Questions: Evaluate the integral by completing the square and using trigonometric substitution. ∫ sqrt(x^2 - 4x - 45) dx

Solution Steps

To evaluate the integral \(\int \sqrt{x^{2}-4 x-45} \, dx\), we can follow these steps:

1. Complete the square for the quadratic expression inside the square root.
2. Use a trigonometric substitution to simplify the integral.
3. Integrate using the appropriate trigonometric identity.

We start with the expression under the square root: \[ x^2 - 4x - 45 \] To complete the square, we rewrite it as: \[ (x - 2)^2 - 49 \] Thus, we have: \[ \sqrt{x^2 - 4x - 45} = \sqrt{(x - 2)^2 - 49} \]

Next, we use the substitution \(x - 2 = 7\sin(\theta)\), which leads to: \[ \sqrt{(x - 2)^2 - 49} = \sqrt{49\sin^2(\theta) - 49} = 7\sqrt{\sin^2(\theta) - 1} = 7\sqrt{-\cos^2(\theta)} = 7\cos(\theta) \] The differential \(dx\) becomes: \[ dx = 7\cos(\theta) d\theta \]

Substituting into the integral, we have: \[ \int \sqrt{x^2 - 4x - 45} \, dx = \int 7\cos(\theta) \cdot 7\cos(\theta) \, d\theta = 49 \int \cos^2(\theta) \, d\theta \] Using the identity \(\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}\), we can integrate: \[ 49 \int \frac{1 + \cos(2\theta)}{2} \, d\theta = \frac{49}{2} \left( \theta + \frac{1}{2}\sin(2\theta) \right) + C \]

Now we substitute back for \(\theta\): \[ \theta = \arcsin\left(\frac{x - 2}{7}\right) \] Thus, the integral evaluates to: \[ \frac{49}{2} \left( \arcsin\left(\frac{x - 2}{7}\right) + \frac{1}{2}\sin\left(2\arcsin\left(\frac{x - 2}{7}\right)\right) \right) + C \]

Final Answer