Question

Evaluate the following integrals. $$\int \frac{x^{2}}{\sqrt{1-9 x^{2}}} d x$$

Short answer

Answer: The evaluated integral is $$\frac{1}{18}\left(\arcsin(3x) - 3x\sqrt{1-9x^2}\right) + C$$.

Step-by-step solution

Identify the substitution

Let \(u = 3x\). This way we can rewrite the integral in terms of \(u\) later on.

Differentiate in terms of x

Differentiate the substitution in terms of x: $$\frac{du}{dx} = 3$$

Find dx in terms of du

To be able to replace \(dx\) in the integral, we need to find it in terms of \(du\). We can do this by writing the differential equation as follows: $$dx = \frac{du}{3}$$

Substitute variables

Now we can substitute our variables into the integral: $$\int \frac{x^2}{\sqrt{1-9x^2}} dx = \int \frac{\left(\frac{u}{3}\right)^2}{\sqrt{1-u^2}} \cdot \frac{du}{3}$$

Simplify the integral

Let's simplify the integral: $$\int \frac{\left(\frac{u^2}{9}\right)}{\sqrt{1-u^2}} \cdot \frac{du}{3} = \frac{1}{9} \int \frac{u^2}{\sqrt{1-u^2}} du$$

Identify integration technique

We will now use integration by substitution again. Notice that the integral looks like the derivative of arcsine function. Let $$v = \arcsin(u)$$ or $$u = \sin(v)$$ then $$du = \cos(v) \, dv$$.

Substitute and integrate

Substituting variables and integrating, we have: $$\frac{1}{9} \int \frac{u^2}{\sqrt{1-u^2}} du = \frac{1}{9} \int \frac{\sin^2(v)}{\sqrt{1-\sin^2(v)}} \cos(v) dv = \frac{1}{9} \int \sin^2(v) dv$$ Now, we can use the double angle formula to rewrite \(\sin^2(v)\): $$\sin^2(v) = \frac{1 - \cos(2v)}{2}$$ So, the integral becomes: $$\frac{1}{9} \int \sin^2(v) dv = \frac{1}{18} \int \left(1 - \cos(2v)\right) dv$$ Now integrating this expression: $$\frac{1}{18} \int \left(1 - \cos(2v)\right) dv = \frac{1}{18}(v - \frac{1}{2}\sin(2v)) + C$$

Substitute back in terms of x

Now, we substitute everything back in terms of \(x\): First, remember that \(u = \sin(v)\), so \(v = \arcsin(u)\). Also, \(u = 3x\), then, $$v = \arcsin(3x)$$. To find \(\sin(2v)\), we use the double angle formula: $$\sin(2v) = 2\sin(v)\cos(v) = 2(3x)\sqrt{1-(3x)^2} = 6x\sqrt{1-9x^2}$$ Finally, the answer is: $$\frac{1}{18}\left(\arcsin(3x) - \frac{1}{2}\cdot 6x\sqrt{1-9x^2}\right) + C$$ So, the integral is: $$\int \frac{x^2}{\sqrt{1-9x^2}} dx = \frac{1}{18}\left(\arcsin(3x) - 3x\sqrt{1-9x^2}\right) + C$$

Key concepts

Definite Integrals

Definite integrals provide the area under a curve within a specific interval. They are different from indefinite integrals, which include a constant of integration, because definite integrals evaluate to a number. If you’re asked to find a definite integral, you’ll typically be given limits of integration, often represented as a and b. For example, the definite integral from a to b of a function f(x) is written as: \[ \int_{a}^{b} f(x) \, dx \]This calculation gives the net area between the function and the x-axis from x = a to x = b. If you're solving a problem around this concept, always remember to evaluate the antiderivative at both the upper and lower bounds and subtract them: \[ F(b) - F(a) \]This result is what you get when using the Fundamental Theorem of Calculus for definite integrals. Keep in mind, definite integrals can result in negative values if the function lies below the x-axis.

Trigonometric Substitution

Trigonometric substitution is a technique useful for integrating expressions involving square roots, like \( \sqrt{a^2 - x^2} \), \( \sqrt{a^2 + x^2} \), or \( \sqrt{x^2 - a^2} \). This method involves substituting \( x \) with a trigonometric function, such as \( x = a \sin(\theta) \), \( x = a \tan(\theta) \), or \( x = a \sec(\theta) \). The substitution transforms the integral into trigonometric terms, often simplifying the square root into a familiar trigonometric identity.

• For \( \sqrt{1-x^2} \), use \( x = \sin(\theta) \).
• For \( \sqrt{1+x^2} \), use \( x = \tan(\theta) \).
• For \( \sqrt{x^2-1} \), use \( x = \sec(\theta) \).

The key is selecting a trigonometric function whose identity simplifies the integration process, such as \( 1 - \sin^2(\theta) = \cos^2(\theta) \). After integration, remember to convert back from the trigonometric function to the original variable using the inverse function.

Integration by Parts

Integration by parts is a method used to integrate the product of two functions. It's particularly handy when one part of the product becomes simpler upon taking its derivative, while another part is easily integrable. The formula for integration by parts is derived from the product rule of differentiation and is given by:\[\int u \, dv = uv - \int v \, du\]When applying this technique, choose \( u \) (which gets differentiated) and \( dv \) (which gets integrated) wisely. The LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) often helps in picking \( u \).

• Choose logs and inverses for \( u \) first, as they become simpler when differentiated.
• Common choices for \( dv \) are exponential or trigonometric functions.

Once \( u \) and \( dv \) are selected, compute \( du \) by differentiating \( u \) and find \( v \) by integrating \( dv \). Substitute these into the formula, simplify if possible, and repeat the process if necessary. Remember, this technique might initially transform the integral into a more complex-looking expression, but it simplifies upon execution.

Substitution Method

The substitution method is an integration technique that simplifies an integral by changing variables. It's akin to the chain rule in differentiation but applied in reverse. The main goal is to convert a difficult integral into an easier one by introducing a new variable. This is done by:

• Selecting a part of the integrand (say, inner function) and setting it equal to \( u \).
• Differentiating \( u \) to find \( du \) and expressing \( dx \) in terms of \( du \).

Once you substitute these into the integral, the expression should look simpler. Solve the new integral in terms of \( u \), and then replace \( u \) with the original variable. An alternative, and often strategic step within, involves identifying parts that resemble the derivative of standard functions. As demonstrated in the example integral of \( \frac{x^2}{\sqrt{1-9x^2}} \), substituting \( u = 3x \) transforms the expression to involve simpler terms, allowing integration with familiar arcsine functions and trigonometric identities. Remember to find the correct substitution by considering factors that will negate complex radicals or terms within the integrand.