Integral rational trigonometric ​
integral rational trigonometric ​ - 1

Answers

Answer 1
Answer:

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.)


Related Questions

How many times larger is 1 x 106 than 5 x 10-5?
Il give brainliest thanks
Which is bigger 0.159 or 1.590​
For the given function, determine consecutive values of x between which each real zero is located. f(x) = –11x^4 – 5x^3 – 9x^2 + 12x + 10
What is the number if 2 is 12.5% of a number?PLS HELP ME!!!

The sum of two rational numbers is (A: Rational) (B: Irrational)
number.

The sum of a rational number and an irrational number is
an (A:a rational) or (B: a irrational)
number.

~~~ZoomZoom44~~~~

Answers

Answer:

Its B

Step-by-step explanation:

Find the angle between the vectors ????=????+???? and ????=−????+????. (Give an exact answer. Use symbolic notation and fractions where needed.)

Answers

Answer:

The angle between them is 60 degrees

Step-by-step explanation:

Given

a = 2i + j -3k

b = 3i - 2j -k

Required

The angle between them

The cosine of the angle between them is:

\cos(\theta) = (a\cdot b)/(|a|\cdot |b|)

First, calculate a.b

a \cdot b =(2i + j -3k) \cdot (3i - 2j -k)

Multiply the coefficients of like terms

a \cdot b =2 * 3 - 1 * 2 - 3 * -1

a \cdot b =7

Next, calculate |a| and |b|

|a| = \sqrt{2^2 + 1^2 + (-3)^2

|a| = \sqrt{14

|b| = √(3^2 + (-2)^2 + (-1)^2)

|b| = √(14)

Recall that:

\cos(\theta) = (a\cdot b)/(|a|\cdot |b|)

This gives:

\cos(\theta) = (7)/(√(14) * √(14))

\cos(\theta) = (7)/(14)

\cos(\theta) = 0.5

Take arccos of both sides

\theta =\cos^(-1)(0.5)

\theta =60^o

1. Which fraction has a value that's equal to 7/8​

Answers

Answer:

to find the right answer multiple by 2

Step-by-step explanation:

and you get 14/16

The ratio of squirrels to chipmunks in the yard was 2:7. How many squirrels were in the yard if there were 49 chipmunks?Explain please.

Answers

Let the number of squirells be 2x and number of chipmunks be 7x ,



Given that number of chipmunks is 49, so

= > 7x = 49

= > x = 49 / 7

= > x = 7




So the number of squirells will be 2x = 2( 7 ) = 14


\bold{There  \: were  \: 14 \:  squirells  \: in \:  the  \: yard}

College Algebra Half Life ProblemRecently, while digging in Chaco Canyon, New Mexico, archaeologists found corn pollen that was 4000 years old. This was evidence that Native Americans had been cultivating crops in the Southwest centuries earlier than scientists had thought.

What percent of the carbon-14 had been lost from the pollen?

(half-life of carbon-14 = 5730)

Answers

I think the correct answer is 56%

A set X is said to be closed under multiplication if for every X1, X2 E X we have X1X2 E X. Let A be the union of all bounded subsets X CR that are closed under multiplication. Does inf(A) exist? If it does, find it.

Answers

Answer:

inf(A) does not exist.

Step-by-step explanation:

As per the question:

We need to prove that A is closed under multiplication,

If for everyX_(1), X_(2)\in X

X_(1)X_(2)\in X

Proof:

Suppose, x, y \in A

Since, both x and y are real numbers thus xy is also a real number.

Now, consider another set B such that:

B = {xy} has only a single element 'xy' and thus [B] is bounded.

Since, [A] represents the union of all the bounded sets, therefore,

B\subset A

⇒ xy \in A

Therefore, from x, y \]in A, we have xy \]in A.

Hence, set a is closed under multiplication.

Now, to prove whether inf(A) exist or not

Proof:

Let us assume that inf(A) exist and inf(A) = \beta

Thus \beta is also a real number.

Let C be another set such that

C = { \beta - 1}

Now, we know that C is a bounded set thus { \beta - 1} is also an element of A

Also, we know:

inf(A) =  \beta

Therefore,

n(A)\geq \beta

But

\beta - 1 is an element of A and  \beta - 1 \leq \beta

This is contradictory, thus inf(A) does not exist.

Hence, proved.