Question

How to integrate? `int dx/(sin^2 x+3sin x cos x-cos^2 x)`

Answer 1

`int 1/(sin^2 x+3sin x cos x-cos^2 x) \ dx = 1/sqrt(13) \ ln |tan(x-1/2arctan(2/3))| + C`

Explanation:

We seek:

`I = int 1/(sin^2 x+3sin x cos x-cos^2 x) \ dx`
`\ \ = int 1/( 3/2(2sin x cos x) - (cos^2 x - sin^2 x) ) \ dx`
`\ \ = int 1/( 3/2sin2x - cos2x ) \ dx`
`\ \ = int 1/( 1/2( 3sin2x - 2cos2x ) ) \ dx`
`\ \ = 2 \ int 1/( 3sin2x - 2cos2x ) \ dx`

Now using the superposition property of sine and cosine, we can put the denominator into the form `Rsin(2x-alpha)`:

Suppose that:

`3sin2x - 2cos2x -= Rsin(2x-alpha)`
`" " = R{sin2xcos alpha - cos2xsinalpha}`
`" " = (Rcos alpha)sin2x - (Rsinalpha)cos2x`

Equating coefficients of `sin2x` and `cos2x` we have:

`sin2x : \ \ \ Rcosalpha = 3 \ \ \` ..... [A]
`cos2x : \ \ \ Rsinalpha = 2 \ \ \` ..... [B]

We can find `R` using: `Eq [A]^2 + Eq[B]^2:`

`=> R^2cos^2alpha + R^2sin^2alpha = 3^2 + 2^2`
`:. R^2(cos^2alpha + sin^2alpha) = 9 + 4`
`:. R^2 = 13 => R=sqrt(13)`

We can find `alpha` using: `Eq [B] divide Eq[B]:`

`=> (Rsinalpha)/(Rcosalpha) = 2/3`
`:. tan alpha = 2/3 => alpha = arctan(2/3)`

Using this we can write the integral as:

`I = 2 \ int 1/( sqrt(13)sin(2x-alpha) ) \ dx`
`\ \ = 2/sqrt(13) \ int 1/( sin(2x-alpha) ) \ dx`
`\ \ = 2/sqrt(13) \ int csc(2x-alpha) \ dx`

Now we can perform a simple substitution:

Let `u=2x-alpha => (du)/dx = 2`

Substituting into the integral, we get:

`I = 2/sqrt(13) \ int \ 1/2 csc(u) \ du`
`\ \ = 1/sqrt(13) \ ln |tan(u/2)|`

Restoring the substitution, we get:

`I = 1/sqrt(13) \ ln |tan(x-alpha/2)|`
`\ \ = 1/sqrt(13) \ ln |tan(x-1/2arctan(2/3))|`

And of course, we shouldn't forget the integration constant; so:

`I = 1/sqrt(13) \ ln |tan(x-1/2arctan(2/3))| + C`