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Math  /  Calculus

QuestionEvaluate the following antiderivatives. a) ∫8cos⁡(x)+sec⁡2(x)dx=□+C\int 8 \cos (x)+\sec ^{2}(x) d x=\square+C b) ∫x5+6xx3dx=□+C\int \frac{x^{5}+6 \sqrt{x}}{x^{3}} d x=\square+C

Studdy Solution

STEP 1

Assumptions
1. We are given two separate antiderivative problems to solve.
2. The first problem is ∫8cos⁡(x)+sec⁡2(x) dx\int 8 \cos (x) + \sec^2(x) \, dx.
3. The second problem is ∫x5+6xx3 dx\int \frac{x^5 + 6 \sqrt{x}}{x^3} \, dx.
4. We assume that CC represents the constant of integration for indefinite integrals.
5. We will solve each integral separately.

STEP 2

For part (a), we need to find the antiderivative of 8cos⁡(x)+sec⁡2(x)8 \cos(x) + \sec^2(x).
The antiderivative of cos⁡(x)\cos(x) is sin⁡(x)\sin(x), and the antiderivative of sec⁡2(x)\sec^2(x) is tan⁡(x)\tan(x).

STEP 3

Apply the linearity of integration to separate the integral:
∫8cos⁡(x)+sec⁡2(x) dx=∫8cos⁡(x) dx+∫sec⁡2(x) dx\int 8 \cos(x) + \sec^2(x) \, dx = \int 8 \cos(x) \, dx + \int \sec^2(x) \, dx

STEP 4

Calculate the antiderivative of each term separately:
1. ∫8cos⁡(x) dx=8∫cos⁡(x) dx=8sin⁡(x)\int 8 \cos(x) \, dx = 8 \int \cos(x) \, dx = 8 \sin(x)
2. ∫sec⁡2(x) dx=tan⁡(x)\int \sec^2(x) \, dx = \tan(x)

STEP 5

Combine the results from STEP_4:
∫8cos⁡(x)+sec⁡2(x) dx=8sin⁡(x)+tan⁡(x)+C\int 8 \cos(x) + \sec^2(x) \, dx = 8 \sin(x) + \tan(x) + C

STEP 6

For part (b), we need to simplify the integrand x5+6xx3\frac{x^5 + 6 \sqrt{x}}{x^3} before integrating.
Rewrite the expression by dividing each term by x3x^3:
x5x3+6xx3=x5−3+6x1/2−3=x2+6x−5/2\frac{x^5}{x^3} + \frac{6 \sqrt{x}}{x^3} = x^{5-3} + 6 x^{1/2 - 3} = x^2 + 6 x^{-5/2}

STEP 7

Now, find the antiderivative of each term separately:
1. The antiderivative of x2x^2 is x2+12+1=x33\frac{x^{2+1}}{2+1} = \frac{x^3}{3}.
2. The antiderivative of 6x−5/26 x^{-5/2} is 6⋅x−5/2+1−5/2+1=6⋅x−3/2−3/2=−4x−3/26 \cdot \frac{x^{-5/2 + 1}}{-5/2 + 1} = 6 \cdot \frac{x^{-3/2}}{-3/2} = -4 x^{-3/2}.

STEP 8

Combine the results from STEP_7:
∫x5+6xx3 dx=x33−4x−3/2+C\int \frac{x^5 + 6 \sqrt{x}}{x^3} \, dx = \frac{x^3}{3} - 4 x^{-3/2} + C

STEP 9

Summarize the solutions for both parts:
a) ∫8cos⁡(x)+sec⁡2(x) dx=8sin⁡(x)+tan⁡(x)+C\int 8 \cos(x) + \sec^2(x) \, dx = 8 \sin(x) + \tan(x) + C
b) ∫x5+6xx3 dx=x33−4x−3/2+C\int \frac{x^5 + 6 \sqrt{x}}{x^3} \, dx = \frac{x^3}{3} - 4 x^{-3/2} + C

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