.
#int1/(a^2cos^2x+b^2sin^2x)^2dx=int1/(b^2sin^2x+a^2cos^2x)^2dx=int1/((b^2sin^2x+a^2(1-sin^2x))^2)dx=int1/((b^2sin^2x+a^2-a^2sin^2x)^2)dx=int1/((b^2-a^2)sin^2x+a^2)^2dx=int(1/cos^4x)/(((b^2-a^2)sin^2x+a^2)^2/cos^4x)dx=intsec^4x/(((b^2-a^2)sin^2x+a^2)/cos^2x)^2dx=intsec^4x/((b^2-a^2)tan^2x+a^2sec^2x)^2dx=intsec^4x/((b^2-a^2)tan^2x+a^2(tan^2x+1))^2dx=intsec^4x/((b^2-a^2)tan^2x+a^2tan^2x+a^2)^2dx=intsec^4x/((b^2-a^2+a^2)tan^2x+a^2)^2dx=intsec^4x/((b^2tan^2x+a^2)^2)dx#
We use #u#-substitution:
Let #u=tanx, :. du=sec^2xdx, :. dx=(du)/sec^2x#
#intsec^4x/((b^2tan^2x+a^2)^2)dx=int(sec^2xsec^2x)/(b^2tan^2x+a^2)^2dx=int(sec^2x(tan^2x+1))/(b^2tan^2x+a^2)^2dx=I#
Let's substitute:
#I=int(sec^2x(u^2+1))/(b^2u^2+a^2)^2*(du)/sec^2x=int(cancelcolor(red)(sec^2x)(u^2+1))/(b^2u^2+a^2)^2*(du)/cancelcolor(red)(sec^2x)#
#I=int(u^2+1)/(b^2u^2+a^2)^2du#
#u^2+1=u^2+1+a^2/b^2-a^2/b^2=1/b^2(b^2u^2+a^2)+1-a^2/b^2#
Now, we can substitute this for #u^2+1#:
#I=int(1/b^2(b^2u^2+a^2)+1-a^2/b^2)/(b^2u^2+a^2)^2du#
#I=int(1/b^2(b^2u^2+a^2))/(b^2u^2+a^2)^2du+int(1-a^2/b^2)/(b^2u^2+a^2)^2du#
#I=1/b^2int1/(b^2u^2+a^2)du+(1-a^2/b^2)int1/(b^2u^2+a^2)^2du#
#I=1/b^2II+(1-a^2/b^2)III#
#II=int1/(b^2u^2+a^2)du#
Let's use substitution:
#p=b/au, :. u=a/bp, :.dp=b/adu, :. du=a/bdp#
#II=int1/(b^2(a^2/b^2p^2)+a^2)*a/bdp=inta/(b(a^2p^2+a^2))dp#
#II=inta/(a^2b(p^2+1))dp=1/(ab)int1/(p^2+1)dp#
This is the integral of an arctangent:
#II=1/(ab)arctanp#
Now, we substitute back for #p=(bu)/a#:
#II=arctan((bu)/a)/(ab)#
#III=int1/(b^2u^2+a^2)^2du#
Here, we can use a reduction formula that says:
#int1/(ct^2+d)^ndt=(2n-3)/(2d(n-1))int1/(ct^2+d)^(n-1)dt+t/(2d(n-1)(ct^2+d)^(n-1))#
In our integral:
#t=u, c=b^2, d=a^2, and n=2#, Therefore:
#III=u/(2a^2(b^2u^2+a^2))+1/(2a^2)int1/(b^2u^2+a^2)du#
The second part is the integral we solved above. So. we plug that in:
#III=u/(2a^2(b^2u^2+a^2))+1/(2a^2)arctan((bu)/a)/(ab)#
Now, we can plug #II# and #III# in:
#I=1/b^2II+(1-a^2/b^2)III#
#I=1/b^2(arctan((bu)/a)/(ab))+(1-a^2/b^2)(u/(2a^2(b^2u^2+a^2))+1/(2a^2)arctan((bu)/a)/(ab))#
#I=((1-a^2/b^2)arctan((bu)/a))/(2a^3b)+(arctan((bu)/a))/(ab^3)+((1-a^2/b^2)u)/(2a^2(b^2u^2+a^2))#
Now that the integral is solved, we can substitute back for #U#:
#int1/(a^2cos^2x+b^2sin^2x)^2dx=((1-a^2/b^2)arctan((btanx)/a))/(2a^3b)+(arctan((btanx)/a))/(ab^3)+((1-a^2/b^2)tanx)/(2a^2(b^2tan^2x+a^2))+C#
We can simplify the above and get:
#int1/(a^2cos^2x+b^2sin^2x)^2dx=((b^2+a^2)arctan((btanx)/a))/(2a^3b^3)+((b^2-a^2)tanx)/(2a^2b^2(b^2tan^2x+a^2))+C#
