int_(-1)^((3)/(2))|x sin pi x|dx=(3)/(pi)+(1)/(pi^(2))

int_(-1)^(3/2) | x sin (pi x ) | dx =

Evaluate int_(-1)^((3)/(2))|x sin(pi x)|dx

Evaluate : int_(-1)^((3)/(2))|x sin pi x|dx

Evaluate int_(-1)^(3//2)|x sin pi x|dx

int _ (- 1) ^ ((3) / (2)) | x sin pi x | dx = [(- x cos pi x) / (pi) + (sin pi x) / (pi ^ (2))] _ (- 1) ^ (1) - [(- x cos pi x) / (pi) + (sin pi x) / (pi ^ (2))] _ (1) ^ ((3) / (2) )

Prove that, int_(0)^(pi)log(1+cos x)dx=-pi log2 , given int_(0)^((pi)/(2))log((sin x))dx=(pi)/(2)"log"(1)/(2) .