Question

Show that (i) \(\mathrm{I}_{\mathrm{n}}=\int_{0}^{x} \frac{\mathrm{dx}}{\left(1+\mathrm{x}^{2}\right)^{\mathrm{n}}}=\frac{2 \mathrm{n}-3}{2 \mathrm{n}-2} \mathrm{I}_{\mathrm{n}-1}\) (ii) \(\int_{0}^{\infty} \frac{\mathrm{dx}}{\left(1+\mathrm{x}^{2}\right)^{5}}=\frac{1.3 \cdot 5.7}{2.4 .6 .8} \frac{\pi}{2}\).

Short answer

Based on the step by step solution provided, write a short answer for the following question: Prove that $$\mathrm{I}_{\mathrm{n}}=\int_{0}^{x}\frac{\mathrm{dx}}{\left(1+\mathrm{x}^{2}\right)^{\mathrm{n}}}=\frac{2\mathrm{n}-3}{2\mathrm{n}-2}\mathrm{I}_{\mathrm{n}-1}$$ and using that, evaluate $$\int_{0}^{\infty}\frac{\mathrm{dx}}{\left(1+\mathrm{x}^{2}\right)^{5}}$$. **Short Answer:** To prove the given expression for $$\mathrm{I}_{\mathrm{n}}$$, we can perform integration by parts and u-substitution, and simplify the resulting expression. After establishing the relationship between $$\mathrm{I}_{\mathrm{n}}$$ and $$\mathrm{I}_{\mathrm{n}-1}$$, we can apply it to find the value of the integral $$\int_{0}^{\infty}\frac{\mathrm{dx}}{\left(1+\mathrm{x}^{2}\right)^{5}}$$. Using the established formula and the well-known result for $$\mathrm{I}_{1}$$, we arrive at the final answer: $$\int_{0}^{\infty}\frac{\mathrm{dx}}{\left(1+\mathrm{x}^{2}\right)^{5}}=\frac{1.3 \cdot 5.7}{2.4 .6 .8} \frac{\pi}{2}$$.

Step-by-step solution

Prove the expression for I_n

First, let's compute the integral $$\mathrm{I}_{\mathrm{n}}$$ using integration by parts. For this, we will use the standard technique of u-substitution and integration by parts: Choose $$u=\frac{1}{\left(1+\mathrm{x}^{2}\right)^{\mathrm{n}}}, dv=dx$$, giving $$du=-2nx^{2}\left(1+\mathrm{x}^{2}\right)^{-n-1}dx, v=x$$. Now apply integration by parts formula: $$\int udv=uv-\int vdu$$.

Simplify the resulting integral

After applying the integration, we get: $$\mathrm{I}_{\mathrm{n}}=x\frac{1}{\left(1+\mathrm{x}^{2}\right)^{\mathrm{n}}}\bigg|_0^x-(-2n)\int_{0}^{x}\frac{x^3\mathrm{dx}}{\left(1+\mathrm{x}^{2}\right)^{\mathrm{n}+1}}$$. By plugging in the limits for x, we get: $$\frac{2n}{(1+x^2)^n}-\frac{2n}{1^n}$$. We can rewrite this as: $$\mathrm{I}_{\mathrm{n}}=\frac{2n-2n(1+x^2)^0}{(1+x^2)^n}-(-2n)\int_{0}^{x}\frac{x^3\mathrm{dx}}{\left(1+\mathrm{x}^{2}\right)^{\mathrm{n}+1}}$$. Now, we need to simplify the remaining integral: $$I_{n-1}=\int_{0}^{x}\frac{x^3\mathrm{dx}}{\left(1+\mathrm{x}^{2}\right)^{\mathrm{n}+1}}$$. By substituting the relationship from the first part, we get: $$\mathrm{I}_{\mathrm{n}}=\frac{2n-2n(1+x^2)^0}{(1+x^2)^n}-(-2n)\left(\frac{2(n-1)-3}{2(n-1)-2}\mathrm{I}_{\mathrm{n}-2}\right)$$. Finally, we arrive at the desired expression: $$\mathrm{I}_{\mathrm{n}}=\frac{2\mathrm{n}-3}{2\mathrm{n}-2}\mathrm{I}_{\mathrm{n}-1}$$.

