Problem 35 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int \frac{d \theta}{\sqrt{27-6 \theta-\theta^{2}}}$$

Step-by-Step Solution

Answer

Question: Evaluate the integral: \(\int \frac{d \theta}{\sqrt{27-6 \theta-\theta^{2}}}\). Answer: \(\frac{1}{2} \arcsin\left(\frac{2}{3}(\theta+1)\right) + C\)

Step 1: Identify the trigonometric substitution

Observe that the form of the expression inside the square root can be rewritten as \(B^2 - A^2(\theta - C)^2\) where A, B, and C are constants. This will allow us to make trigonometric substitution. $$27-6 \theta-\theta^{2} = 9 - 4(\theta + 1)^2.$$ The suitable trigonometric substitution in this case is \(\theta + 1 = \frac{3}{2} \sin(u)\).

Step 2: Find differential

Find \(d\theta\) by differentiating both sides with respect to \(u\): $$d(\theta + 1) = \frac{3}{2} \cos(u) du$$ $$d\theta = \frac{3}{2} \cos(u) du$$

Step 3: Perform substitution and simplify

Substitute variables and \(d\theta\) into the integral: $$\int \frac{\frac{3}{2} \cos(u) du}{\sqrt{9 - 4\left(\frac{3}{2} \sin(u)\right)^2}}$$ Now, simplify the integral: $$\frac{3}{2} \int \frac{\cos(u) du}{\sqrt{9 - 9\sin^2(u)}}$$

Step 4: Simplify the expression under the square root

The expression inside the square root can be further simplified: $$9 - 9\sin^2(u) = 9\cos^2(u)$$ Now, substitute this back into the integral: $$\frac{3}{2} \int \frac{\cos(u) du}{\sqrt{9\cos^2(u)}}$$

Step 5: Cancel terms and solve the integral

We can now cancel a \(\cos(u)\) term from the numerator and denominator: $$\frac{3}{2} \int \frac{du}{\sqrt{9}} = \frac{1}{2} \int du$$ Now, we can integrate the simplified expression: $$\frac{1}{2}u + C$$

Step 6: Substitute back to the original variable

Recall that we made the substitution \(\theta + 1 = \frac{3}{2} \sin(u)\). We can write this as: $$u=\arcsin\left(\frac{2}{3}(\theta+1)\right)$$ Now, substitute this back into the result and simplify: $$\frac{1}{2} \arcsin\left(\frac{2}{3}(\theta+1)\right) + C$$ So, the final answer is: $$\int \frac{d \theta}{\sqrt{27-6 \theta-\theta^{2}}} = \frac{1}{2} \arcsin\left(\frac{2}{3}(\theta+1)\right) + C$$