `csc(x)` functions, we will often have derivatives embedded into an integral. Here, the most important derivative to remember is

`d/dx(cot(x))=-csc^2(x)`

Thus, if we let `u=cot(x)`, then `(du)/dx=-csc^2(x)` and `du=-csc^2(x)dx`.

However, we have `4` different `csc(x)` functions but we only want `2`. We can turn the remaining `csc^2(x)` term into functions of `cot(x)` using the form of the Pythagorean identity:

`1+cot^2(x)=csc^2(x)`

Thus, the integral becomes:

`intcsc^4(x)cot^6(x)dx=intcsc^2(x)csc^2(x)cot^6(x)dx`

`=intcsc^2(x)[(1+cot^2(x))(cot^6(x)]dx`

`=intcsc^2(x)[cot^6(x)+cot^8(x)]dx`

Now, in order to get a `du=-csc^2(x)dx` term, we need to multiply the interior and exterior of the integral by `-1`.

`=-int(-csc^2(x))[cot^6(x)+cot^8(x)]dx`

Now substitute, since we have out `du` term, recalling that `u=cot(x)`:

`=-int(u^6+u^8)du`

Integrating term by term, this gives us

`=-(u^7/7+u^9/9)+C`

`=-cot^9(x)/9-cot^7(x)/7+C`