int log2+bx+cdx

Prove that int log sin x dx = - (pi/2)log2

int(e^(6log x) -e^(5log x ))/(e^(4log x) -e^(3log x)) dx= ax^(3) +bx^(2) +c

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_2^4 log[x]dx is (A) log2 (B) log3 (C) log5 (D) none of these

int_2^4 log[x]dx is (A) log2 (B) log3 (C) log5 (D) none of these

If u=inte^(ax)sin " bx dx" and v=inte^(ax)cos " bx dx then "(u^(2)+v^(2))(a^(2)+b^(2))=

If u=inte^(ax)sin " bx dx" and v=inte^(ax)cos " bx dx then "(u^(2)+v^(2))(a^(2)+b^(2))=

Prove that int_(0)^(pi//2)(2logsinx-logsin2x)dx=(pi)/(2)(log2) .