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Integration Techniques

The main theorem reduces evaluating a definite integral to finding an antiderivative — but for many integrands no antiderivative is immediately visible. This page follows Walz, Foundations of Mathematics, the final part of Section 6.3: it develops integration by parts and the substitution rule, obtained by reading the product and chain rules backwards, and closes with an application to the area of an ellipse.

When No Antiderivative Is Obvious

For integrands such as $x e^x, \quad \ln(x), \quad \sin^2(x), \quad 6x(3x^2+1)^{71}, \quad \frac{\ln(x)}{x},$ the antiderivative is not apparent. Two rules extend our reach considerably. (It should be noted that some continuous integrands, such as $e^{-x^2}$, have no antiderivative expressible in closed form by elementary functions.)

Integration by Parts

Integrating the product rule $(f \cdot g)' = f'g + g'f$ and rearranging gives:

Integration by parts. For functions $f$ and $g$ differentiable on $\lbrack a, b \rbrack$: $\int_a^b f'(x)\,g(x)\,dx = \bigl\lbrack f(x)\,g(x) \bigr\rbrack_a^b - \int_a^b f(x)\,g'(x)\,dx.$

The idea is to choose $f'$ and $g$ so that the remaining integral on the right is easier than the original.

  • A product with $e^x$. For $h(x) = x e^x$, set $f(x) = e^x$ and $g(x) = x$. Then $\int x e^x\,dx = x e^x - \int e^x\,dx = (x - 1)e^x,$ so $\displaystyle\int_1^2 x e^x\,dx = e^2$.
  • The logarithm. Writing $\ln(x) = 1 \cdot \ln(x)$ and integrating by parts gives the antiderivative $F(x) = x\ln(x) - x$, hence $\int_1^7 \ln(x)\,dx = \bigl\lbrack x\ln(x) - x \bigr\rbrack_1^7 = 7\ln(7) - 6.$
  • A trigonometric square. For $\sin^2(x) = \sin(x)\sin(x)$, one integration by parts together with $\cos^2(x) = 1 - \sin^2(x)$ yields $\int \sin^2(x)\,dx = \tfrac{1}{2}\bigl(-\cos(x)\sin(x) + x\bigr).$

The Substitution Rule

Integrating the chain rule $(F \circ g)'(x) = g'(x)\,f(g(x))$, where $F' = f$, gives the substitution rule — the one-dimensional change of variables.

Substitution rule. If $f$ is continuous and $g$ is differentiable on $\lbrack a, b \rbrack$: $\int_a^b g'(x)\,f\bigl(g(x)\bigr)\,dx = \int_{g(a)}^{g(b)} f(\bar{x})\,d\bar{x}.$

A complicated integrand of the form $g' \cdot (f \circ g)$ is replaced by the simpler integrand $f$, with the bounds adjusted accordingly.

  • A high power. For $6x(3x^2+1)^{71}$, take $g(x) = 3x^2 + 1$ (so $g'(x) = 6x$) and $f(\bar{x}) = \bar{x}^{71}$. Then $\int 6x(3x^2+1)^{71}\,dx = \frac{1}{72}(3x^2+1)^{72}.$
  • A logarithm in the integrand. For $\dfrac{\ln(x)}{x}$, take $g(x) = \ln(x)$ (so $g'(x) = \tfrac{1}{x}$) and $f(\bar{x}) = \bar{x}$. Then the antiderivative is $\tfrac{1}{2}\ln^2(x)$, so $\int_1^2 \frac{\ln(x)}{x}\,dx = \frac{\ln^2(2)}{2}.$

Reading the Rule Backwards: the Second Substitution

The substitution rule can also be applied from right to left: to simplify an integrand one introduces a substitution $x = g(t)$, multiplies by the derivative $g'(t)$, and adjusts the bounds. This is sometimes called the second substitution rule.

Application — area of an ellipse. For an ellipse with semi-axes $r$ and $b$, the area is $F = \int_{-r}^{r} b\sqrt{1 - \frac{x^2}{r^2}}\,dx.$ Substituting $x = r\sin(t)$ (so $dx = r\cos(t)\,dt$) and using $\cos(t) = \sqrt{1 - \sin^2(t)}$ together with $\int_0^{\pi/2}\sin^2(t)\,dt = \tfrac{\pi}{4}$ gives $F = \pi r b.$ For the circle $r = b$ this reduces to the familiar $\pi r^2$.

Summary of Integral Calculus

Across these three pages the tools of integration form a coherent whole:

  • The Definite Integral defines the integral as a limit of Riemann sums and gives its basic properties.
  • Antiderivatives and the Main Theorem connects integration to differentiation, so integrals are evaluated via antiderivatives.
  • Integration by parts and substitution, developed here, produce antiderivatives for a wide range of functions.

Together with the differential calculus pages, they complete the core of single-variable calculus as presented in Walz, Chapter 6.