- #1
thereddevils
- 438
- 0
Homework Statement
By using the substitution t = tan x, find
[tex]\int \frac{dx}{\cos^2 x+4\sin^2 x}[/tex]
Homework Equations
The Attempt at a Solution
Well let tan x=t
[tex]\frac{dt}{dx}=\sec^2 x=\tan^2 x+1=1+t^2[/tex]
the integral then becomes
[tex]\int \frac{dx}{\frac{1}{\sqrt{1+t^2}}^2+4\frac{t}{\sqrt{1+t^2}}^2}[/tex]
which simplifies to
[tex]\int \frac{1}{1+4t^2}[/tex]
Then from here i make another substitution **
let t= 1/2 tan b
dt/db = 1/2 sec^2 b
[tex]\int \frac{1}{1+4(\frac{1}{2} \tan b)^2} \cdot \frac{1}{2}\sec^2 b db[/tex]
= b + constant
Back substitute
= [tex]\frac{1}{2}\tan^{-1} (2t)[/tex] + constant
= [tex]\frac{1}{2}\tan^{-1}(2\tan x)[/tex] + constant
Am i correct? Especially this part ** where i made another substitution, is that valid? Or when the question specified the substitution, i have to stick that one substitution only?