Integrate ∫secxtanx3secx+5dx.
Solve the given integral
Let, I=∫secxtanx3secx+5dx
Multiply and divide by 3 on RHS,
⇒I=13∫3secxtanx3secx+5dx
Let, 3secx+5=t;∵dsecxdx=secxtanx,dconstantdx=0
Differentiate both sides,
⇒3secxtanxdx=dt
Replacing value of dx in I,
⇒I=13∫dtt∵∫dtt=lnt+c⇒I=lnt3+c⇒I=ln3secx+53+C
Hence, ∫secxtanx3secx+5dx is integrated as ln3secx+53+C.