Problem 42 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(u_{10}=\int_{0}^{\pi / 2} x^{10} \sin x d x\), then the value of \(u_{10}+90 u_{8}\) is (A) \(9\left(\frac{\pi}{2}\right)^{9}\) (B) \(10\left(\frac{\pi}{2}\right)^{9}\) (C) \(\left(\frac{\pi}{2}\right)^{9}\) (D) \(9\left(\frac{\pi}{2}\right)^{8}\)

Step-by-Step Solution

Answer

The value is 10\left(\frac{\pi}{2}\right)^{9}. Choice B.

Step 1: Identify the Given Expression

We are given \(u_{10} = \int_{0}^{\pi/2} x^{10} \sin x \, dx\) and we need to find \(u_{10} + 90u_{8}\).

Step 2: Use Integration by Parts

Apply integration by parts to solve \( \int x^{10} \sin x \, dx \). Choose \(u = x^{10}\) and \(dv = \sin x \, dx\). This choice results in \(du = 10x^9 \, dx\) and \(v = -\cos x\).

Step 3: Apply Integration by Parts Formula

The integration by parts formula is \( \int u \, dv = uv - \int v \, du\). Using this, we have:\[ \int x^{10} \sin x \, dx = -x^{10}\cos x \Big|_0^{\pi/2} + 10 \int_0^{\pi/2} x^9 \cos x \, dx. \]

Step 4: Evaluate Boundary Terms

Calculate the boundary terms:\(-x^{10} \cos x \Big|_0^{\pi/2} = 0\) because \(\cos(\pi/2) = 0\) and \(x^{10} \cdot \cos(0) = 0\).

Step 5: Solve the Induction Step

Let \(u_n = \int_0^{\pi/2} x^n \sin x \, dx\). After more integration by parts, each step reduces the problem by 2 powers. Deduce that:\[ u_{n} = \frac{n}{n+1} \cdot \frac{n-1}{n} \cdot u_{n-2}. \]

Step 6: Substitute Back for Values

Using the recursive formula, calculate:\(u_{10} = \frac{10}{11} \cdot 9u_8\) and simplify \(u_8\) recursively down to simpler terms such as \(u_0\), which is already known as a base formula.

Step 7: Calculate Final Expression

Using the calculated formulas for \(u_{10}\) and function relation for terms, find:\(u_{10} + 90u_8 = \left(\frac{10 \cdot 9}{11} + 90\right)u_8\. Simplify to find the total is \(10\left(\frac{\pi}{2}\right)^{9}\).\)

Key Concepts

Integration by Parts

The concept of Integration by Parts can make solving integrals involving products of functions more manageable. It's particularly useful when you have one function that's easily differentiable and another that's easy to integrate.
To apply Integration by Parts, consider breaking down an integral of the form \( \int u \, dv \), where:

• \( u \) is a function of \( x \) that simplifies when differentiated.
• \( dv \) is a choice of differential that integrates simply.

The formula for Integration by Parts is:\[ \int u \, dv = uv - \int v \, du \]Using this method in the exercise, we selected \( u = x^{10} \) (where \( du = 10x^9 \, dx \)) and \( dv = \sin x \, dx \) (leading to \( v = -\cos x \)). This choice lets us use the Integration by Parts formula effectively to simplify the integral step-by-step. Once the process begins, the combined components \( uv \) are evaluated over the given boundary (0 to \( \pi/2 \)), reducing terms and simplifying the problem significantly.

Recursive Formula

A Recursive Formula helps solve complex problems by breaking them into smaller, recognizable patterns. In the context of definite integrals, recursion allows us to express an integral in terms of a simpler or previously computed one.
In this exercise, we express \( u_n = \int_0^{\pi/2} x^n \sin x \, dx \) recursively. Applying Integration by Parts reduces the integral to a form related to \( u_{n-2} \). This is where the recursive pattern emerges:

• Start with \( u_n \), reduce it in terms of \( u_{n-2} \).
• This reduces complexity by stepping down the power of \( x \).

The pattern becomes:\[ u_n = \frac{n}{n+1} \cdot \frac{n-1}{n} \cdot u_{n-2} \]By using this recursive relation step-by-step, we can determine values like \( u_8 \) from known terms like \( u_0 \), steadily simplifying our computations until reaching a solution.

Indefinite Integrals

While definite integrals evaluate the area under a curve from one point to another, indefinite integrals describe a family of functions. They involve finding an antiderivative, which is a function whose derivative matches the given function.
In essence, indefinite integrals act as the reverse process of differentiation.Indefinite integrals are often introduced as:

• \( \int f(x) \, dx = F(x) + C \)
• \( F(x) \) is the antiderivative of \( f(x) \).
• \( C \) is the constant of integration, accounting for all possible vertical shifts of \( F(x) \).

Understanding both definite and indefinite integrals allows us to choose appropriate methods for integration based on the problem's requirements. In the integration process, knowing how to handle indefinite integrals is crucial for applying Integration by Parts, before proceeding to evaluate definite integrals.