Substitution (Change of Variable) Rule

Example 1 . Calculate $$$\int{x}{{e}}^{{{{x}}^{{2}}}}{d}{x}$$$.

At first glance, it may appear that this integral is impossible to solve, but the correct substitution will reduce this integral to an easy one.

So, first of all, note that $$${\left({{x}}^{{2}}\right)}'={2}{x}$$$ and there is $$${x}$$$ near the exponent, so let's try the substitution $$${u}={{x}}^{{2}}$$$.

Then, $$${d}{u}={\left({{x}}^{{2}}\right)}'{d}{x}={2}{x}{d}{x}$$$, or $$${x}{d}{x}=\frac{{{d}{u}}}{{2}}$$$.

So,

$$$\int{x}{{e}}^{{{{x}}^{{2}}}}{d}{x}=\int{{e}}^{{u}}\frac{{{d}{u}}}{{2}}=\frac{{1}}{{2}}\int{{e}}^{{u}}{d}{u}=\frac{{1}}{{2}}{{e}}^{{u}}+{C}$$$.

Returning to the old variables gives that

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

Now, just for checking, find the derivative of the result: $$${\left(\frac{{1}}{{2}}{{e}}^{{{{x}}^{{2}}}}\right)}'=\frac{{1}}{{2}}{{e}}^{{{{x}}^{{2}}}}\cdot{\left({{x}}^{{2}}\right)}'={x}{{e}}^{{{{x}}^{{2}}}}$$$ (notice the use of the chain rule).

Example 2 . Calculate $$${\int_{{2}}^{{3}}}{x}{{e}}^{{{{x}}^{{2}}}}{d}{x}$$$.

As was shown above, $$$\int{x}{{e}}^{{{{x}}^{{2}}}}{d}{x}=\frac{{1}}{{2}}{{e}}^{{{{x}}^{{2}}}}+{C}$$$.

So, $$${\int_{{2}}^{{3}}}{x}{{e}}^{{{{x}}^{{2}}}}{d}{x}=\frac{{1}}{{2}}{{e}}^{{{{3}}^{{2}}}}-\frac{{1}}{{2}}{{e}}^{{{{2}}^{{2}}}}=\frac{{1}}{{2}}{\left({{e}}^{{9}}-{{e}}^{{4}}\right)}$$$.

On the other hand, since $$${u}={{x}}^{{2}}$$$ and $$${x}$$$ is changing from $$${2}$$$ to $$${3}$$$, $$${u}$$$ is changing from $$${{2}}^{{2}}={4}$$$ to $$${{3}}^{{2}}={9}$$$.

So,

$$${\int_{{2}}^{{3}}}{x}{{e}}^{{{{x}}^{{2}}}}{d}{x}=\frac{{1}}{{2}}{\int_{{4}}^{{9}}}{{e}}^{{u}}{d}{u}=\frac{{1}}{{2}}{\left({{e}}^{{9}}-{{e}}^{{4}}\right)}$$$.

Example 3 . Find $$${\int_{{0}}^{{3}}}{{e}}^{{{3}{x}}}{d}{x}$$$.

Let $$${u}={3}{x}$$$; then, $$${d}{u}={3}{d}{x}$$$. So, $$${d}{x}=\frac{{{d}{u}}}{{3}}$$$.

$$${x}$$$ is changing from 0 to 3, so $$${u}$$$ is changing from $$${3}\cdot{0}={0}$$$ to $$${3}\cdot{3}={9}$$$.

Therefore, $$${\int_{{0}}^{{3}}}{{e}}^{{{3}{x}}}{d}{x}={\int_{{0}}^{{9}}}{{e}}^{{u}}\frac{{{d}{u}}}{{3}}=\frac{{1}}{{3}}{{e}}^{{u}}{{\mid}_{{0}}^{{9}}}=\frac{{1}}{{3}}{\left({{e}}^{{9}}-{{e}}^{{0}}\right)}=\frac{{1}}{{3}}{\left({{e}}^{{9}}-{1}\right)}$$$.

Example 4. Calculate $$$\int\frac{{1}}{\sqrt{{{1}-{{x}}^{{2}}}}}{d}{x}$$$.

First of all, there is a restriction that $$${\left|{x}\right|}<{1}$$$, otherwise the value under the square root is negative.

Now, let's consider the integral. It is always important to think of how an integral can be reduced to an easy one. In this case, the trigonometric identity $$${{\cos}}^{{2}}{\left({u}\right)}+{{\sin}}^{{2}}{\left({u}\right)}={1}$$$ is very useful.

From this, it follows that $$${\cos{{\left({u}\right)}}}=\sqrt{{{1}-{{\sin}}^{{2}}{\left({u}\right)}}}$$$, which is similar to what is in the denominator.

So, make the substitution $$${x}={\sin{{\left({u}\right)}}}$$$; then, $$${d}{x}={\left({\sin{{\left({u}\right)}}}\right)}'{d}{u}$$$, or $$${d}{x}={\cos{{\left({u}\right)}}}{d}{u}$$$.

