Integration of tangent cubed of x?
Find I = ∫ tan³ x dx. The solution is: I = ½ tan² x - log cos x.
* * * Here is how we can obtain this result: First, let t = tan x,
s = sin x, and c = cos x; then, dI = t³ dx, ds = c dx, dc = -s dx,
and dt = (1 + t²) dx; and, of course, t = s / c. By algebra, t³ =
t(t² + 1) - t; thus, we have dI = t³ dx = t(t² + 1) dx - t dx = t
dt - t dx. Now, d (t²) = 2t dt; thus, t dt = ½ d(t²). On the other
hand, we have d log c = dc / c = -s dx / c = -t dx; thus, t dx = -d
log c. Combining these results, we have dI = t dt - t dx = ½ d(t²)
- d log c. This integrates readily, giving I = ½ t² - log c, which
is the solution we sought. * * * We may check our result, by
differentiating back: dt / dx = 1 + t²; and d(t²) / dt = 2t; thus,
(d/dx)(t²) = 2t dt / dx = 2t (1 + t²). Also, we have d log c / dc =
1 / c; and dc / dx = -s; whence, (d/dx)(log c) = (dc / dx) / c = -s
/ c = -t. Then, dI / dx = ½ (d/dx)(t²) - (d/dx)(log c) = t (1 + t²)
- t = t + t³ - t = t³, re-assuring us that we have integrated
correctly.