Math Snap

STEP 1

What is this asking?
We need to calculate a definite integral involving a tricky-looking function using the Fundamental Theorem of Calculus!
Watch out!
Don't forget to adjust the bounds of integration when making a u-substitution!

STEP 2

1. Substitute
2. Integrate
3. Evaluate

STEP 3

Let's tackle this integral using a u-substitution!
We see a $\sqrt[4]{x}$ term inside the sine function and something similar in the denominator.
Let's set $u = \sqrt[4]{x} = x^{\frac{1}{4}}$.

STEP 4

Now, we need to find $\frac{du}{dx}$.
Remember, the derivative of $x^n$ is $nx^{n-1}$.
So, $\frac{du}{dx} = \frac{1}{4}x^{\frac{1}{4} - 1} = \frac{1}{4}x^{-\frac{3}{4}} = \frac{1}{4x^{\frac{3}{4}}}$.

STEP 5

We want to express the integral in terms of $u$, so we need to solve for $dx$.
Multiplying both sides of the previous equation by $4x^{\frac{3}{4}}$ gives us $4x^{\frac{3}{4}}du = dx$.

STEP 6

Let's rewrite our integral:
$$\int_{1}^{16} \frac{\sin (\sqrt[4]{x})}{\sqrt[4]{x^{3}}} dx = \int_{1}^{16} \frac{\sin (x^{\frac{1}{4}})}{x^{\frac{3}{4}}} dx = \int_{1}^{16} \frac{\sin(u)}{x^{\frac{3}{4}}} \cdot 4x^{\frac{3}{4}} du$$

STEP 7

Notice how the $x^{\frac{3}{4}}$ terms divide to one, which simplifies things nicely!
We're left with:
$$\int 4\sin(u) du$$ Don't forget to change the limits of integration!
When $x = 1$, $u = \sqrt[4]{1} = 1$.
When $x = 16$, $u = \sqrt[4]{16} = 2$.
Our integral becomes:
$$\int_{1}^{2} 4\sin(u) du$$

STEP 8

The integral of $\sin(u)$ is $-\cos(u)$.
So, we have:
$$\int_{1}^{2} 4\sin(u) du = 4 \int_{1}^{2} \sin(u) du = 4[-\cos(u)]_{1}^{2}$$

STEP 9

Let's evaluate the expression at the upper and lower limits of integration:
$$4[-\cos(2) - (-\cos(1))] = 4[-\cos(2) + \cos(1)] = 4[\cos(1) - \cos(2)]$$

The value of the definite integral is $4[\cos(1) - \cos(2)]$.