So I looked at the MIT 2015 integration bee and got almost all of them right however there are 2 that are puzzling me ∫2π01sin4x+cos4xdx
∫40|x−1||x−2|+|x−3|dx
For that first one through intense trig simplification I got it down to ∫2π043+cos4x and then using a variation of Weierstrauss t=tan2x I get that the bounds go from tan0 to tan4π so an integral from 0 to 0 is zero, however the answer is 2π√2. Could this have something to do with tangent being undefined along the interval? As for the second one I thought since the bottoms was never 0 we could just evaluate |∫10x−1x−2+x−3dx|+|∫41x−1x−2+x−3dx|
and the absolute value signs around the integrals is just so I don't have to worry about negatives. The answer to this question is 2+34ln(275). This approach is also incorrect so how am I wrong on both of these and how would I evaluate each of them correctly?
Answer
I=∫2π0dxsin(x)4+cos(x)4=∫2π0dx1−2sin2(x)cos2(x)=∫2π0dx1−12sin2(2x)
leads to:
I=12∫4π0dz1−12sin2(z)=4∫π/20dz1−12sin2(z)=4∫π/20dz1−12cos2(z)
where in the last steps we exploited the symmetry and periodicity of sin2. Since tanx is injective on (0,π/2), we may now substitute z=arctant and get:
I=4∫+∞0dt1+t2−12=2π√2.
Remember: a substitution in a integral is allowed iff it is provided by a diffeomorphism, i.e. a regular and invertible map. If you have to deal with ∫2π0g(u)du, you cannot use the substitution v=tan(u).
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