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3 years ago

Find integral∫cos³x dx) )?

Mathematics

2 answers:

atroni [7]3 years ago

8 0

cos³(x) = cos(x) cos²(x)

… = cos(x) (1 - sin²(x))

Substitute u = sin(x) and du = cos(x) dx. Then in the integral, you get

∫ cos³(x) dx = ∫ (1 - u ²) du

… = u - 1/3 u ³ + C

… = sin(x) - 1/3 sin³(x) + C

son4ous [18]3 years ago

4 0

Answer:

sin(x)−1/3(sin3(x))+C

Step-by-step explanation:

use u substitution method to integrate

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For part A, use the Pythagorean Theorem.

a.) Since we see that the triangle is made up of two right triangles, simply subtract half the leg from the length of the hypotenuse.

10/2 = 5

13² - 5² = 169 - 25 = √144 = 12 centimeters cubed for the height.

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=120 + 660 = 780 cm²

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2. The lengths of three rods are $6 cm, 10 cm, and 11 cm. How can you use these rods to measure a length of 15 cm ?

qwelly [4]

26cm 10cm 11cm

Step-by-step explanation:

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You bought a freezer that is 2 geet wide, 4 feet long and 30 inches deep. What is the capacity or volume of the freezer?

inna [77]

34,560 cubic inches I think,
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3 years ago

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Whats the awnser I need it asap

Dimas [21]

A is the right answer

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3 years ago

Which of the following is an identity? A. sin2x sec2x + 1 = tan2x csc2x B. sin2x - cos2x = 1 C. (cscx + cotx)2 = 1 D. csc2x + co

Ne4ueva [31]

There are three 'Pythagorean' identities that we can look at and they are

sin²(x) + cos²(x) = 1
tan²(x) + 1 = sec²(x)
1 + cot²(x) = csc²(x)

We can start by checking each option to see which one would give us any of the 'Pythagorean' identities as its simplest form

Option A:

sin²(x) sec²(x) + 1 = tan²(x) csc²(x)

Rewriting sec²(x) as 1/cos²(x)
Rewriting tan²(x) as sin²(x)/cos²(x)
Rewriting csc²(x) as 1/sin²(x)

We have

$sin^{2}(x)[ \frac{1}{ cos^{2}(x) }]+1=[ \frac{ sin^{2}( x)}{ cos^{2} (x)}][ \frac{1}{ sin^{2}(x) } ]$
$[\frac{ sin^{2}(x) }{ cos^{2}(x) } ]+1= \frac{1}{ cos^{2}(x) }$
$tan^{2}(x)+1= sec^{2}(x)$

Option B:

sin²(x) - cos²(x) = 1

This expression is already in the simplest form, cannot be simplified further

Option C:

[ csc(x) + cot(x) ]² = 1

Rewriting csc(x) as 1/sin(x)
Rewriting cot(x) as cos(x)/sin(x)

We have

$[ \frac{1}{sin(x)}+ \frac{cos(x)}{sin(x)}] ^{2} =1$
$\frac{1}{sin^2(x)}+2( \frac{1}{sin(x)})( \frac{cos(x)}{sin(x)})+ \frac{cos^2(x)}{sin^2(x)}=1$ $csc^2(x)+2csc^2(x)cos(x)+cot^2(x)=1$

Option D:

csc²(x) + cot²(x) = 1

Rewriting csc²(x) as 1/sin²(x) and cot²(x) as cos²(x)/sin²(x)

$\frac{1}{sin^2(x)}+ \frac{cos^2(x)}{sin^2(x)}=1$
$\frac{1+cos^2(x)}{sin^2(x)} =1$
$1+cos^2(x)=sin^2(x)$
$1=sin^2(x)-cos^2(x)$

from our working out we can see that option A simplified into one of 'Pythagorean' identities, hence the correct answer

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3 years ago

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Find integral∫cos³x dx) )?​

Find integral∫cos³x dx) )?