# Integration by Substitution

Integrals by u-substitution

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1 of 4 videos by turksvids

## Key Questions

• This key question hasn't been answered yet.
• Let $f \left(x\right)$ be defined and continuous in $\left[a , b\right]$ and $g \left(x\right)$ defined and differantiable in $\left[c , d\right]$ with values in $\left[a , b\right]$, such that $g \left(c\right) = a$ and $g \left(d\right) = b$.
Suppose also for simplicity that $g ' \left(x\right) > 0$.

The chain rule states that:

$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) g ' \left(x\right)$

Consider now the definite integral as limit of the Riemann sum:

${\int}_{c}^{d} f ' \left(g \left(x\right)\right) g ' \left(x\right) \mathrm{dx} = {\lim}_{N \to \infty} {\sum}_{n = 0}^{N} f \left(g \left({\xi}_{k}\right)\right) g ' \left({\xi}_{k}\right) \left({x}_{k + 1} - {x}_{k}\right)$

where ${x}_{0} < {x}_{1} < \ldots < {x}_{N} \in \left[c , d\right]$ and ${\xi}_{k} \in \left({x}_{k} , {x}_{k + 1}\right)$.

As the choice of ${\xi}_{k}$ in the interval is arbitrary we can choose them as the points for which, based on Lagrange's theorem:

$g ' \left({\xi}_{k}\right) = \frac{g \left({x}_{k + 1}\right) - g \left({x}_{k}\right)}{{x}_{k + 1} - {x}_{k}}$

so:

${\int}_{c}^{d} f ' \left(g \left(x\right)\right) g ' \left(x\right) \mathrm{dx} = {\lim}_{N \to \infty} {\sum}_{n = 0}^{N} f ' \left(g \left({\xi}_{k}\right)\right) \left(g \left({x}_{k + 1}\right) - g \left({x}_{k}\right)\right)$

Let now: ${y}_{k} = g \left({x}_{k}\right)$. As $g ' \left(x\right) > 0$ the function is strictly increasing so: ${y}_{0} < {y}_{1} < \ldots < {y}_{N}$

Then:

${\int}_{a}^{b} f ' \left(g \left(x\right)\right) g ' \left(x\right) \mathrm{dx} = {\lim}_{N \to \infty} {\sum}_{n = 0}^{N} f ' \left({y}_{k}\right) \left({y}_{k + 1} - {y}_{k}\right)$

but this is a Riemann sum of $f \left(y\right)$ for $y \in \left[a , b\right]$, so:

${\int}_{c}^{d} f ' \left(g \left(x\right)\right) g ' \left(x\right) \mathrm{dx} = {\int}_{a}^{b} f ' \left(y\right) \mathrm{dy} = f \left(b\right) - f \left(a\right)$

• $\sin \left({e}^{x}\right) + c$, where c is a constant

Explanation,

Integration by Substitution,

$\int {e}^{x} \cdot \cos \left({e}^{x}\right) \mathrm{dx}$

let's assume ${e}^{x} = t$

then, ${e}^{x} \cdot \mathrm{dx} = \mathrm{dt}$

$\int \cos \left(t\right) \mathrm{dt}$

$\sin \left(t\right) + c$, where c is a constant

Substituting $t = {e}^{x}$, we finally get

$\sin \left({e}^{x}\right) + c$, where c is a constant

Using integration by substitution, we find the answer to this problem to be $\frac{2}{9} {\left({x}^{3} + 1\right)}^{\frac{3}{2}}$.

#### Explanation:

We need to find a value for $u$ to substitute that will allow us to find this integral more easily. Notice that if we choose our $u$ to be ${x}^{3} + 1$, then our $\mathrm{du}$ (the derivative of our $u$) will be $3 {x}^{2} \mathrm{dx}$. The degree of this answer matches up with what is outside of the radicalâ€”they only differ by a constant, which works for us since we can multiply by a constant as well as we divide by that same constant as well. So with these changes, the integral becomes

$\frac{1}{3} \int 3 {x}^{2} \sqrt{{x}^{3} + 1} \mathrm{dx}$.

Now we can substitute in $u$ for $\sqrt{{x}^{3} + 1}$ and $\mathrm{du}$ for $3 {x}^{2} \mathrm{dx}$ (even though they're on different sides of the integral we can combine them into the $\mathrm{du}$) and we are left with this, in terms of $u$:

$\frac{1}{3} \int \sqrt{u} \mathrm{du}$

Now this is a problem that we can use the anti-power rule on: this integral becomes

$\frac{1}{3} \left[\frac{2}{3} {u}^{\frac{3}{2}}\right] = \frac{2}{9} {u}^{\frac{3}{2}}$.

A very important step that is often overlooked at the end of the problem is that because this is an indefinite integral, we need to substitute back in so that the final answer is in terms of the variable we started with (in this case, $x$). Remember that we substituted $u = {x}^{3} + 1$, so our final answer becomes

$\frac{2}{9} {\left({x}^{3} + 1\right)}^{\frac{3}{2}}$.

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