Integral Problem

Question

Please check the links below.

www.mathportal.org/calculators/calculus/integral-calculator.php?val1=(cos(4x))/(sin(2x)%2Acos(2x))&val2=%5Cdisplaystyle%5Cint+%5Cdfrac%7B%5Ccos%5Cleft(4x%5Cright)%7D%7B%5Csin%5Cleft(2x%5Cright)%7B%5Ccdot%7D%5Ccos%5Cleft(2x%5Cright)%7D%5C%2C+%5Cmathrm+d+x&rb1=indef&val3=pi/6&val4=pi/3

www.mathportal.org/calculators/calculus/integral-calculator.php?val1=2%2Acot(4x)&val2=%5Cdisplaystyle%5Cint+2%7B%5Ccdot%7D%5Ccot%5Cleft(4x%5Cright)%5C%2C+%5Cmathrm+d+x&rb1=indef&val3=pi/6&val4=pi/3

How can integral of (cos(4x))/(sin(2x)*cos(2x)) has different result with integral of 2*cot(4x)?
Whereas both of them are the same, just different form.

Answer 1

They are essentially the same function. I think both answers should be zero. Only the first integral question utilize more complex approximation, but the answer is not that much different. −1.110223024625157×10⁻¹⁶ = -0.0000000000000001≈0

Answer 2

Hi R,

 

They are not the same function. If you are unsure, graph them.

 

Neither integral is difficult to calculate. The first integral, for cos(4x)/(sin(2x)cos(2x)), try looking at ln(sin(2x)cos(2x)), and be sure you remember the trig identity cos(2u)=cos²(u)-sin²(u).

 

The second integral is just a substitution problem. Best luck