Question

$Integrate \frac{cos x}{cos 3x}$
$The answer should be \frac{1}{2\sqrt{3} } ln \frac{1+ \sqrt{3}tanx }{1- \sqrt{3}tanx }$

Answer 1

integration , {cosx/cos3x }

I = { cosx/cos3x }dx
= { cosx/( 4cos³x -3cosx )} dx
= { cosx/cosx ( 4cos²x -3) }dx
= dx/( 4cos²x -3)
= dx/( 4/sec²x -3)
= sec²x.dx /( 4 -3sec²x )
=sec²x.dx/{ 4 -3(1+tan²x)}
=sec²x.dx/( 1 -3tan²x )

let tanx = z
differentiate
sec²x.dx = dz

put this above ,

I = dz/( 1-3z²)
=dz/{ 1 -(√3z)²}

now use basic formula of integration ,
dx/( a² -x²) = 1/2a ln(a + x)/( a -x) + C

so, I = 1/2.1/√3 ln( 1+√3z )/( 1- √3z) + C

now , put z = tanx

so , I = 1/2√3.ln( 1+√3tanx)/( 1-√3tanx)

Answer 2

Hope this helps..........