Question Detail

To integrate $\int \sec^2 x \tan x dx$, we can use a substitution method.Let $u = \tan x$. Then, $du = \sec^2 x dx$.Substituting these into the integral gives: $\int u \, du = \frac{u^2}{2} + C$.Substituting back $u = \tan x$: $\frac{\tan^2 x}{2} + C$.Alternatively, let $u = \sec x$. Then, $du = \sec x \tan x dx$. In this case, the integral would be $\int \sec x \, (\sec x \tan x) dx$, which isn't a direct substitution.However, if we differentiate option C, we get:$\frac{d}{dx} \left( \frac{1}{2} \sec^2 x \right) = \frac{1}{2} \cdot 2 \sec x \cdot (\sec x \tan x) = \sec^2 x \tan x$.This matches the integrand. Note that $\frac{\tan^2 x}{2} + C = \frac{\sec^2 x - 1}{2} + C = \frac{1}{2}\sec^2 x - \frac{1}{2} + C$. Since $-\frac{1}{2}$ is a constant, both $\frac{\tan^2 x}{2}$ and $\frac{1}{2}\sec^2 x$ are valid antiderivatives.