Integral Using U-Substitution

Learning Objectives

By the end of this lesson, you will be able to:

• Understand why u-substitution works.

• Identify when an integral requires substitution.

• Choose an appropriate substitution $u=g(x)$.

• Rewrite the integral entirely in terms of $u$.

• Evaluate integrals involving composite functions.

• Reverse the substitution back to $x$.

What is U-Substitution?

U-substitution is the reverse process of the chain rule.

The chain rule states:

$$\frac{d}{dx}[f(g(x))] = f^\prime(g(x))\cdot g^\prime(x)$$

So the integral version reverses this idea:

$$\int f^\prime(g(x))\cdot g^\prime(x)dx = f(g(x))+C$$

The idea is simple:

• Let $$u=g(x)$$ (inner function)

• Then $$du = g^\prime(x)dx$$

• Rewrite the integral in terms of u

Step-by-Step Method

Step 1: Choose substitution

Select $$u=g(x)$$, the expression inside another function.

Step 2: Compute $du$

Differentiate:

$$ g^\prime(x) = \frac{du}{dx} \longrightarrow g^\prime(x) dx = du $$

Step 3: Rewrite the integral

Replace:

• Every occurrence of $$g(x)$$ with u

• $$g^\prime(x) dx$$ with $$du$$

Step 4: Integrate in terms of u

Step 5: Substitute back to x

Examples

Example 1:

$$\int \sin(2x) dx$$

Let:

$$2x=u \longrightarrow 2 = \frac{du}{dx} \longrightarrow 2\cdot dx=du \longrightarrow dx = \frac{1}{2}du$$ 

Substitute:

$$\int \sin(2x) dx = \int \sin(u) \frac{1}{2}du = \frac{1}{2}\int \sin(u)du=-\frac{1}{2}\cos(u)+C$$

Substitute back:

$$-\frac{1}{2}\cos(2x)+C$$

Example 2:

$$\int 2x(x^{2}+5)^{3} dx$$

Let:

$$u=x^{2}+5 \longrightarrow 1\cdot\frac{du}{dx} = 2x \longrightarrow du = 2x \cdot dx$$

Substitute:

$$\int u^{3} \cdot 2x dx = \int u^{3} du = \frac{1}{4}u^{4}+C$$

Substitute back:

$$\frac{(x+5)^4}{4}+C$$

Example 3: Using a trigonometric inner function

$$\int \cos(x)\sin(x)dx$$ 

Let:

$$u = \sin(x) \longrightarrow 1\cdot\frac{du}{dx} = \cos(x) \longrightarrow du=\cos(x) dx$$

Then:

$$\int u \cdot \cos(x)dx = \int u du = \frac{u^{2}}{2}+C$$

Final answer:

$$\frac{\sin^2(x)}{2}+C$$​

Example 4: Exponential inner function

$$\int e^{3x} dx$$

Let:

$$u=3x \longrightarrow 1\cdot \frac{du}{dx} = 3 \longrightarrow \frac{1}{3}du = dx$$​

Substitute:

$$\frac{1}{3} \int e^{u} du = \frac{1}{3}e^{u}+C$$

Back to x:

$$\frac{1}{3}e^{3x}+C$$​

Example 5: squar root substitution

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

Let:

$$u=x^{2}+1 \longrightarrow 1\cdot\frac{du}{dx} = 2x \longrightarrow \frac{1}{2} du = x dx$$

Substitute:

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

Final answer:

$$\frac{1}{2}\int u^{\frac{1}{2}} du = \frac{1}{2} \frac{2}{3} u^{\frac{3}{2}} +C$$

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

Example 6: Rational substitution

$$\int \frac{1}{xln(x)}dx$$

Let:

$$u=ln(x) \longrightarrow 1\cdot\frac{du}{dx} = \frac{1}{x} \longrightarrow du = \frac{1}{x} dx$$

Substitute:

$$\int\frac{1}{u}\frac{1}{x}dx = \int\frac{1}{u}du = ln|u|+C$$

Final answer:

$$ln|u|+C=ln|ln(x)|+C$$

When Do We Use U-Substitution?

Use u-sub when the integrand looks like:

• A function inside another function (composite function)

• A function multiplied by the derivative of its inside part

Common forms:

• $$(ax+b)^{n}

• $$\sin(ax+b),\cos(ax+b)$$

• $$e^{ax+b}$$

• $$\sqrt{ax+b}$$​​

• U-substitution undoes the chain rule.

• Always convert everything into terms of $u$.

• Don’t forget to substitute back to $x$.

• For definite integrals, you may change limits instead of converting back.