Methods of integration Change of variable

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Sometimes an unfamiliar-looking integral can be made into a familiar one by substituting a new variable for an old one. For example, we know how to integrate 1/x — the answer is $\textrm{In} x$ — but what about

$\int \frac{dx}{2x+1}$

Let u=2x+1 . Differentiating both sides, we have du=2dx , or dx=du/2 ,so

$\begin{align*} \int \frac{dx}{2x+1} &= \int \frac{du/2}{u}\\ &=\frac{1}{2}\textrm{In } u+c \\ &= \frac{1}{2}\textrm{In } (2x+1)+c \end{align*}$

This technique is known as a change of variable or a substitution. (Because the letter u is of- ten employed, you may also see it called -substitution.)

In the case of a definite integral, we have to remember to change the limits of integration to reflect the new variable.

Example

Evaluate $\int_{3}^{4}dx/(2x+1)$ .

As before, let u=2x+1 .

$\begin{align*} \int_{x=3}^{x=4}\frac{dx}{2x+1} &=\int_{u=7}^{u=9}\frac{du/2}{u} \\ &= \frac{1}{2}\textrm{In }u\mid ^{u=9}_{u=7} \end{align*}$

Here the notation $\mid ^{u=9}_{u=7}$ means to evaluate the function at 7 and 9, and sub- tract the former from the latter. The result is

$\begin{align*} \int_{x=3}^{x=4}\frac{dx}{2x+1} &=\frac{1}{2}(\textrm{In }9- \textrm{In }7) \\ &= \frac{1}{2}\textrm{In}\frac{9}{7} \end{align*}$

Sometimes, as in the next example, a clever substitution is the secret to doing a seemingly impossible integral.

Example

Evaluate

$\int \frac{e\sqrt{x}}{\sqrt{x}}dx$

The only hope for reducing this to a form we can do is to let $u=\sqrt{x}$ . Then dx=d(u^2)=2udu , so

$\begin{align*} \int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx &= \int \frac{e^u}{u}\cdot 2udu\\ &= 2\int e^udu\\ &= 2e^u\\ &= 2e^{\sqrt{x}} \end{align*}$

Example really isn't so tricky, since there was only one logical choice for the substitution that had any hope of working. The following is a little more dastardly.

Example

Evaluate

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

The substitution that works is $x=\textrm{ tan }u$ . First let’s see what this does to the expression 1+x^2 . The familiar identity

$\textrm{sin}^2u+\textrm{cos}^2u=1$

when divided by cos^2 u , gives

$\textrm{tan}^2u+1=\textrm{sec}^2u$

so 1+x^2 becomes $\textrm{sec}^2u$ . But differentiating both sides of $x=\textrm{tan }u$ gives

$\begin{align*} dx &=d[\textrm{sin } u(\textrm{cos }u)^{-1}] \\ &=(\textrm{dsin} u)(\textrm{cos }u)^{-1}+(\textrm{sin }u)d\left [ (\textrm{cos }u)^{-1} \right ] \\ &= (1+\textrm{tan}^2u)du\\ &= \textrm{sec}^2u du \end{align*}$

so the integral becomes

$\begin{align*} \int \frac{dx}{1+x^2} &=\int \frac{\textrm{sec}^2u du}{\textrm{sec}^2 u} \\ &= u+c\\ &= \textrm{tan}^{-1}x+c \end{align*}$

Integrate(x) 1/(1+x^2)
ArcTan(x)

Another possible answer is that you can usually smell the possibility of this type of substitution, involving a trig function, when the thing to be integrated contains something reminiscent of the Pythagorean theorem, as suggested by Figure 5.2. The 1+x^2 looks like what you’d get if you had a right triangle with legs 1 and , and were using the Pythagorean theorem to find its hypotenuse.

Figure 5.2 The substitution x = tan u.

Example

Evaluate $\int dx/\sqrt{1-x^2}$

The $\sqrt{1-x^2}$ looks like what you’d get if you had a right triangle with hypotenuse 1 and a leg of length , and were using the Pythagorean theorem to find the other leg, as in Figure 5.3. This motivates us to try the substitution $x= \textrm{cos }u$ , which gives $dx=-\textrm{sin } u du$ and $\sqrt{1-x^2}=\sqrt{1-\textrm{cos}^2 u}=\textrm{sin }u$ . The result is