Question

How do you evaluate the integral of `int (cosx)/(sin^(2)x) dx`?

Answer 1

`intcosx/sin^2xdx=-cscx`

Explanation:

Let `u=sinx`, then `du=cosxdx` and

`intcosx/sin^2xdx`

= `int(du)/u^2`

= `-1/u`

= `-1/sinx`

= `-cscx`

Answer 2

`-csc(x)`

Explanation:

You could do this using `u`-substitution, but there's a simpler way, that makes your life a bit easier.

Here's what we do. First, let's split this expression into the following product:

`cos(x)/sin^2(x) = cos(x)/sin(x) * 1/sin(x)`

Now, let's simplify those. We know that `cos(x)/sin(x) = cot(x)`, and `1/sin(x) = csc(x)`. So, our integral ultimately becomes:

`=> intcsc(x)cot(x) dx`

Now, we'll need to take a peek at our derivative table, and recall that:

`d/dx[csc(x)] = -csc(x)cot(x)`

This is exactly what we have in our integral EXCEPT there's a negative sign we need to take into account. So, we'll need to multiply by -1 twice to take this into account. Note that this does not change the value of the integral, since `-1 * -1 = 1`.

`=> -int-csc(x)cot(x) dx`

And this evaluates to:

`=> -csc(x)`

And that's your answer! You should know how to do this using `u`-sub, but keep an eye out for things like this, since at the very least, it's a way you can quickly check your answer.

Hope that helped :)