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2 years ago

Integral rational trigonometric

Mathematics

1 answer:

abruzzese [7]2 years ago

8 0

Substitute x = 3 - 2 cos(θ) and dx = 2 sin(θ) dθ (where "sin" = "sen"). So we have

∫ sin(θ) / (3 - 2 cos(θ)) dθ = 1/2 ∫ 1/x dx

= 1/2 ln|x| + C

= 1/2 ln(3 - 2 cos(θ)) + C

(We can remove the absolute value because -1 ≤ cos(θ) ≤ 1, so 1 ≤ 3 - 2 cos(θ) ≤ 5, and |x| = x when x ≥ 0.)

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Sketch the region that is common to the graphs of x ≥ 2,

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a. Find the graph of their common region in the attachment

b. The area of the common region of the graphs is 8 units²

<h3>a. How to sketch the region common to the graphs?</h3>

Since we have x ≥ 2, y ≥ 0, and x + y ≤ 6, we plot each graph separately and find their region of intersection.

The graph of x ≥ 2 is the region right of the line x = 2.

The graph of y ≥ 0 is the region above the line y = 0 or x-axis.

To plot the graph of x + y ≤ 6, we first plot the graph of x + y = 6 ⇒ y = -x + 6. Then the graph of x + y ≤ 6 is the graph of y ≤ - x + 6.

So, the graph of x + y ≤ 6 is the region below the line y = - x + 6

From the graph, the regions intersect at (2, 0), (2, 4) and (6, 0)

Find the graph of their common region in the attachment

<h3>b. The area of the common region</h3>

From the graph, we see that the common region is a right angled triangle with

height = 4 units and
base = 4 units

So, its area = 1/2 × height × base

= 1/2 × 4 units × 4 units

= 1/2 × 16 units²

= 8 units²

So, the area of the common region of the graphs is 8 units²

Learn more about region common to graphs here:

brainly.com/question/27932405

#SPJ1

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Integral rational trigonometric ​

Integral rational trigonometric