When would you use u substitution twice?

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When would you use u substitution twice?

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Answer 1 · 2015-07-01T13:05:13

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

Explanation:

$\int {sin}^{4} (7 x) cos (7 x) d x$ $int sin^4(7x)cos(7x)dx$

Let $u = 7 x$ $u=7x$ . This makes $d u = 7 d x$ $du = 7dx$ and our integral can be rewritten:

$\frac{1}{7} \int {sin}^{4} u cos u d u = \frac{1}{7} \int {(sin u)}^{4} cos u d u$ $1/7 int sin^4ucosudu = 1/7int(sinu)^4cosudu$

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

Let $w = sin u$ $w=sinu$ , so we have $d w = cos u d u$ $dw = cosudu$ and our integral becomes:

$\frac{1}{7} \int w^{4} d w$ $1/7intw^4dw$

We the integrate and back-substitute:

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

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

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

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

$\frac{d}{d x} ({(sin (7 x))}^{5}) = 5 {(sin (7 x))}^{4} \cdot \frac{d}{d x} (sin (7 x))$ $d/dx((sin(7x))^5) = 5(sin(7x))^4*d/dx(sin(7x))$

$= 5 {(sin (7 x))}^{4} \cdot cos (7 x) \frac{d}{d x} (7 x)$ $= 5(sin(7x))^4*cos(7x)d/dx(7x)$

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

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When would you use u substitution twice?

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