Use The Indicated Substitution To Evaluate The Integral

Touse the indicated substitution to evaluate the integral, you replace a complicated expression inside the integrand with a simpler variable, typically denoted u. This technique transforms the integral into a form that can be integrated using basic rules, then you revert the substitution to express the antiderivative in terms of the original variable. The method is especially powerful when the integrand contains a function and its derivative, or when a trigonometric, exponential, or polynomial expression can be simplified through a clever change of variables. By following a systematic approach, you can handle a wide range of integrals that would otherwise require obscure tricks or lengthy algebraic manipulations.

Understanding the Core Idea

What is a substitution?

A substitution is a mathematical shortcut that rewrites an integral in terms of a new variable u that captures the inner function of the original integrand. The key insight is that differentials transform accordingly: if u = g(x), then du = g'(x)dx. This relationship allows the dx term to be expressed in terms of du, often canceling out the derivative hidden inside the original integral.

Why does it work?

The substitution method is grounded in the chain rule of differentiation. When you differentiate a composite function f(g(x)), the derivative is f'(g(x))·g'(x). Integration reverses this process: recognizing a derivative inside the integrand lets you “undo” the chain rule by substituting the inner function. This is why the technique is sometimes called u‑substitution.

Step‑by‑Step Procedure

1. Identify the inner function

Look for a part of the integrand that is itself a function of another function, such as sin(x²), e^(3x), or (5x+2)^4. This inner function is the prime candidate for the substitution.

2. Choose u wisely

Set u equal to that inner function. Write the differential du by differentiating u with respect to x. If du contains a factor that matches another part of the integrand, you are on the right track.

3. Rewrite the integral

Replace every occurrence of the inner function with u and replace dx with du/g'(x). Often, the g'(x) factor will cancel with a matching term in the original integrand, leaving a simpler expression in u.

4. Integrate with respect to u

Apply standard integration formulas to the new integrand. This step is usually straightforward because the substitution has simplified the algebraic complexity.

5. Substitute back

Replace u with the original expression in terms of x to obtain the final antiderivative.

Common Substitutions and When to Use Them

Italic emphasis on tricky cases

When the integrand contains a product of a function and its derivative, the substitution often reveals a perfect match. For instance, in ∫x·cos(x²) dx, setting u = x² yields du = 2x dx, and the integral becomes (1/2)∫cos(u) du, which integrates to (1/2)sin(u) + C, then back to (1/2)sin(x²) + C.

Worked Example

Problem

Evaluate the integral ∫(2x+5)^3·2 dx using the indicated substitution.

Solution

1. Identify the inner function: The expression 2x+5 appears inside a power.
2. Choose u: Let u = 2x+5.
3. Compute du: Differentiate to get du = 2 dx → dx = du/2.
4. Rewrite the integral:
   ∫(u)^3·2·(du/2) = ∫u^3 du.
5. Integrate: ∫u^3 du = u^4/4 + C.
6. Substitute back: Replace u with 2x+5: (2x+5)^4/4 + C.

The final antiderivative is (2x+5)^4/4 + C. This example illustrates how a simple linear substitution can transform a seemingly complex integral into a basic power rule problem.

Tips and Common Pitfalls

• Check the differential: Always verify that du contains the exact factor needed to replace dx. If not, you may need to rearrange or multiply/divide by a constant.
• Watch for missing constants: When du includes a constant factor, factor it out of the integral to avoid algebraic errors.
• Avoid circular substitutions: If substituting leads back to the original variable without simplification, reconsider the choice of u.
• Multiple substitutions: Some integrals require more than one substitution step; handle them sequentially, simplifying at each stage.
• Definite integrals: When evaluating a definite integral, also change the limits of integration according to the substitution to maintain correctness.

Frequently Asked Questions (FAQ)

Q1: Can I use substitution with trigonometric identities?
A: Yes. Often you set u = sin(x) or u = cos(x), then use identities like sin²(x) + cos²(x) = 1 to simplify the remaining expression.

Q2: What if the integrand has no obvious inner function?
A: Look for patterns such as f'(x)·f(x)^n or e^{ax}·g(x). Sometimes rewriting the integrand (e.g., multiplying by 1) reveals a hidden derivative.

Q3: Is substitution only for algebraic expressions?
A: No. It applies to exponential, logarithmic, trigonometric,

Advanced Scenarios

When the substitution is not immediately obvious, it can be helpful to manipulate the integrand algebraically before identifying u. For example, consider

[ \int \frac{x}{\sqrt{1+x^{2}}},dx . ]

A quick glance suggests the denominator resembles the derivative of the expression under the radical. By letting

[ u = 1 + x^{2}, ]

we obtain

[ du = 2x,dx \quad\Longrightarrow\quad \frac{1}{2}du = x,dx . ]

Substituting yields

[ \int \frac{x}{\sqrt{1+x^{2}}},dx = \frac{1}{2}\int u^{-1/2},du = \sqrt{u}+C = \sqrt{1+x^{2}}+C . ]

Another useful trick involves trigonometric products. Take

[\int \sin^{3}x,\cos x,dx . ]

Setting

[ u = \sin x ]

gives

[ du = \cos x,dx , ]

and the integral collapses to

[\int u^{3},du = \frac{u^{4}}{4}+C = \frac{\sin^{4}x}{4}+C . ]

These cases illustrate how a modest algebraic rearrangement can expose a clean substitution, even when the original form appears tangled.

Substitution in Definite Integrals

When dealing with limits, the substitution must be applied to the bounds as well. Suppose we need

[ \int_{0}^{1} 3x^{2}e^{x^{3}},dx . ]

Choose

[u = x^{3}\quad\Longrightarrow\quad du = 3x^{2},dx . ]

When (x = 0), (u = 0); when (x = 1), (u = 1). The integral becomes

[ \int_{0}^{1} e^{u},du = e^{u}\Big|_{0}^{1}=e-1 . ]

Notice that the limits transform automatically, eliminating the need to revert to the original variable before evaluating.

Multiple‑Step Substitutions Some integrals demand a chain of substitutions. Consider

[ \int \frac{\ln(\sqrt{x})}{x},dx . ]

First, rewrite the logarithm:

[ \ln(\sqrt{x}) = \tfrac{1}{2}\ln x . ]

Now set

[u = \ln x \quad\Longrightarrow\quad du = \frac{1}{x},dx . ]

The integral transforms to

[ \int \tfrac{1}{2}u,du = \frac{1}{4}u^{2}+C = \frac{1}{4}(\ln x)^{2}+C . ]

If a second substitution were required, it would be handled in the same meticulous fashion, always confirming that the differential matches the remaining factor.

Summary of Key Takeaways

• Identify an inner function whose derivative appears (or can be made to appear) in the integrand. - Compute du and express dx accordingly; factor out constants if necessary. - Replace all occurrences of the original variable with the new u and adjust limits for definite integrals.
• Integrate with respect to u, then back‑substitute to the original variable.
• Verify the result by differentiating the antiderivative; this step often uncovers algebraic slip‑ups.

Conclusion

Integration by substitution is a versatile tool that turns complex-looking integrals into manageable pieces by leveraging the chain rule in reverse. Mastery hinges on recognizing patterns, handling differentials with care, and applying the method consistently — whether the integral is indefinite, definite, or nested within multiple layers of functions. With practice, the substitution step becomes almost instinctive, allowing you to navigate even the most intricate antiderivatives with confidence.