Integration of Circular Functions Revision Notes for VCE SSCE Mathematical Methods

Integration of Circular Functions

Introduction to integrating circular functions

When we differentiate circular functions like sine and cosine, we get predictable results. This knowledge allows us to work backwards and find antiderivatives of these functions through integration.

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The Fundamental Relationship from Differentiation:

The key to understanding integration of circular functions lies in knowing their derivatives:

If $f(x) = \sin(kx + a)$ , then $f'(x) = k \cos(kx + a)$
If $g(x) = \cos(kx + a)$ , then $g'(x) = -k \sin(kx + a)$

By reversing this process, we can determine the integral formulas for circular functions.

Key integration formulas

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Essential Integration Formulas for Circular Functions:

\int \sin(kx + a) \, dx = -\frac{1}{k} \cos(kx + a) + c

\int \cos(kx + a) \, dx = \frac{1}{k} \sin(kx + a) + c

Understanding the formulas:

The coefficient $k$ represents the frequency of the circular function
The constant $a$ represents a phase shift (horizontal translation)
We must divide by k because of the chain rule working in reverse
The constant $c$ represents the constant of integration for indefinite integrals
Notice the negative sign when integrating sine (this comes from the derivative of cosine being negative sine)

Worked example: finding antiderivatives

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Worked Example: Finding Antiderivatives of Circular Functions

Part a: Find an antiderivative of $\sin\left(3x + \frac{\pi}{4}\right)$

Using the formula for integrating sine:

\int \sin(kx + a) \, dx = -\frac{1}{k} \cos(kx + a) + c

Here, $k = 3$ and $a = \frac{\pi}{4}$

Therefore:

\int \sin\left(3x + \frac{\pi}{4}\right) \, dx = -\frac{1}{3} \cos\left(3x + \frac{\pi}{4}\right) + c

Part b: Find an antiderivative of $\frac{1}{4} \sin(4x)$

We can take the constant factor outside the integral:

\int \frac{1}{4} \sin(4x) \, dx = \frac{1}{4} \int \sin(4x) \, dx

Here, $k = 4$ and $a = 0$

Applying the formula:

Worked example: evaluating definite integrals

When evaluating definite integrals, we use the antiderivative and substitute the limits of integration.

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Worked Example: Evaluating Definite Integrals with Exact Values

Part a: Find the exact value of $\int_{0}^{\frac{\pi}{4}} \sin(2x) \, dx$

First, find the antiderivative:

\int \sin(2x) \, dx = -\frac{1}{2} \cos(2x)

Now evaluate at the limits:

\int_{0}^{\frac{\pi}{4}} \sin(2x) \, dx = \left[-\frac{1}{2} \cos(2x)\right]_{0}^{\frac{\pi}{4}}

Part b: Find the exact value of $\int_{0}^{\frac{\pi}{2}} (2 \cos x + 1) \, dx$

Split the integral into two parts:

\int_{0}^{\frac{\pi}{2}} (2 \cos x + 1) \, dx = \int_{0}^{\frac{\pi}{2}} 2 \cos x \, dx + \int_{0}^{\frac{\pi}{2}} 1 \, dx

The antiderivative is:

2 \sin x + x

Evaluating at the limits:

Worked example: finding areas under curves

Finding areas under circular function curves requires careful attention to regions above and below the x-axis.

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Worked Example: Finding Areas Under Circular Function Curves

Part a: Find the exact area under one complete arch of $y = \sin\left(\frac{1}{2}x\right)$ from $x = 0$ to $x = 2\pi$

The graph shows a complete arch of the sine curve above the x-axis. Since the entire region is above the x-axis, we can integrate directly:

\text{Area} = \int_{0}^{2\pi} \sin\left(\frac{1}{2}x\right) \, dx

Therefore, the area of the shaded region is 4 square units.

Part b: Find the total area for regions $A_1$ and $A_2$ of $y = \sin\left(\theta - \frac{\pi}{4}\right)$

This graph shows two regions:

Region $A_1$ : above the θ-axis from $\frac{\pi}{4}$ to $\frac{5\pi}{4}$
Region $A_2$ : below the θ-axis from $\frac{5\pi}{4}$ to $\frac{9\pi}{4}$

Since region $A_2$ is below the axis, we must treat it separately.

Finding Area $A_1$ :

\text{Area } A_1 = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \sin\left(\theta - \frac{\pi}{4}\right) \, d\theta

Finding Area $A_2$ :

Since this region is below the axis, the integral gives a negative value. We take the negative of the integral to get the positive area:

\text{Area } A_2 = -\int_{\frac{5\pi}{4}}^{\frac{9\pi}{4}} \sin\left(\theta - \frac{\pi}{4}\right) \, d\theta

Therefore, the total area of the shaded region is 4 square units.

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Critical Note About Areas Below the Axis:

When finding areas, always check whether the curve lies above or below the x-axis over your interval of integration. For regions below the axis, you must either split the integral or use the absolute value to ensure a positive area result.

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Key Integration Formulas:

\int \sin(kx + a) \, dx = -\frac{1}{k} \cos(kx + a) + c

\int \cos(kx + a) \, dx = \frac{1}{k} \sin(kx + a) + c

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Key Points to Remember:

When integrating $\sin(kx + a)$ , the result is $-\frac{1}{k}\cos(kx + a) + c$ (notice the negative sign and division by k)
When integrating $\cos(kx + a)$ , the result is $\frac{1}{k}\sin(kx + a) + c$ (division by k)
Always divide by the coefficient k when integrating circular functions
For area calculations, be careful with regions below the x-axis—these require separate treatment to ensure positive area values
When evaluating definite integrals, use exact values for common angles like $\frac{\pi}{4}$ , $\frac{\pi}{3}$ , $\frac{\pi}{2}$ , etc.

Integration of Circular Functions (VCE SSCE Mathematical Methods): Revision Notes