How do you use substitution to integrate #(secxtanxdx)/(9+4sec^2x)#?

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Answer 1 · 2015-07-26T06:29:12

The problem is solved below.

Let, #Sec x = t#

#implies Sec x tan x * dx = dt# (Upon differentiating)
Thus, the integral becomes,

#int dt/(9 + 4t^2)#

# = int dt/(3^2 + (2t)^2)#

Now, substituting #2t = z#
We get, #dt = dz/2#

Thus, the integral further becomes,

# 1/2 int dz/(3^2 + z^2)#

# = 1/2*1/3 arctan (z/3) + C#
# = 1/6 arctan ((2Sec x)/3) + C#

I guess I'm right.