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Chapter 7: Problem 35

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

Short Answer

Expert verified
The integral of \(\frac{d\theta}{\sqrt{27 - 6\theta - \theta^2}}\) is \(\arcsin{\left(\frac{\theta + 3}{6}\right)} + C\), where \(C\) is the constant of integration.

Step by step solution

01

Identify the substitution needed

Since the denominator contains a quadratic function under the square root, we will use the substitution \(x = a\sin(\phi)\), \(dx = a\cos(\phi) \, d\phi\), to rewrite the function in a simpler form, where \(a\) is a constant. Let's complete the square in order to find the value of \(a\) and the shift of the angle \(\phi\).
02

Complete the square for the quadratic expression

We first need to complete the square for the expression \(27 - 6\theta -\theta^2\). To do this, factor the coefficient of the \(\theta^2\) term out from the whole expression: $$-1(\theta^2 + 6\theta - 27).$$ Now, we need to add and subtract a suitable constant inside the brackets to form a perfect square: $$-\left(\theta^2 + 6\theta + 9 - 36\right) = -(\theta + 3)^2 + 36.$$ We can see that the constant \(a\) in the substitution \(x = a\sin(\phi)\) should be equal to \(\sqrt{36} = 6\) in order to match the coefficient with the completed square.
03

Apply the substitution

Now, let's apply the substitution \(\theta = -3 + 6\sin(\phi)\) with this constant \(a=6\) and simulate the differential \(d\theta\): $$d\theta = 6\cos(\phi) \, d\phi.$$ Thus, our integral becomes: $$\int \frac{6\cos(\phi)d\phi}{\sqrt{36 - 36\sin^2(\phi)}}.$$
04

Simplify the integral

Reorganize and simplify the expression to make it easier to evaluate: $$\int \frac{6\cos(\phi)d\phi}{\sqrt{36(1 - \sin^2(\phi))}} = \int \frac{\cos(\phi)d\phi}{\sqrt{1-\sin^2(\phi)}}.$$ Notice that since \(\cos^2(\phi) = 1 - \sin^2(\phi)\), we can write the denominator as: $$\int \frac{\cos(\phi)d\phi}{|\cos(\phi)|}.$$
05

Evaluate the integral

Since both \(\cos(\phi)\) and \(|\cos(\phi)|\) are strictly positive in the interval of \(\phi\), we can cancel these terms out in the integral: $$\int \,d\phi.$$ Now, we can directly integrate the remaining expression: $$\int \,d\phi = \phi + C$$ where \(C\) is the constant of integration. Finally, recall the substitution mentioned before: \(\theta = -3 + 6\sin(\phi)\). We need to express our result in terms of \(\theta\). To do this, we should rewrite \(\phi\) as a function of \(\theta\). Since \(\theta = -3 + 6\sin(\phi)\), we have: $$\sin(\phi) = \frac{\theta + 3}{6}. $$ To derive \(\phi\), we can apply the arcsin function to both sides and get: $$\phi = \arcsin{\left(\frac{\theta + 3}{6}\right)}.$$
06

Write the final answer

Now we can replace \(\phi\) with the expression derived from the arcsin function: $$\int \frac{d \theta}{\sqrt{27 - 6\theta - \theta^{2}}} = \arcsin{\left(\frac{\theta + 3}{6}\right)} + C.$$ So the final answer is: $$\int \frac{d \theta}{\sqrt{27 - 6\theta - \theta^{2}}} = \arcsin{\left(\frac{\theta + 3}{6}\right)} + C.$$

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