Question

Evaluate : ∫(tanx + tan³x)/(1 + tan³x) dx

Answer 1

We can rewrite the numerator as:

tanx(1 + tan²x)

We know that tan²x + 1 = sec²x, thus:

\(I = \int \frac{\tan x \sec^2 x}{1 + \tan^3 x} dx\)

Let u = tanx.

Then du = sec²x dx and \(\frac{du}{\sec^2x} \) = dx

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

\(I = \int \frac u{1 + u^3}du\)

We now observe that u³ + 1 = (u + 1)(u² − u + 1).

We can immediately see that A = \(-\frac 13\). This means that B = C = \(\frac 13\). Thus the integral becomes:

Now we let n = u² − u + 1. Then dn = 2u − 1du which means that du = \(\frac{dn}{2u -1}\).

We now have a third integral that needs to be solved. Call it I₃.

We now let \(t = \frac{2u - 1}{\sqrt 3}\). This signifies that \(dt = \frac 2{\sqrt 3} du\) and that \(du = \frac{\sqrt 3}2\) dt.

So putting everything back together:

Answer 2

Given integral

On equation the coefficients of t² and constant terms from both sides in Eq. (ii), we get