Here, ∫sinx+cosx√9+16sin2xdx
∫sinx+cosx√9+16sin2x+16−16dx=∫sinx+cosx√25−16(1−sin2x)dx
=∫sinx+cosx√25−16(sin2x+cos2x−sin2x)dx
=∫sinx+cosx√25−16(sinx−cosx)2dx
Let sinx−cosx=t
(cosx+sinx)dx=dt
∫dt√(5)2−(4t)2=14sin−1(4t5)+c
[∵∫1√a2−x2dx=sin−1(xa)+c]
Putting the value of t, we get
=14sin−1[4(sinx−cosx)5]+c