Chapter 9: Integration Techniques
Section 9.1: Integrations by Substitution

Perform the following indefinite integration.

$$\int x \sqrt{x^2 + 1} dx$$

Answer

$$\int x \sqrt{x^2 + 1} dx = \frac{1}{3} (x^2 + 1)^\frac{3}{2} + C$$

Chain Rule Recalled

$$\frac{d}{dx} (f(g(x)) = \frac{df}{du} \vert_{u = g(x)} \frac{dg}{dx}$$

Inverting the Chain Rule: Integration by Substitution

$$\int f'(g(x)) g'(x) dx = f(g(x)) + C$$

Formalism of Integration by Substitution

Often, one writes the substitution as $u = g(x)$ and $du = g'(x) dx$ and attempts to write the integrand as $f(u) du = f(g(x)) g'(x) dx$. One is then charged with integrating $\int f(u) du = F(u) + C$. We conclude that $\int f(g(x)) g'(x) dx = \int f(u) du = F(u) + C = F(g(x)) + C$.

An integral revisited

Making the substitution $u = x^2 + 1$, so that $du = 2x dx$ we have

$$\int x \sqrt{x^2 + 1} dx = \frac{1}{2} \int \sqrt{u} du$$

Writing $\sqrt{u} = u^\frac{1}{2}$, we find

$$\int \sqrt{u} du = \frac{2}{3} u^\frac{3}{2} + C$$

Thus,

$$\int x \sqrt{x^2 + 1} dx = \frac{1}{3} (x^2 + 1)^\frac{3}{2} + \widetilde{C}$$

Example

Integrate:

$$\int x e^{x^2} dx$$

Solution

Set $u = x^2$. Then $du = 2x dx$.

$$\int x e^{x^2} dx = \frac{1}{2} \int e^{x^2} (2x) dx$$

$$= \frac{1}{2} \int e^u du$$

$$= \frac{1}{2} e^u + C$$

$$= \frac{1}{2} e^{x^2} + C$$

Another Example

Integrate

$$\int \sin(x) \cos(x) dx$$

Solution

Set $u = \sin(x)$ so that $du = \cos(x) dx$.

$$\int \sin(x) \cos(x) dx = \int u du$$

$$= \frac{1}{2} u^2 + C$$

$$= \frac{1}{2} \sin^2(x) + C$$

A third example

Integrate

$$\int \frac{\sqrt{\ln(x)}}{x} dx$$

Solution

Set $u = \ln(x)$ so that $du = \frac{1}{x} dx$.

$$\int \frac{\sqrt{\ln(x)}}{x} dx = \int \sqrt{u} du$$

$$= \int u^\frac{1}{2} du$$

$$= \frac{2}{3} u^\frac{3}{2} + C$$

$$= \frac{2}{3} (\ln(x))^\frac{3}{2} + C$$

A final example

Integrate

$$\int \frac{\sin(x)}{\cos^3(x)} dx$$

Solution

Set $u = \tan(x)$. Then $du = \sec^2 (x) dx$.

$$\int \frac{\sin(x)}{\cos^3(x)} dx = \int \tan(x) \sec^2(x) dx$$

$$= \int u du$$

$$= \frac{1}{2} u^2 + C$$

$$= \frac{1}{2} \tan^2(x) + C$$

An alternate solution

Set $v = \cos(x)$ so that $dv = - \sin(x) dx$.

$$\int \frac{\sin(x)}{\cos^3(x)} dx = -\int v^{-3} du$$

$$= \frac{1}{2} v^{-2} + C'$$

$$= \frac{1}{2} \sec^2(x) + C'$$

Error?

Why are these answers different?

Answer

$$\tan^2(x) + 1 = \sec^2(x)$$

So, $C = \frac{1}{2} + C'$.