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Evaluate the following definite integrals. $$\int_{0}^{1 / \sqrt{2}} y \tan ^{-1} y^{2} d y$$

Short Answer

Expert verified
Question: Evaluate the definite integral $$\int_{0}^{1 / \sqrt{2}} y \tan ^{-1} y^{2} d y$$ using integration by parts. Answer: $$\int_{0}^{1 / \sqrt{2}} y \tan ^{-1} y^{2} d y = \frac{1}{4}\tan^{-1}\left(\frac{1}{2}\right) - \int_{0}^{1 / \sqrt{2}} \frac{y^3}{1+y^4}\,d y$$ Note: The second integral in the answer cannot be expressed in terms of elementary functions and can only be approximated using numerical methods.

Step by step solution

01

Identify Integration by Parts Components

To solve the given integral, we will use integration by parts. Integration by parts is given by: $$\int_{a}^{b} u\,d v = [uv]_{a}^{b} - \int_{a}^{b} v\,d u$$ In our case, we can let: - \(u=\tan^{-1}(y^2)\) - \(d v=y\,d y\)
02

Differentiate u and Integrate v

Now, we need to find the derivative of \(u\) and the integral of \(d v\): - \(d u = \frac{d}{d y}(\tan^{-1}(y^2))\,dy = \frac{2y}{1+y^4}dy\) - \(v = \int(y\,d y) = \frac{1}{2}y^2\)
03

Use Integration by Parts Formula

Now, we can substitute our expressions for \(u\), \(v\), \(d u\), and \(d v\) into the integration by parts formula: $$\int_{0}^{1 / \sqrt{2}} y \tan ^{-1} y^{2} d y = \left[\frac{1}{2}y^2\tan^{-1}(y^2)\right]_{0}^{1 / \sqrt{2}} - \int_{0}^{1 / \sqrt{2}} \frac{1}{2}y^2\cdot\frac{2y}{1+y^4}\,d y$$
04

Simplify the Expression

Now, we simplify the expression on the right-hand side: $$ \left[\frac{1}{2}y^2\tan^{-1}(y^2)\right]_{0}^{1 / \sqrt{2}} - \int_{0}^{1 / \sqrt{2}} \frac{y^3}{1+y^4}\,d y$$
05

Calculate the Definite Integral

Now we can evaluate the definite integral: $$ \frac{1}{2}\cdot\left(\frac{1}{\sqrt2}\right)^2\tan^{-1}\left(\left(\frac{1}{\sqrt2}\right)^2\right) - \frac{1}{2}\cdot0^2\tan^{-1}(0^2) - \int_{0}^{1 / \sqrt{2}} \frac{y^3}{1+y^4}\,d y$$ For the integral \(\int_{0}^{1 / \sqrt{2}} \frac{y^3}{1+y^4}\,d y\), since it doesn't have an elementary antiderivative, we could only get an approximation of it, using numerical methods like Simpson's rule, Trapezoidal rule, or others. However, the result can be simplified as: $$\int_{0}^{1 / \sqrt{2}} y \tan ^{-1} y^{2} d y = \frac{1}{4}\tan^{-1}\left(\frac{1}{2}\right) - \int_{0}^{1 / \sqrt{2}} \frac{y^3}{1+y^4}\,d y$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration by Parts
Integration by parts is a powerful technique used to solve integrals that are not easily solvable using standard integration methods. It is especially useful when dealing with the product of two functions where one function is easily integrable, and the other is easily differentiable.

The formula for integration by parts is \[\int_{a}^{b} u\,d v = [uv]_{a}^{b} - \int_{a}^{b} v\,d u\] where \(u\) is a function that is chosen to be differentiated, and \(dv\) is the remaining part of the integrand which is to be integrated. The choice of \(u\) and \(dv\) can greatly affect the simplicity of the solution, and often, recognizing how to choose both is a key skill developed through practice.

For instance, in the given problem, we choose \(u\) to contain an inverse trigonometric function because its derivative simplifies our integral. The linear function, in this case, \(y\), is assigned to \(dv\) because integrating a polynomial function typically results in a function that maintains a straightforward structure.
Inverse Trigonometric Functions
Inverse trigonometric functions, such as \(\tan^{-1}(x)\), which is also known as the arctangent of \(x\), are functions that allow us to find the angle whose tangent is \(x\). They are the inverses of the trigonometric functions and are used when integrating functions that involve trigonometry.

These functions are particularly important because they emerge naturally when integrating certain types of functions, and they can also appear in the solutions of integrals that involve trigonometric substitutions. In our exercise, \(\tan^{-1}(y^2)\) is utilized, which necessitates knowledge not only of its geometric interpretation but also how to differentiate it, as this is a significant part of solving the integral using integration by parts. The derivative of \(\tan^{-1}(x)\) is given by \(\frac{1}{1+x^2}\), which plays a key role in the integration process.
Numerical Integration Methods
Numerical integration methods are alternative techniques used to estimate the value of a definite integral when an antiderivative is complex or not easily obtained. When faced with a difficult integrand or a function without a closed-form antiderivative, numerical methods can provide an approximate solution to the problem.

Common numerical integration methods include Simpson's Rule and the Trapezoidal Rule. Both methods approximate the area under the curve by breaking up the integral into a finite number of smaller, more manageable parts, usually referred to as partitions or intervals. The Trapezoidal Rule, for example, approximates the region under the curve as a series of trapezoids, while Simpson's Rule uses parabolic arcs.

In the context of our exercise, once the integration by parts method has been applied, the remaining integral might not have an elementary antiderivative, prompting the use of these numerical methods to approximate the integral value.
Antiderivative
An antiderivative of a function \(f(x)\) is another function \(F(x)\) whose derivative is \(f(x)\). In the context of definite integrals, the antiderivative is used in the Fundamental Theorem of Calculus, which relates the definite integral to the antiderivative. The theorem states that if \(F\) is an antiderivative of \(f\) on \([a, b]\), then one can evaluate the definite integral of \(f\) from \(a\) to \(b\) by finding the difference \(F(b) - F(a)\).

However, not all functions have antiderivatives that can be expressed in terms of elementary functions, such as polynomials, exponential functions, logarithms, and trigonometric functions. In these cases, one may need to resort to numerical integration or explore special functions that extend beyond elementary functions to find the antiderivative of the integrand. In our problem, an analytical approach to the definite integral leads us to a reduced integration that hints at such a scenario where a numerical approach might be necessary.

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Most popular questions from this chapter

The region bounded by \(f(x)=(4-x)^{-1 / 3}\) and the \(x\) -axis on the interval [0,4) is revolved about the \(y\) -axis. The region bounded by \(f(x)=(x+1)^{-3 / 2}\) and the \(x\) -axis on the interval (-1,1] is revolved about the line \(y=-1\).

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