# Integration by Substitution

U-substitution

Tip: This isn't the place to ask a question because the teacher can't reply.

## Key Questions

• This key question hasn't been answered yet.
• Integration by substitution (often called "u-substitution") is essentially the reverse of the chain rule just like integration itself is the inverse function of differentiation.

For example, consider this problem:

Find the derivative of $f \left(x\right)$:

$f \left(x\right) = \sqrt{3 {x}^{2} + 2 x - 5}$

The chain rule says that we begin with the "outside function" which is:

$f \left(u\right) = {u}^{\frac{1}{2}}$ where $u = 3 {x}^{2} + 2 x - 5$

$f ' \left(u\right) = \left(\frac{1}{2}\right) {u}^{- \frac{1}{2}}$, but we now must multiply by $u ' = \left(6 x + 2\right)$
giving us the final answer (after substituting for $u$):

f'(x)=(6x+2)/(2sqrt(3x^2+2x-5)

Now consider the integration problem, find the indefinite integral:

$\int \frac{6 x + 2}{2 \sqrt{3 {x}^{2} + 2 x - 5}} \mathrm{dx}$

We recognize this as the exact reverse of the previous problem.

We are being asked to integrate: $\int \frac{\mathrm{du}}{2 \sqrt{u}}$ = $\int \left(\frac{1}{2}\right) {u}^{- \frac{1}{2}} \mathrm{du}$

where $u = 3 {x}^{2} + 2 x - 5$ and $\mathrm{du} = \left(6 x + 2\right) \mathrm{dx}$ which is already present in the problem.

Clearly, the solution is $f \left(x\right) = \sqrt{3 {x}^{2} + 2 x - 5}$.

The key to making this work, and something I have glossed over here by selecting a simplified example where the exact value of $\mathrm{du}$ was already present, is that it only works if we have $\mathrm{du}$ in the integral.

For example, it is not possible to use this method on:

$\int \frac{1}{\sqrt{3 {x}^{2} + 2 x - 5}} \mathrm{dx}$

because the differential,

$\mathrm{du} = 3 {x}^{2} + 2 x - 5$,

does not appear in the problem. This is probably the most common mistake students make when trying to apply integration by $u$-substitution.

• $\sin \left({e}^{x}\right) + c$, 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

• This expression can be integrated using u = substitution where
$u = {x}^{3} + 1$ and du = 3x^2dx^.

We take the $\int \left({x}^{2} \sqrt{{x}^{3} + 1}\right) \mathrm{dx}$ and let $u = \left({x}^{3} + 1\right)$.
This makes $\mathrm{du} = 3 {x}^{2} \mathrm{dx}$. We now see that we are missing a constant of 3 on the ${x}^{2}$ term. We can multiply the term by 3 provided that we multiply the entire expression by $\frac{1}{3}$ so that the expression balances. We now see that the sqrt is enclosing the term ${x}^{3} + 1$. A sqrt is equivalent to the power $\frac{1}{2}$. We take the antiderivative this exponent which is $\frac{3}{2}$ giving us the final answer $\frac{1}{3} {\left({x}^{3} + 1\right)}^{\frac{3}{2}}$.

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