The correct option is B −cosx2+5sinx+C
Let I=∫2sinx+5(2+5sinx)2dx
Dividing numerator and denominator by cos2x, we get
⇒I=∫2tanx⋅secx+5sec2x(2secx+5tanx)2dx
Put 2secx+5tanx=t
⇒(2secx⋅tanx+5sec2x)dx=dt
∴I=∫dtt2=−1t+C
⇒I=−12secx+5tanx+C
∴I=−cosx2+5sinx+C