∫dx5+4cos⁡x=atan−1⁡btan⁡x2+C, then

I=∫dx5+4cos⁡x=∫dx5sin2⁡x2+cos2⁡x2+4cos2⁡x2−sin2⁡x2=∫dx9cos2⁡x2+sin2⁡x2=∫sec2⁡x29+tan2⁡x2dx Now,  I=∫2dt9+t2=23tan−1⁡t3+C∴ I=23tan−1⁡tan⁡x23+C Hence, a=23,b=13