Use The Indicated Substitution To Evaluate The Integral

Use the Indicated Substitution to Evaluate the Integral: A Complete Guide

Integration by substitution is one of the most powerful and widely used techniques in calculus for evaluating integrals. Plus, when you are asked to use the indicated substitution to evaluate the integral, you are being guided toward a specific variable change that simplifies the integrand into a form that is much easier to integrate. Often referred to as u-substitution, this method is essentially the reverse chain rule of differentiation. Whether you are a student encountering calculus for the first time or a seasoned learner reviewing foundational concepts, mastering this technique is essential for solving a vast array of integral problems.

Understanding the Method of Substitution

Before diving into how to use the indicated substitution, it is the kind of thing that makes a real difference. In practice, the substitution method relies on the chain rule for derivatives, applied in reverse. If you have an integral of the form ∫ f(g(x)) · g'(x) dx, the substitution u = g(x) transforms the integral into ∫ f(u) du, which is often far simpler to evaluate.

The key idea is to identify an "inner function" within the integrand and replace it with a single variable, typically denoted as u. In real terms, once the substitution is made, the differential dx must also be expressed in terms of du. This ensures that every part of the original integral is properly converted into the new variable.

When a problem states "use the indicated substitution," it means a specific substitution has already been chosen for you. Your task is to follow that substitution through every step of the evaluation process, from rewriting the integral to expressing the final answer back in terms of the original variable No workaround needed..

The Indicated Substitution: Key Principles

When working with an indicated substitution, there are several principles you should keep in mind to ensure accuracy and efficiency.

• Identify the substitution variable clearly. The problem will typically tell you something like "let u = 2x + 1" or "use the substitution x = 3sin(θ)." Write this down immediately so you do not lose track of what you are doing.

• Compute the differential. Once you have defined u in terms of x ( or another variable), differentiate both sides to find du in terms of dx. Here's one way to look at it: if u = 2x + 1, then du = 2 dx, which means dx = du/2 And that's really what it comes down to..

• Replace every instance of the original variable. After finding du, substitute both the variable expression and the differential into the integral. Every occurrence of x (or the original variable) must be replaced with an expression involving u.

• Adjust the limits of integration if it is a definite integral. For definite integrals, the limits of integration must also be changed to correspond to the new variable. If the original integral runs from x = a to x = b, you must compute u(a) and u(b) and use those as the new limits Simple as that..

• Integrate with respect to the new variable. Once the substitution is complete, perform the integration in terms of u It's one of those things that adds up..

• Back-substitute to return to the original variable. Unless the problem is a definite integral where the limits have already been converted, you must express the final answer in terms of the original variable Small thing, real impact. Which is the point..

Step-by-Step Guide to Evaluating Integrals Using Substitution

Follow these steps systematically whenever you are asked to use the indicated substitution to evaluate the integral.

Step 1: Write Down the Indicated Substitution

Start by clearly writing the substitution given in the problem. Take this: if the problem says "use the substitution u = x² + 3," write:

u = x² + 3

Step 2: Differentiate to Find du

Differentiate both sides of the substitution equation with respect to x:

du/dx = 2x

Therefore:

du = 2x dx

If the integral does not contain a 2x factor, you may need to solve for a portion of the expression. To give you an idea, if you have x dx in the integral, you can write x dx = du/2 Practical, not theoretical..

Step 3: Rewrite the Integral

Substitute u and du (or the appropriate expression) into the original integral. Let us consider a concrete example:

Evaluate ∫ 2x(x² + 3)⁵ dx using the substitution u = x² + 3.

• Let u = x² + 3, so du = 2x dx.
• The integral becomes ∫ u⁵ du.

This is a much simpler integral to evaluate Small thing, real impact..

Step 4: Integrate with Respect to u

Perform the integration using standard integration rules:

∫ u⁵ du = u⁶/6 + C

Step 5: Back-Substitute

Replace u with the original expression in terms of x:

u⁶/6 + C = (x² + 3)⁶/6 + C

And that is your final answer.

