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## Definite integral #int_0^(pi/8)cos^2(2x)dx# ?

Ratnaker Mehta
Featured 3 months ago

$\frac{1}{16} \left(\pi + 2\right)$.

#### Explanation:

We have, $1 + \cos 2 \theta = 2 {\cos}^{2} \theta$.

$\therefore {\int}_{0}^{\frac{\pi}{8}} {\cos}^{2} 2 x \mathrm{dx}$,

$= {\int}_{0}^{\frac{\pi}{8}} \frac{1 + \cos 4 x}{2} \mathrm{dx}$,

$= \frac{1}{2} {\left[x + \frac{1}{4} \sin 4 x\right]}_{9}^{\frac{\pi}{8}}$,

$= \frac{1}{2} \left[\frac{\pi}{8} + \frac{1}{4} \sin \left(4 \cdot \frac{\pi}{8}\right) - 0\right]$,

$= \frac{1}{2} \left(\frac{\pi}{8} + \frac{1}{4} \cdot 1\right)$,

$= \frac{1}{16} \left(\pi + 2\right)$.

## How to expand in Maclaurin series this? #f(x)=int_0^xlog(1-t)/tdt#

Darshan Senthil
Featured 3 months ago

$f \left(x\right) = - \frac{1}{\ln \left(10\right)} \left[x + {x}^{2} / 4 + {x}^{3} / 9 + {x}^{4} / 16 + \ldots + {x}^{n + 1} / {\left(n + 1\right)}^{2}\right]$

Visual: Check out this graph

#### Explanation:

We clearly cannot evaluate this integral as it is using any of the regular integration techniques we've learned. However, since it is a definite integral, we can use a MacLaurin series and do what is called term by term integration.

We'll need to find the MacLaurin series. Since we don't want to find the nth derivative of that function, we'll need to try and fit it into one of the MacLaurin series we already know.

Firstly, we don't like $\log$; we want to make that a $\ln$. To do this, we can simply employ the change of base formula:

$\log \left(x\right) = \ln \frac{x}{\ln} \left(10\right)$

So we have:

${\int}_{0}^{x} \ln \frac{1 - t}{t \ln \left(10\right)} \mathrm{dt}$

Why do we do this? Well, now notice that $\frac{d}{\mathrm{dx}} \ln \left(1 - t\right) = - \frac{1}{1 - t}$ Why is this so special? Well, $\frac{1}{1 - x}$ is one of our commonly used MacLaurin series:

$\frac{1}{1 - x} = 1 + x + {x}^{2} + {x}^{3} + \ldots = {\sum}_{n = 0}^{\infty} {x}^{n}$

...for all $x$ on $\left(- 1 , 1\right]$

So, we can use this relationship to our advantage, and replace $\ln \left(1 - t\right)$ with $\int - \frac{1}{1 - t} \mathrm{dt}$, which allows us to replace that $\ln$ term with a MacLaurin series. Putting this together gives:

$\ln \frac{1 - t}{t \ln \left(10\right)} = - \frac{1}{t \ln \left(10\right)} \int \left[1 + t + {t}^{2} + {t}^{3} + \ldots + {t}^{n}\right] \mathrm{dt}$

Evaluating the integral:
$= - \frac{1}{t \ln \left(10\right)} \left[t + {t}^{2} / 2 + {t}^{3} / 3 + {t}^{4} / 4 + \ldots + {t}^{n + 1} / \left(n + 1\right)\right]$

Cancelling out the $t$ term in the denominator:
$= - \frac{1}{\ln \left(10\right)} \left[1 + \frac{t}{2} + {t}^{2} / 3 + {t}^{3} / 4 + \ldots + {t}^{n} / \left(n + 1\right)\right]$

And now, we take the definite integral we began the problem with:
${\int}_{0}^{x} \left(- \frac{1}{\ln \left(10\right)} \left[1 + \frac{t}{2} + {t}^{2} / 3 + {t}^{3} / 4 + \ldots + {t}^{n} / \left(n + 1\right)\right]\right) \mathrm{dt}$

Note: Observe how we now do not need to worry about dividing by zero in this problem, which is an issue we'd have had in the original integrand due to the $t$ term in the denominator. Since this was cancelled in the previous step, it shows that the discontinuity is removable, which works well for us.

