^{2}

^{ }du]

^{-2 }du]

^{-1}/ -1]

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

Answer: 1/2

Please show all your work.

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Elana G. | A Gator Can Chomp Away Your Accounting FearsA Gator Can Chomp Away Your Accounting F...

You need to use u-substitution.

U=sin(2x)

du/dx=cos(2x) *2

du/2=cos(2x) dx

1/2 [integral of 1/u^{2}^{ }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

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

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