Question

use the method of substitution to find each of the following indefinite integrals. $$ \int x^{-4} \sec ^{2}\left(x^{-3}+1\right) \sqrt[5]{\tan \left(x^{-3}+1\right)} d x $$

Short answer

The integral is \(-\frac{5}{18} \left(\tan(x^{-3}+1)\right)^{6/5} + C\)."

Step-by-step solution

Choose a Substitution

We want to simplify the integral by making a substitution that makes it easier to integrate. Notice that we have an expression within a trigonometric function and a square root involving \(x^{-3} + 1\). Let's choose\(u = x^{-3} + 1\).

Differentiate the Substitution

Differentiate the substitution variable \(u\) with respect to \(x\) to find \(du\):\[\frac{du}{dx} = -3x^{-4}\]Thus, we have:\[du = -3x^{-4} dx\]

Express \(dx\) in terms of \(du\)

Solve for \(dx\) in terms of \(du\) by rearranging the expression:\[dx = -\frac{1}{3}x^{4}du\]

Substitute \(u\) and \(dx\) into the Integral

Replace \(x^{-4}\), the expression \(x^{-3} + 1\), and \(dx\) in the integral:\[\int x^{-4} \sec^2(u) \sqrt[5]{\tan(u)} \left(-\frac{1}{3}\right) x^{4} \, du\]This simplifies to:\[-\frac{1}{3} \int \sec^2(u) \sqrt[5]{\tan(u)} \, du\]

Integrate with Respect to \(u\)

Integrate the simplified integral in terms of \(u\):\[-\frac{1}{3} \int \sec^2(u) \tan^{1/5}(u) \, du\]Using the substitution \(v = \tan(u)\), then \(dv = \sec^2(u) \, du\). The integral becomes:\[-\frac{1}{3} \int v^{1/5} \, dv = -\frac{1}{3} \times \frac{v^{6/5}}{6/5}\]Simplifying, we find:\[-\frac{1}{3} \times \frac{5}{6} v^{6/5} = -\frac{5}{18} v^{6/5} + C\]

Substitute Back in Terms of \(x\)

Replace \(v\) with \(\tan(u)\) and \(u\) with the expression in terms of \(x\) to get back to the original variable:\[-\frac{5}{18} \left(\tan(x^{-3}+1)\right)^{6/5} + C\]

Key concepts

Indefinite Integrals

An indefinite integral, denoted by \[\int f(x) \, dx,\]represents a family of functions whose derivative is the function inside the integral.
This is also known as the antiderivative.The result of evaluating an indefinite integral is not a single value but a general function, plus a constant of integration, usually denoted as C.This constant is crucial because when you take derivatives, constants disappear.Thus, an indefinite integral can represent infinitely many functions.
For example, when calculating \[ \int x^2 \, dx, \]the integral formula gives us \[ \frac{x^3}{3} + C. \]
When solving definite integrals, the constant C is not included since we are not dealing with an evaluation over an interval.But for indefinite integrals, it is always included to show the family of possible solutions.

Trigonometric Functions

Trigonometric functions like sine (\( \sin \)) and cosine (\( \cos \)), play a pivotal role in various branches of mathematics.They help describe relationships within right triangles, oscillations, and circles.In calculus, other trigonometric functions such as tangent (\( \tan \)) and secant (\( \sec \)) are equally important.
In the given example, when we expressed our function as \[\int \sec^2(u) \sqrt[5]{\tan(u)} \, du,\]we utilized the trigonometric identity involving secant and tangent:\[ \frac{d}{du} [\tan(u)] = \sec^2(u). \]This identity is essential for simplifying integrals that involve these functions.
Recognizing such identities and their derivatives helps both in substituting values and subsequently integrating them.

Calculus Integration

Calculus integration involves finding a function given its derivative, which lies at the heart of calculus.Integration is the opposite process of differentiation.It is employed to determine areas under curves, total values, or in finding functions from rates of change.
When tackling an integral with functions or expressions within, the method of substitution is often very useful.In substitution, a part of the integral is replaced with a new variable to simplify the integration process.
For example, in the given integral \( \int x^{-4} \sec^2(x^{-3}+1) \sqrt[5]{\tan(x^{-3}+1)} \, dx \), setting \( u = x^{-3} + 1 \)allows us to replace complex expressions with simpler ones in terms of \( u \).This substitution transforms the integral into a more manageable form, making it easier to integrate.
Ultimately, calculus integration, through techniques like substitution, helps unify and find precise solutions to complex mathematical problems.