I have checked and rechecked my problem the whole way through and I can't figure out what is wrong with it, I would really really like to know what I did wrong.
∫0−π/3(√cosx)(sin3x)dx
Next I just rearranged things:
−1∫0−π/3(cosx)1/2(sin2x)(−sinx)dx
which equals:
−1∫0−π/3(cosx)1/2(1−cos2x)(−sinx)dx
then I defined variables to use for substitution:
u = cos x and du = -sin x dx
then substitute them in:
−1∫cos0cos−π/3(u)1/2(1−u2)du
distribute and find the endpoints:
−∫11/2(u1/2−u)du
then power rule:
−[23u3/2−12u2]and that should have a straight line at the end that means from one half to one, i just dont know how to do it on here.
Next just distribute that -1:
[−23u3/2+12u2]once again there should be a line at the end that means from one half to one.
then use the fundamental theorem of calculus to find the definite integral:
[−23(1)3/2+12(1)2]−[−23(1/2)3/2+12(1/2)2]
then after some simplification I end up with:
−23+12+23(12√2)−18
then I get:
−724+√26
That is my answer, but the correct answer should be:
−821+25√2168
+26396 I have checked and rechecked my problem the whole way through and I can't figure out what is wrong with it, I would really really like to know what I did wrong.
∫0−π/3(√cosx)(sin3x)dx
Until −1∫cos0cos−π/3(u)1/2(1−u2)du it is correct.
distribute and find the endpoints:
−1∫12u12(1−u2) du=−1∫12(u12−u12⋅u2) du|u12∗u2 is not u!!!=−1∫12(u12−u12+2) du=−1∫12(u12−u12+42) du=−1∫12(u12−u52) du=−[23u32−27u72)]112=[−23u32+27u72)]112=−23⋅(1)+27⋅(1)−[−23⋅(12)32+27⋅(12)72]=−23+27+13√2−128√2|1√2=√22=−23+27+√26−√256=−14+621+56−66⋅56⋅√2=−821+50336⋅√2=−821+25168⋅√2
+26396 I have checked and rechecked my problem the whole way through and I can't figure out what is wrong with it, I would really really like to know what I did wrong.
∫0−π/3(√cosx)(sin3x)dx
Until −1∫cos0cos−π/3(u)1/2(1−u2)du it is correct.
distribute and find the endpoints:
−1∫12u12(1−u2) du=−1∫12(u12−u12⋅u2) du|u12∗u2 is not u!!!=−1∫12(u12−u12+2) du=−1∫12(u12−u12+42) du=−1∫12(u12−u52) du=−[23u32−27u72)]112=[−23u32+27u72)]112=−23⋅(1)+27⋅(1)−[−23⋅(12)32+27⋅(12)72]=−23+27+13√2−128√2|1√2=√22=−23+27+√26−√256=−14+621+56−66⋅56⋅√2=−821+50336⋅√2=−821+25168⋅√2
