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andrezito [222]
3 years ago
12

Csc4 x - cot* x = CSc?x + cot? x​

Mathematics
1 answer:
rjkz [21]3 years ago
8 0

Answer:

Step-by-step explanation:

When approaching the integral,

∫

csc

4

(

x

)

cot

6

(

x

)

d

x

, it is helpful to ask about derivatives and integrals of the various functions we see.

d

d

x

(

csc

x

)

=

−

csc

x

cot

x

so perhaps we could split off one of each and rewrite using only

csc

x

. We know that there is a relationship, but it involves squares, not the 3rd and 5th power we have left after separating

csc

x

cot

x

. We'll keep it in mind if we don't get a better idea.

d

d

x

(

cot

x

)

=

−

csc

2

x

. And if we split off a

csc

2

x

, we will have

csc

2

x

remaining and we know that we can rewrite that using

cot

x

, so we'll try that. (with substitution

u

=

cot

(

x

)

(With experience and practice, this reasoning takes place very fast and we know this will work. As students, we have to try something and see if it works.)

∫

csc

4

(

x

)

cot

6

(

x

)

d

x

=

∫

csc

2

(

x

)

cot

6

(

x

)

csc

2

(

x

)

d

x

=

∫

(

cot

2

(

x

)

+

1

)

cot

6

(

x

)

csc

2

(

x

)

d

x

=

∫

(

cot

8

(

x

)

+

cot

6

(

x

)

)

csc

2

x

d

x

=

∫

(

u

8

+

u

6

)

(

−

d

u

)

(

u

=

cot

(

x

)

)

=

−

1

9

u

9

−

1

7

u

7

+

C

=

−

1

9

cot

9

(

x

)

−

1

7

cot

7

(

x

)

+

C

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PolarNik [594]

Answer:

A) Q(x) = (x + 3)² + 5, and the vertex is (-3, 5)

B) R(x) = (x - 3)² + 2, and the vertex is (3, 2)

C) S(x) = (x - 1)² - 5, and the vertex is (1, -5)

Step-by-step explanation:

The given function is P(x) = (x + 3)² + 2

The given function is a parabolic function in vertex form, f(x) = a·(x - h)² + k, and vertex, (h, k)

By comparison, the vertex of the function P(x) = (x + 3)² + 2 is (-3, 2)

A) A function f(x) translated α units UP gives

f(x) (translated α units UP) → f(x) + α

A translation of the function 3 units UP is given by adding 3 to the given function as follows;

Q(x) = P(x) + 3

∴ Q(x) = (x + 3)² + 2 + 3 = (x + 3)² + 5

Q(x) = (x + 3)² + 5, and the vertex by comparison to f(x) = a·(x - h)² + k, and vertex, (h, k) is (-3, 5)

B) A function f(x) translated <em>b</em> units RIGHT gives;

f(x) translated <em>b</em> units right → f(x - b)

∴ P(x) = (x + 3)² + 2 translated 6 units RIGHT gives;

P(x) = (x + 3)² + 2 (translated 6 units RIGHT) → R(x) = (x + 3 - 6)² + 2 = (x - 3)² + 2

R(x) = (x - 3)² + 2, and the vertex by comparison is (3, 2)

C) A function translated α units DOWN and <em>b</em> units RIGHT is given as follows;

f(x) \ translated \ by\  \dbinom{b}{a} \rightarrow f(x - b) - a

Therefore, the given function, P(x) = (x + 3)² + 2, translated 7 units DOWN and 4 units RIGHT gives;

P(x) = (x + 3)^2 + 5 \ translated \ by\  \dbinom{4}{-7} \rightarrow P(x - 4) - 7 = S(x)

S(x) = P(x - 4) - 7 = (x + 3 - 4)² + 2 - 7 = (x - 1)² - 5

P(x) = (x + 3)^2 + 5 \ translated \ by\  \dbinom{4}{-7} \rightarrow (x - 1)^2 - 5= S(x)

S(x) = (x - 1)² - 5, and the vertex by comparison is (1, -5)

7 0
3 years ago
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