$= - \frac{1}{\ln \left(10\right)} \left[t + {t}^{2} / 4 + {t}^{3} / 9 + {t}^{4} / 16 + \ldots + {t}^{n + 1} / {\left(n + 1\right)}^{2}\right]$ evaluated from $0$ to $x$

$= - \frac{1}{\ln \left(10\right)} \left[x + {x}^{2} / 4 + {x}^{3} / 9 + {x}^{4} / 16 + \ldots + {x}^{n + 1} / {\left(n + 1\right)}^{2}\right] - 0$

$= - \frac{1}{\ln \left(10\right)} \left[x + {x}^{2} / 4 + {x}^{3} / 9 + {x}^{4} / 16 + \ldots + {x}^{n + 1} / {\left(n + 1\right)}^{2}\right]$

Make sure you realize, though, that this series is only good on the interval $\left(1 , 1\right]$, since the MacLaurin series we used above is only convergent on this interval. Check out this graph I made to get a better idea of what this looks like.

Hope that helped :)

## Find the mass of an object which is in the form of a cuboid#[0,1]×[2,4]×[1,3]#.The density at any point #(x,y,z)# on the cuboid is given by #delta(x,y,z)=x^2+y^2+z^2# ?

Cesareo R.
Featured 3 months ago

$56$

#### Explanation:

The mass is obtained by calculating

${\int}_{1}^{3} \left({\int}_{2}^{4} \left({\int}_{0}^{1} \delta \left(x , y , z\right) \mathrm{dx}\right) \mathrm{dy}\right) \mathrm{dz} =$
$= {\int}_{1}^{3} \left({\int}_{2}^{4} \left({\int}_{0}^{1} \left({x}^{2} + {y}^{2} + {z}^{2}\right) \mathrm{dx}\right) \mathrm{dy}\right) \mathrm{dz} =$
$= {\int}_{1}^{3} \left({\int}_{2}^{4} {\left(\frac{1}{3} {x}^{3} + {y}^{2} x + {z}^{2} x\right)}_{x = 0}^{x = 1} \mathrm{dy}\right) \mathrm{dz} =$
$= {\int}_{1}^{3} \left({\int}_{2}^{4} \left(\frac{1}{3} + {y}^{2} + {z}^{2}\right) \mathrm{dy}\right) \mathrm{dz} =$
$= {\int}_{1}^{3} {\left(\frac{1}{3} y + \frac{1}{3} {y}^{3} + {z}^{2} y\right)}_{y = 2}^{y = 4} \mathrm{dz} =$
$= {\int}_{1}^{3} \left(\frac{4 - 2}{3} + \frac{1}{3} \left({4}^{3} - {2}^{3}\right) + \left(4 - 2\right) {z}^{2}\right) \mathrm{dz} = 56$

## Calculate #lim_(x ->x_o) (x^2- x_o x)/(x^2-x_o^2)# for every value of #x_o in RR# ?

Andrea S.
Featured 3 months ago

${\lim}_{x \to {x}_{0}} \frac{{x}^{2} - {x}_{0} x}{{x}^{2} - {x}_{0}^{2}} = \left\{\begin{matrix}\frac{1}{2} \text{ for " x_0 !=0 \\ 1 " for } {x}_{0} = 0\end{matrix}\right.$

#### Explanation:

Simplify the function:

$\frac{{x}^{2} - {x}_{0} x}{{x}^{2} - {x}_{0}^{2}} = \frac{x \left(x - {x}_{0}\right)}{\left(x + {x}_{0}\right) \left(x - {x}_{0}\right)} = \frac{x}{x + {x}_{0}}$

so for ${x}_{0} \ne 0$:

