Question

Trigonometric substitutions Evaluate the following integrals using trigonometric substitution. $$\int \sqrt{9-4 x^{2}} d x$$

Short answer

Question: Evaluate the integral $\int \sqrt{9 - 4x^2} dx$. Answer: $\frac{9}{4}\left(\arcsin\left(\frac{2}{3} x\right) + \left(\frac{4}{3} x\right)\left(\sqrt{1 - \left(\frac{2}{3} x\right)^2}\right)\right) + C$

Step-by-step solution

Determine the appropriate substitution

Since we have the square-root of the difference of squares, we can consider using the following trigonometric substitution: $$ x = \frac{3}{2} \sin(\theta) $$ This substitution is based on the identity \(\sin^2(\theta) + \cos^2(\theta) = 1\). With this substitution, the term inside the square-root becomes: $$ 9 - 4 x^2 = 9 - 4 \left(\frac{3}{2} \sin(\theta)\right)^2 = 9 - 9 \sin^2(\theta) $$

Calculate the differential

To switch from \(dx\) to \(d\theta\), we need to find \(\frac{d x}{d \theta}\). Using our substitution, we get: $$ \frac{d x}{d \theta} = \frac{3}{2} \cos(\theta) $$ Therefore, \(dx = \frac{3}{2} \cos(\theta) d\theta\).

Perform the substitution

Now we can substitute our expression for \(x\) and \(dx\) into the integral: $$ \int \sqrt{9 - 4x^2} dx = \int \sqrt{9 - 9 \sin^2(\theta)} \left(\frac{3}{2} \cos(\theta) d\theta\right) $$

Simplify the integral

We can simplify the integral by replacing the term inside the square-root with the appropriate trigonometric identity: $$ \int \sqrt{9 \left(1 - \sin^2(\theta)\right)} \left(\frac{3}{2} \cos(\theta) d\theta\right) = \int 3\sqrt{1 - \sin^2(\theta)} \left(\frac{3}{2} \cos(\theta) d\theta\right) $$ Using the identity, \(\cos^2(\theta) = 1 - \sin^2(\theta)\), we have: $$ \int 3\cos(\theta) \left(\frac{3}{2} \cos(\theta) d\theta\right) $$

Evaluate the simplified integral

Now we can easily integrate: $$ \int 3\cos(\theta) \left(\frac{3}{2} \cos(\theta) d\theta\right) = \int \frac{9}{2} \cos^2(\theta) d\theta $$ To integrate this, we use the identity: \(\cos^2(\theta) = \frac{1}{2}\left(1 + \cos(2\theta)\right)\): $$ \int \frac{9}{2} \cos^2(\theta) d\theta = \frac{9}{4}\int \left(1 + \cos(2\theta)\right)d\theta $$ Now, we can integrate each term separately: $$ \frac{9}{4} \int \left(1 + \cos(2\theta)\right) d\theta = \frac{9}{4}\left(\theta + \frac{1}{2}\sin(2\theta)\right) + C $$

Back-substitute

Finally, we need to substitute back in terms of x by replacing \(\theta\). From step 1, we have \(x = \frac{3}{2} \sin(\theta)\), so: $$ \sin(\theta) = \frac{2}{3} x $$ Using the inverse sine function, we get: $$ \theta = \arcsin\left(\frac{2}{3} x\right) $$ Now, we can find \(\sin(2\theta)\) using the double-angle identity, \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\): $$ \sin(2\theta) = 2\left(\frac{2}{3} x\right)\cos(\theta) $$ Since \(\cos^2(\theta) = 1 - \sin^2(\theta) = 1 - \left(\frac{2}{3} x\right)^2\), we have: $$ \cos(\theta) = \sqrt{1 - \left(\frac{2}{3} x\right)^2} $$ Therefore, \(\sin(2\theta) = 2\left(\frac{2}{3} x\right)\left(\sqrt{1 - \left(\frac{2}{3} x\right)^2}\right)\).

Final solution

Substituting our expressions for \(\theta\) and \(\sin(2\theta)\) back into our integral result, we have: $$ \frac{9}{4}\left(\arcsin\left(\frac{2}{3} x\right) + \frac{1}{2}\left(2\left(\frac{2}{3} x\right)\left(\sqrt{1 - \left(\frac{2}{3} x\right)^2}\right)\right)\right) + C $$ Simplifying, we get the final solution: $$ \frac{9}{4}\left(\arcsin\left(\frac{2}{3} x\right) + \left(\frac{4}{3} x\right)\left(\sqrt{1 - \left(\frac{2}{3} x\right)^2}\right)\right) + C $$

Key concepts

Integration Techniques

Integration techniques can significantly simplify seemingly complicated integrals. In this instance, trigonometric substitution transforms a challenging integral into one that is much easier to manage. Trigonometric substitution is particularly useful for integrals involving the square root of a sum or difference of squares. Using trigonometric substitution involves:

• Identifying the form of the integrand.
• Selecting an appropriate trigonometric substitution based on trigonometric identities.
• Replacing variables and differentials to simplify the integral.

The substitution technique used here is to replace a variable, in this case, \(x\), with a trigonometric function of another variable, often \(\theta\). This move simplifies the square root term in the integral. In our problem, since the expression under the square root is \(9 - 4x^2\), we use the substitution \(x = \frac{3}{2} \sin(\theta)\). This choice aligns the expression with a standard trigonometric identity, \(\sin^2(\theta) + \cos^2(\theta) = 1\), which simplifies further calculations.

Trigonometric Identities

Trigonometric identities are the foundation of solving integrals using trigonometric substitution. In the given exercise, some powerful identities come into play:

• The Pythagorean identity: \(\sin^2(\theta) + \cos^2(\theta) = 1\).
• Expression rearrangement: \(1 - \sin^2(\theta)\) is equivalent to \(\cos^2(\theta)\).
• Double angle identity for cosine: \(\cos^2(\theta) = \frac{1}{2}(1 + \cos(2\theta))\).

These identities transform expressions to a simpler form and enable the integration process. In our example, once the substitution \(x = \frac{3}{2} \sin(\theta)\) is applied, the expression \(9 - 4x^2\) simplifies using \(\cos^2(\theta)\). This step effectively reduces the complexity, allowing us to focus on the integral of a cosine function instead of the more complicated radical expression. The use of the double angle formula further simplifies the integration by breaking down \(\cos^2(\theta)\) into easily integrable components.

Inverse Trigonometric Functions

Inverse trigonometric functions play a vital role in converting solutions back to the variable of interest after integration. After applying trigonometric substitution and completing the integration, one must revert to the original variable, \(x\), which involves using inverse trigonometric functions. In the solution, after integrating with respect to \(\theta\), the function \(\theta = \arcsin\left(\frac{2}{3} x\right)\) is used to express the original variable \(x\) back in terms of \(\theta\).
This use of the inverse sine function helps transition from the angles and trigonometric relationships back to algebraic expressions in terms of the variable \(x\). It also allows us to apply the original limits of integration, if any, in terms of \(x\). Hence, the substitution and its reversal are both essential parts of solving integrals using this method. The entire approach is aimed at taking the expression back from \(\theta\) to \(x\), ensuring that the end result, represented by a definite or indefinite integral, accurately reflects the initial function's behavior. These steps, coupled with careful handling of inverse trigonometric functions, complete the solution process for the given integral.