# When would you use u substitution twice?

Jul 1, 2015

When we are reversing a differentiation that had the composition of three functions. Here is one example.

#### Explanation:

$\int {\sin}^{4} \left(7 x\right) \cos \left(7 x\right) \mathrm{dx}$

Let $u = 7 x$. This makes $\mathrm{du} = 7 \mathrm{dx}$ and our integral can be rewritten:

$\frac{1}{7} \int {\sin}^{4} u \cos u \mathrm{du} = \frac{1}{7} \int {\left(\sin u\right)}^{4} \cos u \mathrm{du}$

To avoid using $u$ to mean two different things in one discussion, we'll use another variable ($t , v , w$ are all popular choices)

Let $w = \sin u$, so we have $\mathrm{dw} = \cos u \mathrm{du}$ and our integral becomes:

$\frac{1}{7} \int {w}^{4} \mathrm{dw}$

We the integrate and back-substitute:

$\frac{1}{7} \int {w}^{4} \mathrm{dw} = \frac{1}{35} {w}^{5} + C$

$= \frac{1}{35} {\sin}^{5} u + C$

$= \frac{1}{35} {\sin}^{5} 7 x + C$

If we check the answer by differentiating, we'll use the chain rule twice.

$\frac{d}{\mathrm{dx}} \left({\left(\sin \left(7 x\right)\right)}^{5}\right) = 5 {\left(\sin \left(7 x\right)\right)}^{4} \cdot \frac{d}{\mathrm{dx}} \left(\sin \left(7 x\right)\right)$

$= 5 {\left(\sin \left(7 x\right)\right)}^{4} \cdot \cos \left(7 x\right) \frac{d}{\mathrm{dx}} \left(7 x\right)$

$= 5 {\left(\sin \left(7 x\right)\right)}^{4} \cdot \cos \left(7 x\right) \cdot 7$