Question

Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating. $$\int \frac{3}{\sqrt{1-25 x^{2}}} d x$$

Short answer

Question: Evaluate the indefinite integral $$\int \frac{3}{\sqrt{1-25 x^{2}}} d x$$ Answer: $$\int \frac{3}{\sqrt{1-25 x^{2}}} d x = \pm \frac{3}{5}\arcsin(5x) + C$$

Step-by-step solution

Perform the substitution

In order to get a simplified form for the integrand \(\frac{3}{\sqrt{1-25 x^{2}}}\), we can use the substitution \(x = \frac{1}{5}\sin u\). With this substitution, note that when \(u=0\), \(x=0\), and the substitution makes sense. Additionally, we have \(d x = \frac{1}{5}\cos u \, d u\). Now, we'll substitute \(x\) and \(dx\) in the integral: $$ \int \frac{3}{\sqrt{1-25 x^{2}}} d x = \int \frac{3}{\sqrt{1-25 (\frac{1}{5}\sin u)^{2}}} \cdot \frac{1}{5}\cos u \, d u $$

Simplify the integrand

Let's simplify the integrand in the above integral: $$ \int \frac{3}{\sqrt{1-25 (\frac{1}{5}\sin u)^{2}}} \frac{1}{5}\cos u \, d u = \int \frac{3\cos u}{\sqrt{1-\sin^{2} u}} \frac{1}{5} \, d u = \int \frac{3\cos u}{\sqrt{\cos^{2} u}} \frac{1}{5} \, d u $$ We can further simplify the expression under the square root: $$ = \int \frac{3\cos u}{|\cos u|} \frac{1}{5} \, d u $$ This integral can now be evaluated by taking into account the sine of \(\cos u\).

Evaluate the integral

Because the cosine function can be positive or negative, we break the integral into two cases: Case 1: \(\cos u \geq 0\) Here, the integral becomes: $$ \int \frac{3\cos u}{\cos u} \frac{1}{5} \, d u = \int \frac{3}{5} \, d u = \frac{3}{5} u + C_1 $$ where \(C_1\) is the constant of integration. Case 2: \(\cos u < 0\) In this case, the integral turns into: $$ \int \frac{3\cos u}{-\cos u} \frac{1}{5} \, d u = -\int \frac{3}{5} \,d u= -\frac{3}{5} u + C_2 $$ where \(C_2\) is the constant of integration. Now, we can combine Cases 1 and 2 into a single function F(u): $$ F(u) = \begin{cases} \frac{3}{5}u + C_1, & \text{if}\ \cos u \geq 0 \\ -\frac{3}{5}u + C_2, & \text{if}\ \cos u < 0 \end{cases} $$

Reverse the substitution

Now, we need to substitute back for \(x\). Since \(x = \frac{1}{5}\sin u\), \(u = \arcsin (5x)\). Therefore, we obtain: $$ F(x) = \begin{cases} \frac{3}{5}\arcsin(5x) + C_1, & \text{if}\ \cos u \geq 0 \\ -\frac{3}{5}\arcsin(5x) + C_2, & \text{if}\ \cos u < 0 \end{cases} $$ As \(C_1\) and \(C_2\) are constants, we can combine them into a single constant \(C\): $$ F(x) = \pm \frac{3}{5}\arcsin(5x) + C $$

Check by differentiating

Finally, we'll check our work by differentiating our result: $$ \frac{dF(x)}{dx} = \pm \frac{3}{5} \cdot \frac{d}{dx}\arcsin(5x) = \pm \frac{3}{5} \cdot \frac{5}{\sqrt{1 - 25x^2}} = \frac{3}{\sqrt{1 - 25x^2}} $$ As the differentiation result matches the original integrand, the solution is correct: $$ \int \frac{3}{\sqrt{1-25 x^{2}}} d x = \pm \frac{3}{5}\arcsin(5x) + C $$

Key concepts

Integration by Substitution

Integration by substitution is a fundamental technique used in calculus to solve integrals. It's especially helpful when dealing with complex functions that appear challenging to integrate directly. This method involves changing the variable of integration, making the integral easier to evaluate.
When you employ this method, you typically follow these steps:

• Identify a part of the integrand that can be replaced with a single variable. This usually simplifies the expression.
• Express the original variable in terms of the new variable. You'll also need the derivative of this expression, known as the differential.
• Substitute the expressions into the integral, transforming it into an easier form.
• Integrate with respect to the new variable.
• Finally, substitute back the original variable to return to the form of the original problem.

This process is demonstrated in the example of evaluating the integral \( \int \frac{3}{\sqrt{1-25x^{2}}} dx \). Here, a substitution like \( x = \frac{1}{5} \sin u \) is used, simplifying the calculation significantly. After using this substitution, the integration becomes manageable and can be solved without much difficulty. Once the integration is complete, reversing the substitution gives the solution in terms of the original variable \( x \).

Trigonometric Substitution

Trigonometric substitution is a specific case of integration by substitution, particularly useful for integrals involving expressions like \( \sqrt{a^2 - x^2} \), \( \sqrt{a^2 + x^2} \), or \( \sqrt{x^2 - a^2} \). The technique leverages trigonometric identities to simplify integral expressions.
In the example of \( \int \frac{3}{\sqrt{1-25 x^{2}}} d x \), the substitution \( x = \frac{1}{5} \sin u \) was chosen. This is effective because it transforms \( \sqrt{1-25x^2} \) into a trigonometric expression that can be simplified using the identity \( 1 - \sin^2 u = \cos^2 u \).
Here's how the process unfolds:

• Select a trigonometric function substitution to eliminate the square root. In this case, \( x = \frac{1}{5} \sin u \) is used.
• Rewrite the differential \(dx\) in terms of \(du\) and other necessary trigonometric functions.
• Substitute into the integral, leveraging trigonometric identities to simplify the expression to a basic integral form.

This technique not only simplifies the integration process but also reveals deeper connections between algebraic and trigonometric expressions.

Arc Sine Function

The arc sine function, denoted as \( \arcsin(x) \), is the inverse function of the sine function over a specific domain. It is instrumental in integration, especially when using trigonometric substitutions that involve the sine function.
In the provided exercise, after substituting \( x = \frac{1}{5} \sin u \), the variable \( u \) was expressed in terms of \( x \) using the arc sine function: \( u = \arcsin(5x) \). This transformation helps revert the expression back to the variable originally presented in the problem.
When dealing with the inverse sine function, keep in mind a few key properties:

• \( \arcsin(x) \) is defined for \( -1 \leq x \leq 1 \).
• Its range is \( -\frac{\pi}{2} \leq \arcsin(x) \leq \frac{\pi}{2} \).
• The derivative of \( \arcsin(x) \) is \( \frac{1}{\sqrt{1-x^2}} \), which is crucial for back-checking integration results.

These properties ensure that under the correct circumstances, the integration process and its subsequent checks coordinate seamlessly, as evidenced by a successful verification involving differentiation in this exercise.