Find the integral?

Question

Find the integral of (cos(2x) dx)/(sin^2 (2x)) from pi/12 to pi/4. 

 

Answer: 1/2

 

Please show all your work.

Answer 1

You need to use u-substitution. 

 

U=sin(2x)

du/dx=cos(2x) *2

 

du/2=cos(2x) dx

 

1/2 [integral of 1/u² du]

 

1/2[integral of u^-2 du]

 

integrate 

 

1/2[u^-1 / -1]

 

1/2[-1/sin(2x)] from pi/12 to pi/4

 

1/2[-1/sin(2*pi/4) - -1/sin(2*pi/12)]

 

1/2[-1/sin(pi/2) + 1/sin(pi/6)]

1/2[-1/1 + 1/(1/2)]

 

1/2[-1+2]

 

1/2[1]

 

1/2

 

 

 

 

 

 

 

Answer 2

Find the integral of (cos(2x) dx)/(sin^2(2x)) from pi/12 to pi/4. Answer: 1/2

I = ∫{pi/12,pi/4}(cos(2x) dx)/(sin^2(2x))

Let u = sin(2x)
u(pi/12)=sin(pi/6)=1/2
u(pi/4)=sin(pi/2)=1

du = 2 cos(2x) dx

I = ∫{1/2,1}((du/2)/u^2)

I = 1/2 ∫{1/2,1}(u^(-2) du)

I = 1/2 [–u^(-1)]{1/2,1}

I = 1/2 [–(1)^(-1) + (1/2)^(-1)]

I = 1/2 [–1 + 2]

I = 1/2