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# Calculus

Integration of Functions

# Change of Variable

Page 1
Problems 1-2
Page 2
Problems 3-5

Let $$F\left( x \right)$$ be an indefinite integral or antiderivative of $$f\left( x \right).$$ To evaluate an integral, it is often necessary to make a substitution. Suppose that

${\int {f\left( x \right)dx} } = {\int {f\left( {g\left( u \right)} \right)g’\left( u \right)du} } = {F\left( u \right) } = {F\left( {{g^{ – 1}}\left( x \right)} \right),}$

where $$x = g\left( u \right)$$ is the new substitution. Respectively, the inverse function $$u = {g^{ – 1}}\left( x \right)$$ determines the new variable depending on the old one.

Making a substitution, we should also replace the differential in the old variable $$dx$$ with the new differential $$du.$$ For a definite integral, we should also change the limits of integration.

## Solved Problems

Click on problem description to see solution.

### ✓Example 1

Calculate the integral $$\int {\large\frac{{dx}}{{\sqrt {{a^2} – {x^2}} }}\normalsize} .$$

### ✓Example 2

Find the integral $$\int {{\large\frac{{x + 1}}{{{x^2} + 2x – 5}}\normalsize} dx}.$$

### ✓Example 3

Calculate the integral $$\int {{2^x}{e^x}dx} .$$

### ✓Example 4

Calculate the integral $$\int {\cot \left( {3x + 5} \right)dx}.$$

### ✓Example 5

Find the integral $$\int {{\large\frac{{\sin 2x}}{{\sqrt {1 + {{\cos }^2}x} }}\normalsize} dx}.$$

### Example 1.

Calculate the integral $$\int {\large\frac{{dx}}{{\sqrt {{a^2} – {x^2}} }}\normalsize} .$$

#### Solution.

Let $$u = \large\frac{x}{a}\normalsize.$$ Then $$x = au,$$ $$dx = adu.$$ Hence, the integral is

$\require{cancel} {\int {\frac{{dx}}{{\sqrt {{a^2} – {x^2}} }}} } = {\int {\frac{{adu}}{{\sqrt {{a^2} – {{\left( {au} \right)}^2}} }}} } = {\int {\frac{{adu}}{{\sqrt {{a^2}\left( {1 – {u^2}} \right)} }}} } = {\int {\frac{{\cancel{a}du}}{{\cancel{a}\sqrt {1 – {u^2}} }}} } = {\int {\frac{{du}}{{\sqrt {1 – {u^2}} }}} } = {\arcsin u + C } = {\arcsin \frac{x}{a} + C.}$

### Example 2.

Find the integral $$\int {{\large\frac{{x + 1}}{{{x^2} + 2x – 5}}\normalsize} dx}.$$

#### Solution.

We make the substitution $$u = {x^2} + 2x – 5.$$ Then $$du = 2xdx + 2dx$$ $$= 2\left( {x + 1} \right)dx$$ or $$\left( {x + 1} \right)dx = {\large\frac{{du}}{2}\normalsize}.$$ The integral is easy to calculate with the new variable:

${\int {\frac{{x + 1}}{{{x^2} + 2x – 5}}dx} } = {\int {\frac{{\frac{{du}}{2}}}{u}} } = {\frac{1}{2}\int {\frac{{du}}{u}} } = {\frac{1}{2}\ln \left| u \right| + C } = {\frac{1}{2}\ln \left| {{x^2} + 2x – 5} \right| }+{ C.}$
Page 1
Problems 1-2
Page 2
Problems 3-5