Question

How we integrate this, ?

`int dx/((a^2cos^2x + b^2sin^2x)^2)`

Answer 1

`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`

Explanation:

.

`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`