Question

Evaluate each integral. $$\int \frac{d x}{x \sqrt{4-x^{8}}}$$

Short answer

Question: Evaluate the integral of the function \(\frac{1}{x\sqrt{4-x^8}}\). Answer: \(\frac{1}{2}\arcsin\frac{x^4}{2}+C\)

Step-by-step solution

Identify a substitution

We will make the following substitution: $$u = x^4.$$ Now we have $$u = x^4 \quad$$$=>$$$\quad x^8 = u^2$$ Therefore, $$4-x^8 = 4-u^2$$

Calculate the differential

We now need to find the differential of our substitution. So, differentiating both sides with respect to x, we get: $$\frac{du}{dx}=4x^3$$ Thus we have the differential as follows: $$dx = \frac{1}{4x^3}du$$

Rewrite the integral in terms of the new variable

Now we can substitute our expressions for both u and dx in the original integral: $$\int\frac{1}{x\sqrt{4-x^8}}dx=\int\frac{1}{x\sqrt{4-u^2}}\frac{1}{4x^3}du$$ Our integral now simplifies to $$\frac{1}{4}\int\frac{1}{\sqrt{4-u^2}}du$$

Perform the integration

We recognize the integral as that of the arcsin function: $$\int\frac{1}{\sqrt{4-u^2}}du = \frac{1}{2} \arcsin\frac{u}{2}+C$$ where C is the constant of integration.

Write the result in terms of the original variable

Now, we will substitute the expression for u back in terms of x: $$\frac{1}{2}\arcsin\frac{u}{2}+C = \frac{1}{2}\arcsin\frac{x^4}{2}+C$$ Thus, the evaluated integral is: $$\int\frac{1}{x\sqrt{4-x^8}}dx = \frac{1}{2}\arcsin\frac{x^4}{2}+C$$

Key concepts

U-Substitution

U-substitution is a widely-used technique in differential calculus, particularly when dealing with integrals. Its main purpose is to simplify an integral by changing the variable of integration to something more manageable. To utilize this method effectively, one typically identifies a portion of the original integrand that, when substituted with a new variable (usually denoted as 'u'), makes the integral easier to solve.

Consider the exercise at hand: evaluating the integral of a function involving a radical and a variable raised to a high power. Choosing the appropriate substitution is crucial. In the provided example, the substitution is made by setting \( u = x^4 \), which simplifies the integrand's denominator and subsequently the entire integral. From this point, the differential \( du \) and the new integral in terms of \( u \) are found. This process transforms the original complex expression into a more recognizable form, allowing for straightforward integration. It's a powerful tool that can turn an intimidating problem into a solvable one by recognizing patterns and applying the appropriate change of variables.

Differential Calculus

Differential calculus deals with the study of rates at which quantities change. It is the foundation of many techniques in calculus, including the method of u-substitution. One key aspect of differential calculus is the computation of differentials. These are small changes in a function's output relative to small changes in its input.

In performing u-substitution, identifying the differential \( du \) of the new variable \( u \) with respect to the original variable \( x \) is essential. This step is part of differential calculus. In the exercise, the differential of \( u = x^4 \) is determined by differentiating both sides with respect to \( x \) to get \( du = 4x^3dx \). This relationship allows for the substitution of \( dx \) in the integral. Understanding how to compute differentials is a fundamental skill in calculus that facilitates the manipulation and evaluation of complex integrals.

Arcsin Function Integration

The arcsin function, also known as the inverse sine function, has a specific integral form which becomes extremely handy when appearing in calculus problems. When faced with the integral of \( 1 / \sqrt{a^2 - u^2} \), where \( a \) is a constant, this form can be directly recognized as an arcsin function. This is a part of what is sometimes referred to as inverse trigonometric integration.

In our given problem, after applying u-substitution and simplifying the integral, we encounter the form \( 1 / \sqrt{4 - u^2} \), which aligns with the arcsin function pattern. By identifying this, you can directly integrate to receive \( \arcsin(u / a) + C \), where \( C \) is the integration constant. Converting back to the original variable, the final solution includes the arcsin function in terms of \( x \) instead of \( u \). Recognizing and understanding how to integrate the arcsin function can greatly simplify solving integrals that may at first appear to be challenging.