Evaluate the integral: I = integral tan³ x dx

Evaluate the integral: $$ I = \int \tan^3 x\,dx. $$

Step 1 · Rewrite the integrand

Using $\tan^2x=\sec^2x-1$:

$$ \tan^3x = \tan x\,(\sec^2x-1). $$

Thus, the integral splits as:

$$ I = \int \tan x\sec^2x\,dx - \int \tan x\,dx. $$

Step 2 · Evaluate the first integral

Let $u=\tan x$, so $du=\sec^2x\,dx$:

$$ \int \tan x\sec^2x\,dx = \int u\,du = \frac{u^2}{2} = \frac{\tan^2x}{2}. $$

Step 3 · Evaluate the second integral

Recall the standard result:

$$ \int \tan x\,dx = -\ln|\cos x|. $$

Step 4 · Combine the results

$$ I = \frac{\tan^2x}{2} - \left(-\ln|\cos x|\right) + C = \frac{\tan^2x}{2} + \ln|\cos x| + C. $$

Summary Table

Additional Notes

• The domain excludes $x=\tfrac\pi2+k\pi$ for integer $k$, where $\tan x$ is undefined.
• This splitting technique — using $\tan^2x=\sec^2x-1$ to reduce a higher odd power of tangent by two — can be applied repeatedly (a reduction formula) for any odd power $\tan^{2n+1}x$.
• The result can be checked by differentiating $\tfrac{\tan^2x}{2}+\ln|\cos x|$ and confirming it returns $\tan^3x$.