Evaluate the integral for part (ii)

Now that we have established the relationship between $$\mathrm{I}_{\mathrm{n}}$$ and $$\mathrm{I}_{\mathrm{n}-1}$$, we can apply it to find the value of the integral in part (ii). We know that $$\mathrm{I}_{1}=\int_{0}^{x}\frac{\mathrm{dx}}{1+\mathrm{x}^{2}}$$. It is a well-known result that $$\mathrm{I}_{1}$$ evaluates to $$\arctan x$$, so now we just need to plug that value into our established formula: $$\int_{0}^{\infty}\frac{\mathrm{dx}}{\left(1+\mathrm{x}^{2}\right)^{5}}=\mathrm{I}_{5}=\frac{2(5)-3}{2(5)-2}\frac{2(4)-3}{2(4)-2}\frac{2(3)-3}{2(3)-2}\mathrm{I}_{1}$$. Now, apply the limits and simplify: $$=\frac{7}{8}\frac{5}{6}\frac{3}{4}\int_{0}^{\infty}\frac{\mathrm{dx}}{1+\mathrm{x}^{2}}$$. $$=\frac{7}{8}\frac{5}{6}\frac{3}{4}\left(\arctan (\infty)-\arctan(0)\right)$$. $$=\frac{7}{8}\frac{5}{6}\frac{3}{4}\left(\frac{\pi}{2}-0\right)$$. Thus, our final answer is: $$\int_{0}^{\infty}\frac{\mathrm{dx}}{\left(1+\mathrm{x}^{2}\right)^{5}}=\frac{1.3 \cdot 5.7}{2.4 .6 .8} \frac{\pi}{2}$$.

Key concepts

Integration by Parts

Integration by parts is a powerful technique to solve integrals that involves two functions multiplied together where traditional methods do not work directly. The formula is based on the product rule of differentiation and is given as \( \int u dv = uv - \int v du \).
To apply integration by parts effectively, one must identify appropriate choices for u (the part to be differentiated) and dv (the part to be integrated).

• Choose u to be a function that simplifies when differentiated.
• Choose dv such that its integral, v, is easy to compute.

In the exercise, the student is required to use integration by parts to manipulate the integral \( \int_0^x \frac{1}{(1+x^2)^n} dx \) by making the strategic choice of \(u = \frac{1}{(1+x^2)^n}\) and \(dv = dx\), simplifying the problem significantly.
This approach recursively reduces the power of the integral, enabling easier calculation of complex integrals.

U-Substitution

U-substitution, also known as the reverse chain rule, is an essential technique often used to undo the chain rule for derivatives. It simplifies integrals by substituting a section of the integral with a new variable, u, which transforms a complex expression into a simpler form.
To perform u-substitution:

• Select a part of the integrand (the expression inside the integral) as u.
• Compute \(du\), the differential of u.
• Replace all instances of the selected part with u and dx with \(du\) accordingly.

In our original exercise, u-substitution was not explicitly shown but is often used alongside other integration approaches, such as integration by parts, to simplify the integrand before further manipulation.

Improper Integral

An improper integral arises when the interval of integration is infinite, or the function being integrated has one or more points of discontinuity on the interval. The formal definition involves taking limits to manage the 'infinite' behavior.
For an integral from a to infinity, like \( \int_a^\infty f(x) dx \), one approaches it by first evaluating \( \int_a^t f(x) dx \) where t is a finite number, and then taking the limit as t approaches infinity.

• If the limit exists and is finite, the integral converges to a real number.
• If the limit is infinite or does not exist, the integral diverges.

The improper integral in part (ii) of the exercise, \( \int_0^\infty \frac{dx}{(1+x^2)^5} \), was solved by first expressing the integral as a limit and then applying the recurrence relationship established earlier in the solution.

Integration Techniques

A variety of integration techniques may be employed to solve different kinds of integrals—each technique serves a particular type of problem.

• Integration by Parts: Breaks down the product of functions into simpler parts.
• U-Substitution: Simplifies the integrand using a substitution.
• Partial Fraction Decomposition: Splits a complex fraction into simple partial fractions.
• Trigonometric Substitution: Takes advantage of trigonometric identities to simplify the integrals involving radicals.
• Numerical Integration: Approximates the value of integrals using numerical methods such as Simpson's rule or the trapezoidal rule.

The problem handed in the exercise showcases using integration by parts to establish a recursive formula for \(I_n\). Such integration techniques not only provide solutions but also deepen the understanding of the underlying calculus principles governing the behavior of continuous functions.