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

Question

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

Emily Ammerman · Accepted Answer

#intcosx/sin^2xdx=-cscx# Let #u=sinx#, then #du=cosxdx# and #intcosx/sin^2xdx# = #int(du)/u^2# = #-1/u# = #-1/sinx# = #-cscx#

Aurora Alvarado · Answer

# -csc(x)#

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 :)

William Abdalla · Answer

undefined To evaluate the integral ∫(cos(x))/(sin^2(x)) dx, you can use the substitution method. Let u = sin(x), then du = cos(x) dx. The integral becomes ∫du/u^2, which is straightforward to solve. Integrating ∫du/u^2 gives -1/u + C. Now, replace u with sin(x) to get the final answer: -1/sin(x) + C.