The correct option is C π3
We have,Let I=∫π015+4 cos xdx=∫π015+4 ⎛⎜⎝1−tan2 x21+tan2 x2⎞⎟⎠dx=∫π01+tan2 x25(1+tan2 x2)+4 (1−tan2 x2)dx=∫π0sec2 x29+tan2 x2dxLet tan (x2)=t⇒12sec2 x2dx=dtAlso, x=0⇒t=0 and x=π⇒t=∞∴I=∫∞0sec2 x29+t2×2sec2 x2dt⇒I=2∫∞019+t2dt=2∫∞0132+t2dt⇒I=2 (tan−1(t3)3]∞0=23(tan−1(t3)]∞0⇒I=23(tan−1(∞)−tan−1(0)]⇒I=23(π2−0)⇒I=π3