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Chapter 7: Problem 16

Use a table of integrals to determine the following indefinite integrals. $$\int \frac{d x}{225-16 x^{2}}$$

Short Answer

Expert verified
Answer: The indefinite integral of the function $$\frac{1}{225 - 16x^2}$$ with respect to x is $$\frac{1}{60} \arctan \left(\frac{4x}{15}\right) + C$$.

Step by step solution

01

Identify the standard form of the given integral

Let's first rewrite the given integral in a recognizable standard form. Notice that the given integral looks like the standard form of arctangent integral, which is given by: $$\int \frac{d x}{a^{2} + x^{2}} = \frac{1}{a} \arctan \left(\frac{x}{a}\right) + C$$ Rewrite the given integral as follows: $$\int \frac{d x}{225 - 16 x^{2}} = \int \frac{d x}{15^2 - (4x)^2}$$
02

Use the arctangent integral formula

Now that the integral is in the recognizable form $$\int \frac{d x}{a^2 - x^2}$$, we can use the arctangent integral formula. In this case, we have \(a = 15\) and the integral becomes: $$\int \frac{d x}{15^2 - (4x)^2} = \frac{1}{15} \int \frac{d x}{1 - \frac{(4x)^2}{15^2}}$$ Apply the arctangent integral formula, where \(a = 15\) and \(u = 4x\). Then, $$\frac{1}{15} \int \frac{d x}{1 - \frac{(4x)^2}{15^2}} = \frac{1}{15} \cdot \frac{1}{4} \arctan \left(\frac{4x}{15}\right) + C$$
03

Simplify the result

Finally, simplify the result obtained in the previous step: $$\frac{1}{15} \cdot \frac{1}{4} \arctan \left(\frac{4x}{15}\right) + C = \frac{1}{60} \arctan \left(\frac{4x}{15}\right) + C$$ The indefinite integral of the given function is: $$\int \frac{d x}{225 - 16 x^{2}} = \frac{1}{60} \arctan \left(\frac{4x}{15}\right) + C$$

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