Plugging these results into the integral gives the following:

$$$\int\frac{{1}}{\sqrt{{{1}-{{x}}^{{2}}}}}{d}{x}=\int\frac{{1}}{\sqrt{{{1}-{{\sin}}^{{2}}{\left({u}\right)}}}}{\cos{{\left({u}\right)}}}{d}{u}=\int\frac{{1}}{{\cos{{\left({u}\right)}}}}{\cos{{\left({u}\right)}}}{d}{u}=\int{d}{u}={u}+{C}$$$.

Now, since $$${x}={\sin{{\left({u}\right)}}}$$$, we have that $$${u}={\operatorname{asin}{{\left({x}\right)}}}$$$.

So,

$$$\int\frac{{1}}{\sqrt{{{1}-{{x}}^{{2}}}}}{d}{x}={\operatorname{asin}{{\left({x}\right)}}}+{C}$$$, $$${\left|{x}\right|}<{1}$$$.

Example 5. Find $$$\int{\tan{{\left({2}{t}+{9}\right)}}}{d}{t}$$$.

There is only one term under the integral, namely $$${\tan{}}$$$, so it is not very clear what substitution to make. However, recall that $$${\tan{{\left({x}\right)}}}=\frac{{\sin{{\left({x}\right)}}}}{{\cos{{\left({x}\right)}}}}$$$. So, the integral can be rewritten as follows:

$$$\int{\tan{{\left({2}{t}+{9}\right)}}}{d}{t}=\int\frac{{{\sin{{\left({2}{t}+{9}\right)}}}}}{{{\cos{{\left({2}{t}+{9}\right)}}}}}{d}{t}$$$.

As can be seen, the substitution $$${u}={\cos{{\left({2}{t}+{9}\right)}}}$$$ is perfect, because it will remove the numerator and simplify the denominator. In this case, $$${d}{u}={\left({\cos{{\left({2}{t}+{9}\right)}}}\right)}'{d}{t}=-{\sin{{\left({2}{t}+{9}\right)}}}\cdot{\left({2}{t}+{9}\right)}'{d}{t}=-{2}{\sin{{\left({2}{t}+{9}\right)}}}{d}{t}$$$; so, $$${\sin{{\left({2}{t}+{9}\right)}}}{d}{t}=-\frac{{{d}{u}}}{{2}}$$$.

This simplifies the integral to:

$$$\int\frac{{\sin{{\left({2}{t}+{9}\right)}}}}{{\cos{{\left({2}{t}+{9}\right)}}}}{d}{t}=\int\frac{{-\frac{{{d}{u}}}{{2}}}}{{{u}}}=-\frac{{1}}{{2}}\int\frac{{{d}{u}}}{{u}}=-\frac{{1}}{{2}}{\ln{{\left({u}\right)}}}+{C}$$$.

Returning to the old variable gives that

$$$\int{\tan{{\left({2}{t}+{9}\right)}}}=-\frac{{1}}{{2}}{\ln{{\left({\cos{{\left({2}{t}+{9}\right)}}}\right)}}}+{C}$$$.

It is very important to 'read' the integral to define the better substitution.

Example 6. Find $$$\int\frac{{{\cos{{\left({x}\right)}}}}}{{{1}+{{\sin}}^{{2}}{\left({x}\right)}}}{d}{x}$$$.

Let's 'read' this integral: in the denominator, we have something like $$$\int\frac{{1}}{{{1}+{{t}}^{{2}}}}{d}{t}={\operatorname{atan}{{\left({t}\right)}}}$$$.

In the numerator, we have $$${\cos{{\left({x}\right)}}}$$$, which is the derivative of $$${\sin{{\left({x}\right)}}}$$$. Therefore, the substitution $$${t}={\sin{{\left({x}\right)}}}$$$ will work.

So, $$${d}{t}={\cos{{\left({x}\right)}}}{d}{x}$$$, and the integral can be rewritten as $$$\int\frac{{1}}{{{1}+{{t}}^{{2}}}}={\operatorname{atan}{{\left({t}\right)}}}+{C}$$$.

Returning to the old variables gives that $$$\int\frac{{{\cos{{\left({x}\right)}}}}}{{{1}+{{\sin}}^{{2}}{\left({x}\right)}}}{d}{x}={\operatorname{atan}{{\left({\sin{{\left({x}\right)}}}\right)}}}+{C}$$$.

Example 7. Find $$$\int\frac{{{2}{x}}}{{{{x}}^{{2}}+{1}}}{d}{x}$$$.

In the denominator, we have something similar to the derivative of inverse tangent, but the numerator is not the same. However, since $$${\left({{x}}^{{2}}+{1}\right)}'={2}{x}$$$, we can make the substitution $$${u}={{x}}^{{2}}+{1}$$$. In this case, $$${d}{u}={2}{x}{d}{x}$$$, and the integral can be rewritten as follows:

$$$\int\frac{{{d}{u}}}{{u}}={\ln}{\left|{u}\right|}+{C}$$$.

Returning to the old variables gives that $$$\int\frac{{{2}{x}}}{{{{x}}^{{2}}+{1}}}{d}{x}={\ln{{\left({{x}}^{{2}}+{1}\right)}}}+{C}$$$.