Problem 140
Medium Trigonometric Integrals

Evaluate the integral: I = integral dx/(cos^4x)

Evaluate the integral: $$ I = \int \frac{dx}{\cos^4x}. $$

Explanation of Method

Rewriting the integrand as $\sec^4x=\sec^2x\cdot\sec^2x=(1+\tan^2x)\sec^2x$ prepares it for the substitution $u=\tan x$, since $\sec^2x\,dx$ is exactly $du$.

  1. Rewrite $\sec^4x$ as $(1+\tan^2x)\sec^2x$.
  2. Substitute $u=\tan x$.
  3. Integrate the resulting polynomial.
  4. Substitute back and add the constant of integration.

Step 1 · Rewrite the integrand

The integrand is $\sec^4x$. Using $\sec^2x=1+\tan^2x$ on one factor:

$$ \sec^4x = \sec^2x\cdot\sec^2x = (1+\tan^2x)\sec^2x. $$

Thus, the integral becomes:

$$ I = \int (1+\tan^2x)\sec^2x\,dx. $$

Step 2 · Substitution

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

$$ I = \int (1+u^2)\,du. $$

Step 3 · Integrate

$$ I = u + \frac{u^3}{3} + C. $$

Step 4 · Substitute back

Substituting back $u=\tan x$:

$$ I = \tan x + \frac{\tan^3x}{3} + C. $$


Summary Table

Quantity Value
Substitution$u = \tan x$
$\sec^2x\,dx$ in terms of $du$$du$
Antiderivative$u+u^3/3$

Additional Notes

Result

The solution to the integral is:

$$ I = \tan x + \frac{\tan^3x}{3} + C. $$