Thursday, 4 June 2015

integration - Puzzling Definite Integrals



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|x1||x2|+|x3|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 2π2. 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 |10x1x2+x3dx|+|41x1x2+x3dx|
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=2π0dx12sin2(x)cos2(x)=2π0dx112sin2(2x)


leads to:
I=124π0dz112sin2(z)=4π/20dz112sin2(z)=4π/20dz112cos2(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+t212=2π2.

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).


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...