Answer
$2\sqrt{1+tant} +C$
Work Step by Step
Evaluate the Integral using substitution: $\int \frac{dt}{cos^2t\sqrt{1+tant}}$
Substitution Rule: $\int f(g(x))g’(x)dx = \int f(u)du$
$u= 1+tant$
$du =sec^2t = \frac{1}{cos^2t}$
Solve the integral in terms of $u$:
$\int \frac{1}{\sqrt u}du$
$2\sqrt u +C$
Substitute for $u$:
$2\sqrt{1+tant} +C$
