Evaluate the integral: I = integral (sqrt(tan x))/(cos²

Evaluate the integral: $$ I = \int \frac{\sqrt{\tan x}}{\cos^2 x}\,dx. $$

Step 1 · Rewrite the integrand

Recall that $\dfrac{1}{\cos^2x}=\sec^2x$. Thus, the integral becomes:

$$ I = \int \sqrt{\tan x}\,\sec^2x\,dx. $$

Step 2 · Substitution

Let:

$$ u = \tan x \quad \text{so that} \quad du = \sec^2x\,dx. $$

Substituting into the integral:

$$ I = \int \sqrt{u}\,du. $$

Step 3 · Apply the power rule

Using $\int u^n\,du = \dfrac{u^{n+1}}{n+1}$ with $n=\tfrac12$:

$$ I = \frac{u^{3/2}}{3/2} + C = \frac{2}{3}u^{3/2} + C. $$

Step 4 · Substitute back

Substituting back $u=\tan x$:

$$ I = \frac{2}{3}\tan^{3/2}x + C. $$

Summary Table

Additional Notes

• The integrand requires $\tan x\geq0$ for the square root to be real, restricting $x$ to intervals such as $[k\pi,\,k\pi+\tfrac\pi2)$.
• Recognizing $\sec^2x$ as the derivative of $\tan x$ is one of the most common substitution triggers in trigonometric integrals — it turns any expression $f(\tan x)\sec^2x$ into a plain algebraic integral $\int f(u)\,du$.
• The result can be checked by differentiating $\tfrac23\tan^{3/2}x$ with the chain rule and confirming it returns $\sqrt{\tan x}\,\sec^2x$.