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Math  /  Calculus

QuestionEvaluate the integral. Use CC (upper case) for the constant of integration. ∫8x+916−x2dx\int \frac{8 x+9}{\sqrt{16-x^{2}}} d x

Studdy Solution

STEP 1

1. The integral is of the form ∫8x+916−x2 dx\int \frac{8x + 9}{\sqrt{16 - x^2}} \, dx.
2. The expression inside the square root, 16−x216 - x^2, suggests a trigonometric substitution.
3. We will use CC for the constant of integration.

STEP 2

1. Simplify the integral by splitting it into two separate integrals.
2. Use trigonometric substitution for each integral.
3. Evaluate each integral.
4. Combine the results and include the constant of integration.

STEP 3

Split the integral into two separate integrals:
∫8x+916−x2 dx=∫8x16−x2 dx+∫916−x2 dx\int \frac{8x + 9}{\sqrt{16 - x^2}} \, dx = \int \frac{8x}{\sqrt{16 - x^2}} \, dx + \int \frac{9}{\sqrt{16 - x^2}} \, dx

STEP 4

For the first integral, ∫8x16−x2 dx\int \frac{8x}{\sqrt{16 - x^2}} \, dx, use the substitution x=4sin⁡θx = 4\sin\theta, which implies dx=4cos⁡θ dθdx = 4\cos\theta \, d\theta and 16−x2=4cos⁡θ\sqrt{16 - x^2} = 4\cos\theta.
Substitute into the integral:
∫8x16−x2 dx=∫8(4sin⁡θ)4cos⁡θ⋅4cos⁡θ dθ=∫32sin⁡θ dθ\int \frac{8x}{\sqrt{16 - x^2}} \, dx = \int \frac{8(4\sin\theta)}{4\cos\theta} \cdot 4\cos\theta \, d\theta = \int 32\sin\theta \, d\theta

STEP 5

Evaluate the integral ∫32sin⁡θ dθ\int 32\sin\theta \, d\theta:
∫32sin⁡θ dθ=−32cos⁡θ+C1\int 32\sin\theta \, d\theta = -32\cos\theta + C_1
Convert back to xx using cos⁡θ=1−sin⁡2θ=1−(x4)2=16−x24\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \left(\frac{x}{4}\right)^2} = \frac{\sqrt{16 - x^2}}{4}:
−32cos⁡θ=−32⋅16−x24=−816−x2-32\cos\theta = -32 \cdot \frac{\sqrt{16 - x^2}}{4} = -8\sqrt{16 - x^2}

STEP 6

For the second integral, ∫916−x2 dx\int \frac{9}{\sqrt{16 - x^2}} \, dx, use the same substitution x=4sin⁡θx = 4\sin\theta, which implies 16−x2=4cos⁡θ\sqrt{16 - x^2} = 4\cos\theta.
Substitute into the integral:
∫916−x2 dx=∫94cos⁡θ⋅4cos⁡θ dθ=∫9 dθ\int \frac{9}{\sqrt{16 - x^2}} \, dx = \int \frac{9}{4\cos\theta} \cdot 4\cos\theta \, d\theta = \int 9 \, d\theta

STEP 7

Evaluate the integral ∫9 dθ\int 9 \, d\theta:
∫9 dθ=9θ+C2\int 9 \, d\theta = 9\theta + C_2
Convert back to xx using θ=arcsin⁡(x4)\theta = \arcsin\left(\frac{x}{4}\right):
9θ=9arcsin⁡(x4)9\theta = 9\arcsin\left(\frac{x}{4}\right)

STEP 8

Combine the results from both integrals and include the constant of integration:
∫8x+916−x2 dx=−816−x2+9arcsin⁡(x4)+C\int \frac{8x + 9}{\sqrt{16 - x^2}} \, dx = -8\sqrt{16 - x^2} + 9\arcsin\left(\frac{x}{4}\right) + C
The evaluated integral is:
−816−x2+9arcsin⁡(x4)+C\boxed{-8\sqrt{16 - x^2} + 9\arcsin\left(\frac{x}{4}\right) + C}

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