Question

Integrate (tan x +tan^3 x)/(1+tan^3 x)?

Answer 1

`I = 1/6ln|tan^2x - tanx + 1| + 1/sqrt(3)arctan((2tanx - 1)/sqrt(3)) - 1/3ln|tanx+ 1| +C`

Explanation:

We can rewrite the numerator as:

`tanx(1 + tan^2x)`

We know that `tan^2x + 1 = sec^2x`, thus:

`I = int(tanxsec^2x)/(1 + tan^3x)dx`

Let `u = tanx`. Then `du = sec^2xdx` and `(du)/sec^2x = dx`.

`I = int (tanxsec^2x)/(1 + tan^3x) * (du)/sec^2x`

`I = int tanx/(1 + tan^3x) du`

Now we need to change into `u`. This is simple to do because `u = tanx`.

`I =int u/(1 + u^3) du`

We now observe that `u^3 + 1 = (u + 1)(u^2 - u + 1)`.

`I = int u/((u + 1)(u^2 - u + 1)) du`

Now by partial fractions, we have:

`A/(u + 1) + (Bu+ C)/(u^2 - u + 1) = u/((u + 1)(u^2 - u + 1))`

`A(u^2 - u + 1) + (Bu + C)(u + 1) = u`

`Au^2 - Au + A + Bu^2 + Cu + B u + C = u`

Now write a system of equations in 3 variables.

`{(A + B = 0), (B + C - A = 1), (A + C = 0):}`

We can immediately see that `A = -1/3`. This means that `B = C = 1/3`. Thus the integral becomes:

`I = int -1/(3(u + 1)) + (1/3u + 1/3)/(u^2 - u + 1)du`

`I = -1/3ln|u + 1| + 1/3int (u + 1)/(u^2 - u + 1) du`

Now we have another integral to solve. Call it `I_2`.

`I_2 = 1/3int (u + 1)/(u^2 - u + 1)`

Now we can rewrite as

`I_2 = 1/3int (1/2(2u - 1) + 3/2)/(u^2 - u + 1) du`

We can now separate into two integrals.

`I_2 = 1/3int(1/2(2u - 1))/(u^2- u + 1) + 3/(2(u^2 - u + 1))`

`I_2 = 1/6 int (2u -1)/(u^2 - u + 1) + 3/(2(u^2 - u + 1)`

Now we let `n = u^2 - u +1`. Then `dn = 2u - 1du` which means that `du= (dn)/(2u - 1)`.

`I_2 = 1/6int (2u - 1)/(u^2 - u + 1) * (dn)/(2u - 1) +int 3/(2(u^2 - u + 1))du`

`I_2 = 1/6int 1/n dn + int 3/(2(u^2 - u + 1))du`

`I_2 = 1/6ln|n| + int 3/(2(u^2 - u + 1))du`

We now have a third integral that needs to be solved. Call it `I_3`.

`I_3 = 3/2int 1/(1(u^2 - u + 1/4 - 1/4) + 1) du`

`I_3 = 3/2int 1/((u - 1/2)^2 + 3/4) du`

`I = 3/2int 1/(((2u - 1)/2)^2 + 3/4)du`

We now let `t = (2u -1)/sqrt(3)`. This signifies that `dt = 2/sqrt(3)du` and that `du = sqrt(3)/2dt`.

`I_3 =(3)/(4sqrt(3))int 1/((sqrt(3)/2t)^2 + 3/4) dt`

`I_3= 3/(4sqrt(3))int 1/(3/4(t^2 + 1)) dt`

`I_3 = 1/sqrt(3)int 1/(t^2 + 1) dt`

`I_3 = 1/sqrt(3)arctan(t)`

We now must reverse the substitutions.

`I_3 = sqrt(3)arctan((2u - 1)/sqrt(3))`

So putting everything back together:

`I = 1/6ln|u^2 - u + 1| + 1/sqrt(3)arctan((2u - 1)/sqrt(3)) - 1/3ln|u + 1| + C`

`I = 1/6ln|tan^2x - tanx + 1| + 1/sqrt(3)arctan((2tanx - 1)/sqrt(3)) - 1/3ln|tanx+ 1| +C`

Hopefully this helps!