Evaluate `int_(-1)^(1)[x+[x+[x]]]dx` , where [.] denotes the greatest integer function

Answer

`atos;btos;ctor;dtoq`

Explanation

`int_(-1)^(1)[x+[x+[x]]]dx` (use property `[x+n]=[x]+n` if `n` is integer)
`=int_(-1)^(1)3[x]dx=3int_(-1)^(1)[x]dx=3int_(0)^(1)([x]+[-x])dx`
`=3` (as `[x]+[-x]=-1`)
b. `int_(2)^(5)([x]+[-x])dx=int_(2)^(5)-1dx=-3`
c. `sgn(x-[x])={(1,"if",x"is not an integer"),(0,"if",x "is an integer"):}`
Hence `int_(-1)^(3) sgn(x-[x])dx=4(1-0)=4`
d. Let `I=25int_(0)^(pi//4) (tan^(6)(x-[x])+tan^(4)(x-[x]))dx { :' 0 lt x le (pi)/4implies[x]}`
`=25int_(0)^(pi//4)(tan^(6)x+tan^(4)x)dx`
`=25int_(0)^(pi//4) tan^(4)x(tan^(2)x+1)dx`
`=25int_(0)^(pi//4)tan^(4)xsec^(2)xdx`
`=25((tan^(5)x)/5)_(0)^(pi//4)=25xx1/5=5`