Question

Integrate `(sin2x)/("p"cos^2x+"q"sin^2x)` ?

`(sin2x)/("p"cos^2x+"q"sin^2x)`

Answer 1

`-1/(p-q) ln| p cos^2 x+q sin^2 x |+C`

Explanation:

Use the trigonometric identities

`cos^2 x = 1/2(1+cos 2x)`,
`sin^2 x = 1/2(1-cos 2x)`

to rewrite the denominator in the form

`p cos^2 x+q sin^2 x = 1/2{p(1+cos 2x)+q(1-cos 2x)} = {p+q}/2+{p-q}/2 cos 2x`

Substitute `{p+q}/2+{p-q}/2 cos 2x = u`. Then `-(p-q) sin 2x = du` Thus

`int (sin2x)/("p"cos^2x+"q"sin^2x) dx = -1/(p-q) int du/u`
`qquad = -1/(p-q) ln u+C = -1/(p-q) ln| p cos^2 x+q sin^2 x |+C`

Answer 2

`1/(q-p)ln(p(cosx)^2+q(sinx)^2)+C`

Explanation:

`int (sin2x*dx)/[p(cosx)^2+q(sinx)^2]`

=`int (2sinx*cosx*dx)/[p(cosx)^2+q(sinx)^2]`

=`int (2tanx*dx)/[q(tanx)^2+p]`

=`int (2tanx*[(tanx)^2+1]*dx)/([q(tanx)^2+p][(tanx)^2+1])`

After using `y=tanx` and `dy=[(tanx)^2+1]*dx` transforms, this integral became

`int (2y*dy)/[(y^2+1)*(qy^2+p)]`

After using `z=y^2` and `dz=2y*dy` transforms, it became

`int (dz)/[(z+1)*(qz+p)]`

Now, I decomposed integrand into basic fractions,

`1/[(z+1)*(qz+p)]=A/(z+1)+B/(qz+p)`

After expanding denominator,

`A*(qz+p)+B*(z+1)=1`

Setting `z=-1`, `A*(p-q)=1`, so `A=-1/(q-p)`

Setting `z=-p/q`, `B*(q-p)/q=1`, so `B=q/(q-p)`

Thus,

`int (dz)/[(z+1)*(qz+p)]`

=`q/(q-p)int (dp)/(qz+p)-1/(q-p)int (dp)/(z+1)`

=`1/(q-p)ln(qz+p)-1/(q-p)ln(z+1)+C`

=`1/(q-p)ln(qy^2+p)-1/(q-p)ln(y^2+1)+C`

=`1/(q-p)ln(q(tanx)^2+p)-1/(q-p)ln((tanx)^2+1)+C`

=`1/(q-p)ln(q(tanx)^2+p)-1/(q-p)ln((secx)^2)+C`

=`1/(q-p)ln(p(cosx)^2+q(sinx)^2)+C`