Question

∫(√tanx + √cotx)dx is equal to

(A) √2sin^-1(sinx + cosx) + c

(B) 1/√2sin^-1(sinx + cosx) + c

(C) √2sin^-1(sinx - cosx) + c

(D) 1/√2sin^-1(sinx + cosx) + c  

Answer 1

Answer is (C) √2sin^-1(sinx - cosx) + c