Worked Examples with Indicated Substitutions

Example 1: Algebraic Substitution

Evaluate ∫ 6x²(x³ + 2)⁴ dx using the substitution u = x³ + 2.

1. Given substitution: u = x³ + 2
2. Differentiate: du = 3x² dx, so 2 du = 6x² dx
3. Rewrite the integral: ∫ (u⁴)(2 du) = 2∫ u⁴ du
4. Integrate: 2 · (u⁵/5) + C = 2u⁵/5 + C
5. Back-substitute: 2(x³ + 2)⁵/5 + C

Example 2: Trigonometric Substitution

Evaluate ∫ x√(9 − x²) dx using the substitution x = 3sin(θ).

1. Given substitution: x = 3sin(θ), so dx = 3cos(θ) dθ
2. Rewrite the radical: √(9 − x²) = √(9 − 9sin²(θ)) = 3√(cos²(θ)) = 3cos(θ)
3. Rewrite the integral: ∫ (3sin(θ))(3cos(θ))(3cos(θ) dθ) = 27∫ sin(θ)cos²(θ) dθ
4. Use another substitution: Let w = cos(θ), so dw = −sin(θ) dθ
5. Integrate: −27∫ w² dw = −27 · w³/3 + C = −9w³ + C
6. Back-substitute: Since cos(θ) = √(1 − sin²(θ)) = √(1 − x²/9) = √(9 − x²)/3, the answer becomes −9 · (√(9 − x²)/3)³ + C = −(9 − x²)^(3/2)/3 + C

Example 3: Definite Integral with Substitution

Evaluate ∫₀¹ 2x·e^(x²) dx using the substitution u = x² Most people skip this — try not to..

1. Given substitution: u = x²
2. Differentiate: du = 2x dx
3. Change limits: When x = 0

Example 3 (continued): Definite Integral with Substitution

1. Given substitution: (u = x^{2})

2. Differentiate: (du = 2x,dx)

3. Adjust the limits of integration

   • When (x = 0), (u = 0^{2} = 0).
   • When (x = 1), (u = 1^{2} = 1).

4. Rewrite the integral in terms of (u). Because (2x,dx = du), the integrand becomes (e^{u}):
   [ \int_{0}^{1} 2x,e^{x^{2}},dx = \int_{u=0}^{1} e^{u},du . ]

5. Integrate with respect to (u):
   [ \int e^{u},du = e^{u}+C . ]
   Applying the new limits:
   [ \bigl[e^{u}\bigr]_{0}^{1}=e^{1}-e^{0}=e-1 . ]

6. Result:
   [ \boxed{\displaystyle \int_{0}^{1} 2x,e^{x^{2}},dx = e-1 } . ]

Example 4: Rational Substitution

Evaluate (\displaystyle \int \frac{2x+1}{x^{2}+x+1},dx) by letting (u = x^{2}+x+1) Worth knowing..

1. Substitution: (u = x^{2}+x+1).

2. Differential: (du = (2x+1),dx). Notice that the numerator matches (du) exactly.

3. Rewrite the integral:
   [ \int \frac{2x+1}{x^{2}+x+1},dx = \int \frac{du}{u}. ]

4. Integrate: (\displaystyle \int \frac{du}{u}= \ln|u|+C) Worth keeping that in mind..

5. Back‑substitute: (\displaystyle \ln!\bigl(x^{2}+x+1\bigr)+C).

Example 5: Trigonometric Substitution with a Linear Factor

Evaluate (\displaystyle \int \frac{x}{\sqrt{4-x^{2}}},dx) using (x = 2\sin\theta).

1. Substitution: (x = 2\sin\theta) (\Rightarrow) (dx = 2\cos\theta,d\theta) Simple, but easy to overlook..

2. Express the radical: (\sqrt