Question #68567

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Question #68567

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Answer 1 · 2017-04-15T18:55:40

$- \frac{1}{2}$ $-1/2$

Explanation:

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

$u = 2 x$ $u=2x$

$d u = 2$ $du=2$

This changes our integrand to:

$u_{1} = 2 (\frac{π}{12}) = \frac{π}{6}$ $u_1=2(pi/12)=pi/6$

$u_{2} = 2 (\frac{π}{4}) = \frac{π}{2}$ $u_2=2(pi/4)=pi/2$

We lack a $2$ $2$ to make this substitution so multiply by $\frac{2}{2}$ $2/2$ . Leave the $\frac{1}{2}$ $1/2$ on the outside and put the numerator inside:

$(\frac{1}{2}) \int_{\frac{π}{6}}^{\frac{π}{2}} 2 csc (2 x) cot (2 x) d x$ $(1/2)int_(pi/6)^(pi/2) 2csc(2x)cot(2x)dx$

Make the substitution:

$(\frac{1}{2}) \int_{\frac{π}{6}}^{\frac{π}{2}} csc (u) cot (u) d u$ $(1/2)int_(pi/6)^(pi/2) csc(u) cot(u) du$

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

$tutorial.math.lamar.edu$

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

$(\frac{1}{2}) (- csc u)$ $(1/2) (-cscu)$

Now plug in $\frac{π}{2}$ $pi/2$ and subtract it from the result that you get if you plug in $\frac{π}{6}$ $pi/6$ :

$(\frac{1}{2}) (1 - 2) = - \frac{1}{2}$ $(1/2)(1-2)=-1/2$

Answer 2 · 2017-04-15T19:02:56

Using well-known derivative:

$\frac{d}{d z} csc z = - csc z cot z$ $d/(dz) csc z = - csc z cot z$

You have:

$\int_{\frac{π}{12}}^{\frac{π}{4}} csc 2 x cot 2 x d x$ $int_(pi/12)^(pi/4)csc2xcot2xdx$

$= \int_{\frac{π}{12}}^{\frac{π}{4}} \frac{d}{d x} (- \frac{1}{2} csc 2 x) d x$ $= int_(pi/12)^(pi/4) d/dx (- 1/2 csc2x ) dx$

$= \frac{1}{2} {[csc 2 x]}_{\frac{π}{4}}^{\frac{π}{12}} = \frac{1}{2}$ $= 1/2 [ csc2x ]_(pi/4)^(pi/12) = 1/2$

[ 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 :)]

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Question #68567

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