# Integration by Substitution

In this topic we shall see an important method for evaluating many complicated integrals.

Substitution for integrals corresponds to the chain rule for derivatives.

Suppose that $$F\left( u \right)$$ is an antiderivative of $$f\left( u \right):$$

${\int {f\left( u \right)du} = F\left( u \right) + C.}$

Assuming that $$u = u\left( x \right)$$ is a differentiable function and using the chain rule, we have

${\frac{d}{{dx}}F\left( {u\left( x \right)} \right) }={ F^\prime\left( {u\left( x \right)} \right)u^\prime\left( x \right) }={ f\left( {u\left( x \right)} \right)u^\prime\left( x \right).}$

Integrating both sides gives

${\int {f\left( {u\left( x \right)} \right)u^\prime\left( x \right)dx} }={ F\left( {u\left( x \right)} \right) + C.}$

Hence

${\int {{f\left( {u\left( x \right)} \right)}{u^\prime\left( x \right)}dx} }={ \int {f\left( u \right)du},\;\;}\kern0pt{\text{where}\;\;{u = u\left( x \right)}.}$

This is the substitution rule formula for indefinite integrals.

Note that the integral on the left is expressed in terms of the variable $$x.$$ The integral on the right is in terms of $$u.$$

The substitution method (also called $$u-$$substitution) is used when an integral contains some function and its derivative. In this case, we can set $$u$$ equal to the function and rewrite the integral in terms of the new variable $$u.$$ This makes the integral easier to solve.

Do not forget to express the final answer in terms of the original variable $$x!$$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Compute the integral $$\int {{e^{\large{\frac{x}{2}}\normalsize}}dx}.$$

### Example 2

Find the integral $$\int {{{\left( {3x + 2} \right)}^5}dx}.$$

### Example 3

Find the integral $$\int {\large{\frac{{dx}}{{\sqrt {1 + 4x} }}}\normalsize}.$$

### Example 4

Evaluate the integral $$\int {\large{\frac{{xdx}}{{\sqrt {1 + {x^2}} }}}\normalsize}.$$

### Example 5

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

### Example 6

Evaluate the integral $${\int {{\large{\frac{{{x^2}}}{{{x^3} + 1}}}\normalsize}dx} }$$ using an appropriate substitution.

### Example 7

Find the integral $$\int {\sqrt[3]{{1 – 3x}}dx}.$$

### Example 8

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

### Example 9

Compute the integral $$\int {\large{\frac{{xdx}}{{1 + {x^4}}}}\normalsize}.$$

### Example 10

Evaluate the integral $$\int {\large{\frac{{xdx}}{{{x^4} + 2{x^2} + 1}}}\normalsize}.$$

### Example 11

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

### Example 12

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

### Example 13

Evaluate the integral $$\int {\large{\frac{{\sin x}}{{1 – \cos x}}}\normalsize dx}.$$

### Example 14

Evaluate the integral $$\int {x\sqrt {x + 1} dx}.$$

### Example 15

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

### Example 16

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

### Example 1.

Compute the integral $$\int {{e^{\large{\frac{x}{2}}\normalsize}}dx}.$$

Solution.

Let $$u = \large{\frac{x}{2}}\normalsize.$$ Then

${du = \frac{{dx}}{2},}\;\; \Rightarrow {dx = 2du.}$

So now we can easily integrate:

${\int {{e^{\frac{x}{2}}}dx} = \int {{e^u} \cdot 2du} }={ 2\int {{e^u}du} }={ 2{e^u} + C }={ 2{e^{\frac{x}{2}}} + C.}$

### Example 2.

Find the integral $$\int {{{\left( {3x + 2} \right)}^5}dx}.$$

Solution.

We make the substitution $$u = 3x + 2.$$ Then

$du = d\left( {3x + 2} \right) = 3dx.$

So the differential $$dx$$ is given by

$dx = \frac{{du}}{3}.$

Plug all this in the integral:

${\int {{{\left( {3x + 2} \right)}^5}dx} = \int {{u^5}\frac{{du}}{3}} }={ \frac{1}{3}\int {{u^5}du} }={ \frac{1}{3} \cdot \frac{{{u^6}}}{6} + C }={ \frac{{{u^6}}}{{18}} + C }={ \frac{{{{\left( {3x + 2} \right)}^6}}}{{18}} + C.}$

### Example 3.

Find the integral $$\int {\large{\frac{{dx}}{{\sqrt {1 + 4x} }}}\normalsize}.$$

Solution.

We can try to use the substitution $$u = 1 + 4x.$$ Hence

$du = d\left( {1 + 4x} \right) = 4dx,$

so

$dx = \frac{{du}}{4}.$

This yields

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

### Example 4.

Evaluate the integral $$\int {\large{\frac{{xdx}}{{\sqrt {1 + {x^2}} }}}\normalsize}.$$

Solution.

Let $$u = 1 + {x^2}.$$ Then

$du = d\left( {1 + {x^2}} \right) = 2xdx.$

We see that

$xdx = \frac{{du}}{2}.$

Hence

${\int {\frac{{xdx}}{{\sqrt {1 + {x^2}} }}} }={ \int {\frac{{\frac{{du}}{2}}}{{\sqrt u }}} }={ \int {\frac{{du}}{{2\sqrt u }}} }={ \sqrt u + C }={ \sqrt {1 + {x^2}} + C.}$

### Example 5.

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 6.

Evaluate the integral $${\int {{\large{\frac{{{x^2}}}{{{x^3} + 1}}}\normalsize}dx} }$$ using an appropriate substitution.

Solution.

We try the substitution $$u = {x^3} + 1.$$

Calculate the differential $$du:$$

${du = d\left( {{x^3} + 1} \right) = 3{x^2}dx.}$

We see from the last expression that

${{x^2}dx = \frac{{du}}{3},}$

so we can rewrite the integral in terms of the new variable $$u:$$

${I = \int {\frac{{{x^2}}}{{{x^3} + 1}}dx} }={ \int {\frac{{\frac{{du}}{3}}}{u}} }={ \int {\frac{{du}}{{3u}}} .}$

Now we can easily evaluate this integral:

${I = \int {\frac{{du}}{{3u}}} }={ \frac{1}{3}\int {\frac{{du}}{u}} }={{\frac{1}{3}\ln \left| u \right|} + C.}$

Express the result in terms of the variable $$x:$$

${I = \frac{1}{3}\ln \left| u \right| + C }={{ \frac{1}{3}\ln \left| {{x^3} + 1} \right| + C}}.$

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Problems 1-6
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Problems 7-16