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3 years ago

True or false help please pre cal

Mathematics

1 answer:

geniusboy [140]3 years ago

4 0

Answer:

True

Step-by-step explanation:

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<img src="https://tex.z-dn.net/?f=%20%20%5Cdisplaystyle%20%5Cint%20%5Climits_%7B0%7D%5E%7B%20%5Cfrac%7B%20%5Cpi%7D%7B2%7D%20%7D%

murzikaleks [220]

Let $x = \arcsin(y)$ , so that

$\sin(x) = y$

$\tan(x)=\dfrac y{\sqrt{1-y^2}}$

$dx = \dfrac{dy}{\sqrt{1-y^2}}$

Then the integral transforms to

$\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_{y=\sin(0)}^{y=\sin\left(\frac\pi2\right)} \frac{y}{\sqrt{1-y^2}} \ln(y) \frac{dy}{\sqrt{1-y^2}}$

$\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy$

Integrate by parts, taking

$u = \ln(y) \implies du = \dfrac{dy}y$

$dv = \dfrac{y}{1-y^2} \, dy \implies v = -\dfrac12 \ln|1-y^2|$

For 0 < y < 1, we have |1 - y²| = 1 - y², so

$\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = uv \bigg|_{y\to0^+}^{y\to1^-} + \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy$

It's easy to show that uv approaches 0 as y approaches either 0 or 1, so we just have

$\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy$

Recall the Taylor series for ln(1 + y),

$\displaystyle \ln(1+y) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n y^n$

Replacing y with -y² gives the Taylor series

$\displaystyle \ln(1-y^2) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n (-y^2)^n = - \sum_{n=1}^\infty \frac1n y^{2n}$

and replacing ln(1 - y²) in the integral with its series representation gives

$\displaystyle -\frac12 \int_0^1 \frac1y \sum_{n=1}^\infty \frac{y^{2n}}n \, dy = -\frac12 \int_0^1 \sum_{n=1}^\infty \frac{y^{2n-1}}n \, dy$

Interchanging the integral and sum (see Fubini's theorem) gives

$\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy$

Compute the integral:

$\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy = -\frac12 \sum_{n=1}^\infty \frac{y^{2n}}{2n^2} \bigg|_0^1 = -\frac14 \sum_{n=1}^\infty \frac1{n^2}$

and we recognize the famous sum (see Basel's problem),

$\displaystyle \sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}6$

So, the value of our integral is

$\displaystyle \int_0^{\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \boxed{-\frac{\pi^2}{24}}$

6 0

3 years ago

If a substance is transported across a cell membrane from an area of low concentration to an area of high concentration, you kno

Ronch [10]

It is a type of active transport because it will require energy to move against the concentration gradient (low to high)
- option A

6 0

3 years ago

Read 2 more answers

Ar if and Keisha go to a restaurant for dinner.Their meals total $13.75 the tax is 5% how much tax is added to the bill?

Jlenok [28]

0.6875 if i am wrong i am sorry

8 0

3 years ago

Rolando has $120,000 in a CD at Big Bucks Bank, which just failed. If the FDIC insurance limit per depositor, per bank, is $250,

damaskus [11]

Answer:

120,000

Step-by-step explanation:

Rolando would get back 120,000. Because the insurance limit is 250k, Rolando only put in 120k, so he should get everything back.

8 0

3 years ago

PLEASE HELP ON 5,6,8,9<br> DUE TOMORROW

Elis [28]

5. You have correctly shown the decimal equivalents of the numbers in selection C. Those are in order. The decimal equivalents for selection D would be

... -1.33, -2.00, -1.00, -0.08, -0.07

The first of these is greater than the second of these, so this list is not in order. The correct choice is D.

6. √169 = 13, the only rational number in the lot. The correct choice is C.

8. All the numbers except the ones with radicals are rational numbers, so the appropriate choice is the one that lists the radicals only: G.

9. You have correctly plotted the points. When you connect them to make a geometric figure, you get a triangle that is 6 units high and 6 units wide. Its area will be half that of a square that is 6×6, so is 18 square units.

_____

The formula for the area of a triangle is

... A = (1/2)bh

where b is the base length of the triangle (6 units), and h is the height (6 units). Filling in the numbers you have, this is

... A = (1/2)·6·6 = 18 . . . . square units.

When you have a plot like this on a graph, it is usually pretty easy to see that the triangle area is half the area of a rectangle (or square) with the same height and width.

6 0

3 years ago