When do you use u-substitution

U‑substitution is the integration analogue of the chain rule for differentiation. You look for a composite function where an inner function’s derivative appears elsewhere in the integrand. When that pattern exists, setting $u$ equal to the inner function simplifies the integral, often reducing it to a standard form you already know how to integrate.

Recognizing the pattern

The key is spotting a factor that is the derivative of another factor. For example, in $\int 3x^2\sin(x^3)\,dx$, the inner function $x^3$ has derivative $3x^2$, which is present as a multiplicative factor. If the derivative is off by a constant, you can adjust it by factoring that constant out. This recognition step saves time and prevents trial‑and‑error attempts with integration by parts.

Typical forms that suggest u‑substitution:

• A product of a function and its derivative, e.g., $f'(x)\,g(f(x))$
• A rational function where the denominator’s derivative appears in the numerator
• Trigonometric integrals like $\sin(u)\,u'$ or $\cos(u)\,u'$
• Exponential expressions of the form $e^{u}\,u'$

Apply u‑substitution with these steps:

1. 1Identify $u$ as a function whose derivative appears in the integrand
2. 2Compute $du = u'\,dx$ and solve for $dx$
3. 3Rewrite the integral entirely in terms of $u$ and $du$
4. 4Integrate with respect to $u$
5. 5If the original integral is definite, change the limits to $u$-values; otherwise, substitute back $u = \text{original expression}$

Worked example: evaluate $\int 2x\cos(x^2)\,dx$. Let $u = x^2$; then $du = 2x\,dx$. The integral becomes $\int \cos(u)\,du$, whose antiderivative is $\sin(u) + C$. Substituting back gives $\sin(x^2) + C$. This demonstrates how the derivative $2x$ matches the extra factor, making the substitution straightforward.

Before and after substitution:

After you finish the substitution, always verify by differentiating your result; you should recover the original integrand. For definite integrals, adjust the limits before integrating, which avoids the extra back‑substitution step. Mastering the pattern‑recognition stage makes u‑substitution a quick tool for many seemingly complex integrals.