Evaluate the integral (Remember the constant of integration): $\int \sqrt{75 - 10x - x^2} \, dx$

Neetesh Kumar | December 10, 2024

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This is the solution to Math 132
Assignment: 7.5 Question Number 6
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Step-by-step solution:

To evaluate $\int \sqrt{75 - 10x - x^2} \, dx$, we will simplify the expression inside the square root and use a trigonometric substitution.

Step 1: Simplify the quadratic expression

The quadratic inside the square root is $75 - 10x - x^2$.

Rewrite it in standard form:

$-x^2 - 10x + 75$

Factor out $-1$ to make the leading term positive:

$- (x^2 + 10x - 75)$

Complete the square for $x^2 + 10x - 75$:

1. Take half the coefficient of $x$, square it: $\left(\frac{10}{2}\right)^2 = 25$.

2. Add and subtract $25$ inside the parentheses:

$x^2 + 10x - 75 = (x^2 + 10x + 25) - 25 - 75 = (x + 5)^2 - 100$

So the quadratic becomes:

$75 - 10x - x^2 = -\left((x + 5)^2 - 100\right) = 100 - (x + 5)^2$

Thus, the integral is:

$\int \sqrt{75 - 10x - x^2} \, dx = \int \sqrt{100 - (x + 5)^2} \, dx$

Step 2: Substitution for the circle

The expression $\sqrt{100 - (x + 5)^2}$ represents a semicircle with radius $10$.

Use the substitution:

$x + 5 = 10 \sin(\theta), \quad dx = 10 \cos(\theta) \, d\theta$

Substitute into the integral:

$\int \sqrt{100 - (x + 5)^2} \, dx = \int \sqrt{100 - 100 \sin^2(\theta)} \cdot 10 \cos(\theta) \, d\theta$

Simplify the square root using the Pythagorean identity $1 - \sin^2(\theta) = \cos^2(\theta)$:

$\sqrt{100 - 100 \sin^2(\theta)} = \sqrt{100 \cos^2(\theta)} = 10 \cos(\theta)$

The integral becomes:

$\int 10 \cos(\theta) \cdot 10 \cos(\theta) \, d\theta = 100 \int \cos^2(\theta) \, d\theta$

Step 3: Simplify $\cos^2(\theta)$

Use the half-angle identity:

$\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$

Substitute into the integral:

$100 \int \cos^2(\theta) \, d\theta = 100 \int \frac{1 + \cos(2\theta)}{2} \, d\theta = 50 \int (1 + \cos(2\theta)) \, d\theta$

Split the integral:

$50 \int (1 + \cos(2\theta)) \, d\theta = 50 \int 1 \, d\theta + 50 \int \cos(2\theta) \, d\theta$

1. For $\int 1 \, d\theta$:

$\int 1 \, d\theta = \theta$

2. For $\int \cos(2\theta) \, d\theta$:

$\int \cos(2\theta) \, d\theta = \frac{\sin(2\theta)}{2}$

Combine the results:

$50 \int (1 + \cos(2\theta)) \, d\theta = 50 \left(\theta + \frac{\sin(2\theta)}{2}\right)$

Step 4: Back-substitute $\theta$

Recall that $x + 5 = 10 \sin(\theta)$, so:

$\sin(\theta) = \frac{x + 5}{10}, \quad \theta = \arcsin\left(\frac{x + 5}{10}\right)$

Also, use the double-angle formula for $\sin(2\theta)$:

$\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$

From the substitution, $\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(\frac{x + 5}{10}\right)^2}$.

Substitute back into the result:

$50 \theta + 25 \sin(2\theta) = 50 \arcsin\left(\frac{x + 5}{10}\right) + 25 \cdot 2 \cdot \frac{x + 5}{10} \cdot \sqrt{1 - \left(\frac{x + 5}{10}\right)^2}$

Simplify:

$50 \arcsin\left(\frac{x + 5}{10}\right) + 5(x + 5) \sqrt{1 - \left(\frac{x + 5}{10}\right)^2}$

Final Answer:

$\int \sqrt{75 - 10x - x^2} \, dx = \boxed{50 \arcsin\left(\frac{x + 5}{10}\right) + 5(x + 5) \sqrt{1 - \left(\frac{x + 5}{10}\right)^2} + C}$

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