Source: Extra Practice

Evaluate the indefinite integral: (sinx+sec2x)dx\int (\sin x + \sec^2 x) \, dx.

Options

Option A is correct

cosx+tanx+C-\cos x + \tan x + C

Option B

cosx+tanx+C\cos x + \tan x + C

Option C

cosxtanx+C-\cos x - \tan x + C

Option D

cosxtanx+C\cos x - \tan x + C

Explanation

We can solve this indefinite integral by integrating each term separately: (sinx+sec2x)dx=sinxdx+sec2xdx\int (\sin x + \sec^2 x) \, dx = \int \sin x \, dx + \int \sec^2 x \, dx

Using standard trigonometric integration formulas:

  1. sinxdx=cosx+C1\int \sin x \, dx = -\cos x + C_1 (since the derivative of cosx-\cos x is sinx\sin x)
  2. sec2xdx=tanx+C2\int \sec^2 x \, dx = \tan x + C_2 (since the derivative of tanx\tan x is sec2x\sec^2 x)

Combining these results and introducing a single constant of integration CC: (sinx+sec2x)dx=cosx+tanx+C\int (\sin x + \sec^2 x) \, dx = -\cos x + \tan x + C

  • Option A is correct.
  • Option B is incorrect because sinxdx=cosx\int \sin x \, dx = -\cos x, not cosx\cos x (confusing integration with differentiation).
  • Option C is incorrect because the integral of sec2x\sec^2 x is +tanx+\tan x, not tanx-\tan x.
  • Option D has incorrect signs for both terms.