Working with substitution

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We explore more difficult problems involving substitution.

We begin by restating the substitution formula.

Integral Substitution Formula If $u$ is differentiable on the interval $[a,b]$ and $f$ is differentiable on the interval $[u(a),u(b)]$ , then $\int _a^b f'(u(x)) u'(x) \d x =\int _{u(a)}^{u(b)} f'(u) \d u.$

We spend pretty much this entire section working out examples.

Compute: $\int _{5}^{8}\sqrt {x-4} \d x$

Let $u =\answer [given]{x-4},$ computing $\d u$ , we find $\d u =\answer [given]{1}\d x$

Now

$\begin{align*} \int_{5}^{8}\sqrt{x-4} \d x &= \int_{u(5)}^{u(8)} \sqrt{u}\d u \\ &= \int_{\answer[given]{1}}^{\answer[given]{4}} \answer[given]{\sqrt{u}} \d u\\ &= \eval{\answer[given]{\frac{2}{3}}u^{3/2}}_{\answer[given]{1}}^{\answer[given]{4}}\\ &= \frac{2}{3}\left(\answer[given]{4}^{3/2}-\answer[given]{1}^{3/2}\right)\\ &= \frac{2}{3}(\answer[given]{8}-\answer[given]{1})\\ &= \answer[given]{14/3}. \end{align*}$

The next example requires a new technique.

Compute: $\int x(3+x)^{14}\d x$

Here it is not apparent that the chain rule is involved. However, if it was involved, perhaps a good guess for $u$ would be $u = \answer [given]{3+x}$ and then we can write $\begin{align*} \int x(3+x)^{14}\d x &= \int(u-3)(u)^{14}\d u\\ &= \int u^{\answer[given]{15}}-3u^{\answer[given]{14}}\d u \end{align*}$

Now we can use the power rule.

$\begin{align*} \int x(3+x)^{14}\d x &= \answer[given]{1/(16)}u^{\answer[given]{16}}\\ &-\answer[given]{3/(15)}u^{\answer[given]{15}}\\ &= \frac{1}{16}(\answer[given]{3+x})^{\answer[given]{16}}\\ &-\frac{1}{5}(\answer[given]{3+x})^{\answer[given]{15}}+C. \end{align*}$

Compute: $\int \frac {\sin (\sqrt {x})}{\sqrt {x}}\d x$

What should we let $u$ be? $u = \answer [given]{\sqrt {x}}$

So we get the derivative of $u$ is $\d u = \answer [given]{\frac {1}{2\sqrt {x}}} \Rightarrow \answer [given]{2}du \\$ $\phantom {\d u = \frac {1}{2\sqrt {x}} \Rightarrow 2 du} =\answer [given]{\frac {1}{\sqrt {x}}} \d x$

The integral resulting after substitution is

$\begin{align*} \int \frac{\sin(\sqrt{x})}{\sqrt{x}}\d x &= \int\answer[given]{2}\sin(\answer[given]{u})\d u\\ &= \answer[given]{-2}\cos(u)\\ &=-2\cos(\answer[given]{\sqrt{x}})+C \end{align*}$