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Find the integral?

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2 Answers

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
 
 
 
 
 
 
 

Comments

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

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