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Math

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PROBLEM

Evaluate the following antiderivatives.
a) 8cos(x)+sec2(x)dx=+C\int 8 \cos (x)+\sec ^{2}(x) d x=\square+C

STEP 1

Assumptions
1. We are given the integral (8cos(x)+sec2(x))dx\int (8 \cos(x) + \sec^2(x)) \, dx.
2. We need to find the antiderivative of the given function.
3. The constant of integration is denoted by CC.

STEP 2

The integral can be split into two separate integrals:
(8cos(x)+sec2(x))dx=8cos(x)dx+sec2(x)dx\int (8 \cos(x) + \sec^2(x)) \, dx = \int 8 \cos(x) \, dx + \int \sec^2(x) \, dx

STEP 3

Evaluate the first integral 8cos(x)dx\int 8 \cos(x) \, dx.
The antiderivative of cos(x)\cos(x) is sin(x)\sin(x), so:
8cos(x)dx=8cos(x)dx=8sin(x)\int 8 \cos(x) \, dx = 8 \int \cos(x) \, dx = 8 \sin(x)

STEP 4

Evaluate the second integral sec2(x)dx\int \sec^2(x) \, dx.
The antiderivative of sec2(x)\sec^2(x) is tan(x)\tan(x), so:
sec2(x)dx=tan(x)\int \sec^2(x) \, dx = \tan(x)

SOLUTION

Combine the results of the two integrals and add the constant of integration CC:
(8cos(x)+sec2(x))dx=8sin(x)+tan(x)+C\int (8 \cos(x) + \sec^2(x)) \, dx = 8 \sin(x) + \tan(x) + C The antiderivative is 8sin(x)+tan(x)+C8 \sin(x) + \tan(x) + C.

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