Question

\(\int\frac{\cos(\tan x)}{\cos^2x}dx=\)

• A. (tanx)sin(tanx)+C
• B. sin(tanx)+C
• C. sec(tanx)+C
• D. (cosx)sin(tanx)+C
• E. cos²(tanx)+C

Answer 1

The Correct Option is B

We are given the integral \( \int \frac{\cos(\tan x)}{\cos^2 x} \, dx \) and are asked to find the solution. 

We can solve this by using substitution. Let:

\( u = \tan x \),

so that \( du = \sec^2 x \, dx \).

Now, rewrite the integral in terms of \( u \):

\( \int \frac{\cos(\tan x)}{\cos^2 x} \, dx = \int \cos(u) \cdot \frac{du}{\cos^2 x} \).

Since \( \sec^2 x = 1/\cos^2 x \), this becomes:

\( \int \cos(u) \cdot \sec^2 x \, du \),

which simplifies to:

\( \int \cos(u) \, du \).

The integral of \( \cos(u) \) is \( \sin(u) \), so we have:

\( \sin(\tan x) + C \).

The correct answer is \( \sin(\tan x) + C \).