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4 years ago

Given

Mathematics

1 answer:

Nadya [2.5K]4 years ago

4 0

Answer:

im sorry idk the answer

Step-by-step explanation:

idk

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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

HELP HELP PLEASE DUE TONIGHT

Serjik [45]

Answer:

Gurl... I GOT CHU!!

Step-by-step explanation:

1. x = -9

2. y = -42

3. x = -15

4. z = 15

5. x = -3

Here ya go!!

7 0

4 years ago

What is the same number like 0.07

Dafna11 [192]

Answer:

we need more information

Step-by-step explanation:

5 0

3 years ago

The time t required to empty a tank varies inversely as

vesna_86 [32]

If $t$ and $r$ are inverse proportional to one another, then for some constant volume $V$ we have

$V = tr$

It takes 45 min to empty a tank containing at 300 gal/min, so the tank contains

$V = (45\,\mathrm{min}) \left(300\dfrac{\rm gal}{\rm min}\right) = 13500\,\mathrm{gal}$

of liquid.

If it's emptied at 500 gal/min, it would take

$13500\,\mathrm{gal} = t \left(500\dfrac{\rm gal}{\rm min}\right) \implies t = \boxed{27\,\mathrm{min}}$

4 0

2 years ago

Order the fraction to least to greatest 7\9 13\18 5\6 2\3

Olegator [25]

First, you need to have a common denominator for all of them, in this case, they can all go in 18 so that will be the common denominator.

$\frac{7}{9} x \frac{2}{2} = \frac{14}{18}$
$\frac{5}{6} x \frac{3}{3} = \frac{15}{18}$
$\frac{2}{3} x \frac{6}{6} = \frac{12}{18}$
$\frac{13}{18}$

Now you can tell which ones are the biggest or smallest. So from least to greatest you would have this.

$\frac{2}{3} , \frac{13}{18} , \frac{7}{9} , \frac{5}{6}$

7 0

3 years ago