Question

Question #68567

Answer 1

`-1/2`

Explanation:

The `2x` bothers me so I'm going to do a `u` substitution:

`u=2x`

`du=2`

This changes our integrand to:

`u_1=2(pi/12)=pi/6`

`u_2=2(pi/4)=pi/2`

We lack a `2` to make this substitution so multiply by `2/2`. Leave the `1/2` on the outside and put the numerator inside:

`(1/2)int_(pi/6)^(pi/2) 2csc(2x)cot(2x)dx`

Make the substitution:

`(1/2)int_(pi/6)^(pi/2) csc(u) cot(u) du`

It helps to know a few trig derivatives for this integral:

Since the derivative of csc(u) is -csc(u)cot(u), the integral of csc(u)cot(u) must be -csc(u).

`(1/2) (-cscu)`

Now plug in `pi/2` and subtract it from the result that you get if you plug in `pi/6`:

`(1/2)(1-2)=-1/2`

Answer 2

Using well-known derivative:

`d/(dz) csc z = - csc z cot z`

You have:

`int_(pi/12)^(pi/4)csc2xcot2xdx`

`= int_(pi/12)^(pi/4) d/dx (- 1/2 csc2x ) dx`

`= 1/2 [ csc2x ]_(pi/4)^(pi/12) = 1/2`

[ PS : Always a good idea with the inverted trig functions (and tangent/cotangent) to check there are no singularities in the interval. Here, there aren't :)]