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

Apr 12, 2018

$\int \cos \frac{x}{\sin} ^ 2 x \mathrm{dx} = - \csc x$

#### Explanation:

Let $u = \sin x$, then $\mathrm{du} = \cos x \mathrm{dx}$ and

$\int \cos \frac{x}{\sin} ^ 2 x \mathrm{dx}$

= $\int \frac{\mathrm{du}}{u} ^ 2$

= $- \frac{1}{u}$

= $- \frac{1}{\sin} x$

= $- \csc x$

Apr 12, 2018

$- \csc \left(x\right)$

#### 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:

$\cos \frac{x}{\sin} ^ 2 \left(x\right) = \cos \frac{x}{\sin} \left(x\right) \cdot \frac{1}{\sin} \left(x\right)$

Now, let's simplify those. We know that $\cos \frac{x}{\sin} \left(x\right) = \cot \left(x\right)$, and $\frac{1}{\sin} \left(x\right) = \csc \left(x\right)$. So, our integral ultimately becomes:

$\implies \int \csc \left(x\right) \cot \left(x\right) \mathrm{dx}$

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

$\frac{d}{\mathrm{dx}} \left[\csc \left(x\right)\right] = - \csc \left(x\right) \cot \left(x\right)$

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 $- 1 \cdot - 1 = 1$.

$\implies - \int - \csc \left(x\right) \cot \left(x\right) \mathrm{dx}$

And this evaluates to:

$\implies - \csc \left(x\right)$

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 :)