MA005: Calculus I

Successful integration is mostly a matter of recognizing patterns. The "change of variable" technique can make some underlying patterns of an integral easier to recognize. Basically, the technique involves rewriting an integral which is originally in terms of one variable, say $x$, in terms of another variable, say $u$. The hope is that after doing the rewriting, it will be easier to find an antiderivative of the new integrand.

For example, $\int \cos (5 x+1) \mathrm{dx}$ can be rewritten by setting $u=5 x+1$. Then $\cos (5 x+1)$ becomes $\cos (u)$ and $\mathbf{d u}=\mathrm{d}(5 x+1)=\mathbf{5} \mathrm{d} \mathrm{x}$ so the differential $dx$ is $\frac{1}{5} \mathrm{du}$. Finally $\int \cos (5 x+1) \mathrm{dx}=\int \cos (u) \cdot \frac{\mathbf{1}}{\mathbf{5}} \mathrm{du} =\frac{1}{5} \sin (u)+C=\frac{1}{5} \sin (5 x+1)+\mathrm{C}$.

The steps of the "change of variable" method can be summarized as

(1) set a new variable, say $u$, equal to some function of the original variable $x$ (usually $u$ is set equal to some part of the original integrand function)

(2) calculate the differential $du$ as a function of $dx$

(3) rewrite the original integral in terms of $u$ and $du$

(4) integrate the new integral to get an answer in terms of $u$

(5) replace the u in the answer to get an answer in terms of the original variable

Changing the Variable

Fig. 1 shows the general flow of this algorithm and another example.

Example 3: Make the suggested change of variable and rewrite each integral in terms of $u$ and $du$.

(a) $\int \cos (x) \mathrm{e}^{\sin (x)} \mathrm{dx} \text { with } u=\sin (x)(\mathrm{b})$

(b) $\int \frac{2 x}{5+x^{2}} \mathrm{dx} \text { with } u=5+x^{2}$

Solution:

(a) Since $u=\sin (x)$, then $\mathrm{du}=\mathrm{d}(\sin (x))=\cos (x) \mathrm{d} \mathrm{x}$ so $\mathrm{e}^{\sin (x)}=\mathrm{e}^{u}$ and $\cos (x) d x=d u$. Then $\int \cos (x) \mathrm{e}^{\sin (x)} \mathrm{dx}=\int \mathrm{e}^{u} \mathrm{du}=\mathrm{e}^{u}+\mathrm{C}=\mathrm{e}^{\sin (x)}+\mathrm{C}$.

(b) If $u=5+x^{2}$, then $\mathbf{d u}=\mathrm{d}\left(5+x^{2}\right)=\mathbf{2} x \mathbf{d} \mathbf{x}$, so $\int \frac{2 x}{5+x^{2}} \mathbf{d x}=\int \frac{1}{u} \mathbf{d u}=\ln |u|+C=\ln 5+x^{2} \mid+C$

In each example, the change of variable did not find the antiderivative, but it did make the pattern of the integrand more obvious so it was easier to determine an antiderivative.

Practice 3: Make the suggested change of variable and rewrite each integral in terms of $u$ and $du$.

(a) $\int(7 x+5)^{3} \mathrm{dx}$ with $u=7 x+5$

(b) $\int \sin \left(x^{3}-1\right) 3 x^{2} \mathrm{dx}$ with $u=x^{3}-1$

In the future, you will need to decide what u should equal, and there are no rules which guarantee that your choice will lead to an easier integral. There is, however, a "rule of thumb" which frequently results in easier integrals. Even though the following suggestion is not guaranteed, it is often worth trying.

A "Rule of thumb" for changing the variable

If part of the integrand is a composition of functions, $\mathrm{f}(\mathrm{g}(x))$,

then try setting $u=g(x)$, the "inner" function.

If part of the integrand is being raised to a power, try setting u equal to the part being raised to the power. For example, if the integrand includes $(3+\sin (x))^{5}$, try $u=3+\sin (x)$. If part of the integrand involves a trigonometric (or exponential or logarithmic) function of another function, try setting $u$ equal to the "inside" function. If the integrand includes the function $\sin \left(3+x^{2}\right), \text { try } u=3+x^{2}$.

The key to becoming skilled at selecting a good u and correctly making the substitution is practice.

Example 4: Select a function for u for each integral and rewrite the integral in terms of $u$ and $du$.

Solution:

(a) Put $u=2+\sin (3 x)$. Then $\mathrm{du}=3 \cos (3 x) \mathrm{dx}$, and the integral becomes $\int \frac{1}{3} \sqrt{u} \mathrm{du}$.

(b) Put $u=2+\mathrm{e}^{x}$. Then $\mathrm{du}=\mathrm{e}^{\boldsymbol{x}} \mathrm{d} \mathbf{x}$, and the integral becomes $\int \frac{5}{u} \mathrm{du}$.

(c) Put $u=\mathrm{e}^{x}$. Then $\mathrm{du}=\mathrm{e}^{\boldsymbol{x}} \mathrm{dx}$, and the integral becomes $\int \sin (u) \mathrm{du}$