Question

\[ \int \frac{\tan^2 \sqrt{x}}{\sqrt{x}} \, dx \text{ is equal to:} \]

• A. \(\sec \sqrt{x} + C\)
• B. \(2\sqrt{x} \tan x - x + C\)
• C. \(2\left( \tan \sqrt{x} - \sqrt{x} \right) + C\)
• D. \(2 \tan \sqrt{x} - x + C\)

Hint:

If the integrand contains a composite function like \(\tan^2(\sqrt{x})\), use substitution \(t = \sqrt{x}\) to simplify. Then apply trigonometric identities to integrate easily.

Answer 1

The Correct Option is C

We are given: \[ \int \frac{\tan^2 \sqrt{x}}{\sqrt{x}} \, dx \] Let us use substitution. Put: \[ t = \sqrt{x} \Rightarrow x = t^2 \Rightarrow dx = 2t \, dt \] Now substitute in the integral: \[ \int \frac{\tan^2 t}{t} \cdot 2t \, dt = \int 2 \tan^2 t \, dt \] We know the identity: \[ \tan^2 t = \sec^2 t - 1 \Rightarrow \int 2 \tan^2 t \, dt = \int 2(\sec^2 t - 1) \, dt = 2 \int \sec^2 t \, dt - 2 \int dt = 2 \tan t - 2t + C \] Now revert back to the original variable \(x\) using \(t = \sqrt{x}\): \[ f(x) = 2 \tan \sqrt{x} - 2\sqrt{x} + C = 2(\tan \sqrt{x} - \sqrt{x}) + C \] \[ \boxed{f(x) = 2(\tan \sqrt{x} - \sqrt{x}) + C} \]