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

The given integral is:

\[\int \frac{\tan^2 \sqrt{x}}{\sqrt{x}} \, dx\]

To solve this, we'll use substitution. Set \( u = \sqrt{x} \), then \( x = u^2 \), and \( dx = 2u \, du \). The integral becomes:

\[\int \frac{\tan^2 u}{u} \cdot 2u \, du = 2 \int \tan^2 u \, du\]

We know the identity:

\[\tan^2 u = \sec^2 u - 1\]

Using this identity, the integral becomes:

\[2 \int (\sec^2 u - 1) \, du = 2 \left(\int \sec^2 u \, du - \int 1 \, du\right)\]

The integral of \(\sec^2 u\) is \(\tan u\), and the integral of \(1\) is \(u\). So,

\[= 2 (\tan u - u) + C\]

Substitute back \( u = \sqrt{x} \):

\[= 2 \left(\tan \sqrt{x} - \sqrt{x}\right) + C\]

Therefore, the solution to the given integral is:

\[2\left( \tan \sqrt{x} - \sqrt{x} \right) + C\]

Answer 2

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 \(t = \sqrt{x}\): \[ f(x) = 2 \tan \sqrt{x} - 2\sqrt{x} + C = 2(\tan \sqrt{x} - \sqrt{x}) + C \] \[ \boxed{f(x) = 2(\tan \sqrt{x} - \sqrt{x}) + C} \]