Evaluate the integral∫π/20√1+sinx dx
∫π20√1+sinxdx
∫√1+sinxdx=∫(cosx2+sinx2)dx
=2+(sinx2−cosx2)+c
=2(sinx2−cosx2)+c
∫π20√1+sinxdx=[2+(sinx2−cosx2)+c]π20
=[2(sinπ4−cosπ4)+c]−[2(sin0−cos0)+c]
=2(0)−2(−1)=2
∫π20√1+sinxdx=2