int_(0)^(1)(sin^(-1)x)/(x)dx=(pi)/(2)log2

Prove that : int_(0)^(1)(sin^(-1)x)/(x) dx = (pi)/(2) log 2

int_(0)^(1) sin^(-1) x dx =(pi)/(2) -1

int_(0)^(1) sin^(-1) x dx =(pi)/(2) -1

int_(0)^(pi/2) sin^(-1)(sin x)dx

int_(0)^(1)(tan^(-1)x)/(x)dx is equals to int_(0)^((pi)/(2))(sin x)/(x)dx(b)int_(0)^((pi)/(2))(x)/(sin x)dx(1)/(2)int_(0)^((pi)/(2))(sin x)/(x)dx(d)(1)/(2)int_(0)^((pi)/(2))(x)/(sin x)dx

Statement-1: int_(0)^(pi//2) x cot x dx=(pi)/(2)log2 Statement-2: int_(0)^(pi//2) log sin x dx=-(pi)/(2)log2

int_(0)^(1)(log|1+x|)/(1+x^(2))dx=(pi)/(8)log2

Show that int_(0)^((pi)/(2))log(sin2x)dx=-(pi)/(2)(log2)

int_(0)^(2)(|sin x|)/([(x)/(pi)]+(1)/(2))dx