The correct option is B −ln∣∣∣(cosx+12)+√cos2x+cosx∣∣∣+C
I=∫√secx−1 dx =∫√1−cosxcosxdx =∫√(1−cosx)cosx×(1+cosx)(1+cosx)dx =∫√1−cos2xcosx+cos2xdx =∫sinx√cos2x+cosxdx
Putting cosx=t
⇒−sinx dx=dt∴I=∫−dt√t2+t =−∫dt√(t+12)2−(12)2 =−ln∣∣
∣∣(t+12)+√(t+12)2−(12)2∣∣
∣∣+C =−ln∣∣∣(t+12)+√t2+t∣∣∣+C⇒I=−ln∣∣∣(cosx+12)+√cos2x+cosx∣∣∣+C