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

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

2 Answers
Mar 22, 2018

# -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#

Mar 22, 2018

#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#