${\lim}_{x \to {x}_{0}} \frac{{x}^{2} - {x}_{0} x}{{x}^{2} - {x}_{0}^{2}} = {\lim}_{x \to {x}_{0}} \frac{x}{x + {x}_{0}} = {x}_{0} / \left({x}_{0} + {x}_{0}\right) = \frac{1}{2}$

while for ${x}_{0} = 0$:

${\lim}_{x \to {x}_{0}} \frac{{x}^{2} - {x}_{0} x}{{x}^{2} - {x}_{0}^{2}} = {\lim}_{x \to 0} {x}^{2} / {x}^{2} = 1$

## What is the value of #lim_(x->0) (1-cos(mx))/(xsinx)#?

Ratnaker Mehta
Featured 2 months ago

${m}^{2} / 2$.

#### Explanation:

$\text{The Reqd. Lim.=} {\lim}_{x \to 0} \frac{1 - \cos m x}{x \sin x}$,

$= {\lim}_{x \to 0} \frac{2 {\sin}^{2} \left(\frac{m x}{2}\right)}{x \sin x}$,

$= \lim 2 {\left\{\sin \frac{\frac{m x}{2}}{\frac{m x}{2}} \cdot \left(\frac{m x}{2}\right)\right\}}^{2} \div \left\{x \left(\frac{\sin x}{x} \cdot x\right)\right\}$,

$= \lim 2 {\left\{\sin \frac{\frac{m x}{2}}{\frac{m x}{2}}\right\}}^{2} \cdot \frac{{m}^{2} {x}^{2}}{4} \div \left\{{x}^{2} \left(\sin \frac{x}{x}\right)\right\}$,

$= \left(2 \cdot {m}^{2} / 4\right) \cdot {\lim}_{\frac{m x}{2} \to 0} {\left\{\sin \frac{\frac{m x}{2}}{\frac{m x}{2}}\right\}}^{2} \div {\lim}_{x \to 0} \left\{\left(\sin \frac{x}{x}\right)\right\}$,

$= {m}^{2} / 2 \cdot {\left(1\right)}^{2} \div 1$.

$\Rightarrow \text{The Reqd. Lim.=} {m}^{2} / 2$.

## Solve: Lim #x->1# #(sqrt(x)-sqrt(2-x^2))/(2x-sqrt(2+2x^2)# ?

Cesareo R.
Featured 1 month ago

$\frac{3}{2}$

#### Explanation:

#lim_(x->1)(sqrt(x)-sqrt(2-x^2))/(2x-sqrt(2+2x^2)#

#(sqrt[x] - sqrt[2 - x^2])/(2 x - sqrt[2 + 2 x^2]) *(sqrt[x] + sqrt[ 2 - x^2])/(2 x + sqrt[2 + 2 x^2])#

= $\frac{x - 2 + {x}^{2}}{4 {x}^{2} - 2 - 2 {x}^{2}} = \frac{\left(x + 2\right) \left(x - 1\right)}{2 \left(x + 1\right) \left(x - 1\right)}$

= $\frac{1}{2} \left(\frac{x + 2}{x + 1}\right)$ and then

$\frac{\sqrt{x} - \sqrt{2 - {x}^{2}}}{2 x - \sqrt{2 + 2 {x}^{2}}} = \frac{1}{2} \left(\frac{x + 2}{x + 1}\right) \frac{2 x + \sqrt{2 + 2 {x}^{2}}}{\sqrt{x} + \sqrt{2 - {x}^{2}}}$

and then

${\lim}_{x \to 1} \frac{\sqrt{x} - \sqrt{2 - {x}^{2}}}{2 x - \sqrt{2 + 2 {x}^{2}}} = {\lim}_{x \to 1} \frac{1}{2} \left(\frac{x + 2}{x + 1}\right) \frac{2 x + \sqrt{2 + 2 {x}^{2}}}{\sqrt{x} + \sqrt{2 - {x}^{2}}}$

= $\frac{1}{2} \cdot \frac{3}{2} \cdot \frac{4}{2} = \frac{3}{2}$

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