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3

## How do you integrate ln(x+sqrt(x^2+1)) ?

George C.
Featured 1 month ago

$\int \textcolor{w h i t e}{.} \ln \left(x + \sqrt{{x}^{2} + 1}\right) \textcolor{w h i t e}{.} \mathrm{dx} = x {\sinh}^{- 1} x - \sqrt{{x}^{2} + 1} + C$

$\textcolor{w h i t e}{\int \ln \left(x + \sqrt{{x}^{2} + 1}\right) \mathrm{dx}} = x \ln \left(x + \sqrt{{x}^{2} + 1}\right) - \sqrt{{x}^{2} + 1} + C$

#### Explanation:

Let us try a hyperbolic substitution.

Let $x = \sinh \theta$

Then:

$\int \textcolor{w h i t e}{.} \ln \left(x + \sqrt{{x}^{2} + 1}\right) \textcolor{w h i t e}{.} \mathrm{dx} = \int \textcolor{w h i t e}{.} \ln \left(\sinh \theta + \sqrt{{\sinh}^{2} \theta + 1}\right) \frac{\mathrm{dx}}{d \theta} d \theta$

$\textcolor{w h i t e}{\int \textcolor{w h i t e}{.} \ln \left(x + \sqrt{{x}^{2} + 1}\right) \textcolor{w h i t e}{.} \mathrm{dx}} = \int \textcolor{w h i t e}{.} \ln \left(\sinh \theta + \cosh \theta\right) \cosh \theta \textcolor{w h i t e}{.} d \theta$

$\textcolor{w h i t e}{\int \textcolor{w h i t e}{.} \ln \left(x + \sqrt{{x}^{2} + 1}\right) \textcolor{w h i t e}{.} \mathrm{dx}} = \int \textcolor{w h i t e}{.} \ln \left({e}^{\theta}\right) \cosh \theta \textcolor{w h i t e}{.} d \theta$

$\textcolor{w h i t e}{\int \textcolor{w h i t e}{.} \ln \left(x + \sqrt{{x}^{2} + 1}\right) \textcolor{w h i t e}{.} \mathrm{dx}} = \int \textcolor{w h i t e}{.} \theta \cosh \theta \textcolor{w h i t e}{.} d \theta$

$\textcolor{w h i t e}{\int \textcolor{w h i t e}{.} \ln \left(x + \sqrt{{x}^{2} + 1}\right) \textcolor{w h i t e}{.} \mathrm{dx}} = \theta \sinh \theta - \cosh \theta + C$

$\textcolor{w h i t e}{\int \textcolor{w h i t e}{.} \ln \left(x + \sqrt{{x}^{2} + 1}\right) \textcolor{w h i t e}{.} \mathrm{dx}} = x {\sinh}^{- 1} x - \sqrt{{x}^{2} + 1} + C$

If we would prefer not to have an inverse hyperbolic function in the answer, let's find another expression for it:

Let:

$y = {\sinh}^{- 1} x$

Then:

$x = \sinh y = \frac{1}{2} \left({e}^{y} - {e}^{- y}\right)$

Hence:

${e}^{y} - 2 x - {e}^{- y} = 0$

So:

${\left({e}^{y}\right)}^{2} - \left(2 x\right) \left({e}^{y}\right) - 1 = 0$

${e}^{y} = \frac{2 x \pm \sqrt{{\left(2 x\right)}^{2} + 4}}{2}$

$\textcolor{w h i t e}{{e}^{y}} = x \pm \sqrt{{x}^{2} + 1}$

If $y$ is real then ${e}^{y} > 0$ and we need the $+$ sign here.

So:

${e}^{y} = x + \sqrt{{x}^{2} + 1}$

and:

$y = \ln \left(x + \sqrt{{x}^{2} + 1}\right)$

That is:

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

So:

$\int \textcolor{w h i t e}{.} \ln \left(x + \sqrt{{x}^{2} + 1}\right) \textcolor{w h i t e}{.} \mathrm{dx} = x \ln \left(x + \sqrt{{x}^{2} + 1}\right) - \sqrt{{x}^{2} + 1} + C$

1

## How do you sketch the graph f(x)=2x^3-12x^2+18x-1?

HSBC244
Featured 1 month ago

Step $1$: Determine the first derivative

$f ' \left(x\right) = 6 {x}^{2} - 24 x + 18$

Step $2$: Determine the critical numbers

These will occur when the derivative equals $0$.

$0 = 6 {x}^{2} - 24 x + 18$

$0 = 6 \left({x}^{2} - 4 x + 3\right)$

$0 = \left(x - 3\right) \left(x - 1\right)$

$x = 3 \mathmr{and} 1$

Step $3$: Determine the intervals of increase/decrease

We select test points.

Test point 1: $x = 0$

$f ' \left(0\right) = 6 {\left(0\right)}^{2} - 24 \left(0\right) + 18 = 18$

Since this is positive, the function is uniformly increasing on $\left(- \infty , 1\right)$.

Test point 2: $x = 2$

$f ' \left(2\right) = 6 {\left(2\right)}^{2} - 24 \left(2\right) + 18 = - 6$

Since this is negative, the function is decreasing on $\left(1 , 3\right)$.

I won't select a test point for $\left(3 , \infty\right)$ because I know the function is increasing on the interval. The point $x = 3$ is referred to as a turning point because it goes from decreasing to increasing or vice versa.

Step $4$: Determine the second derivative

This is the derivative of the first derivative.

$f ' ' \left(x\right) = 12 x - 24$

Step $5$: Determine the points of inflection

These will occur when $f ' ' \left(x\right) = 0$.

$0 = 12 x - 24$

$0 = 12 \left(x - 2\right)$

$x = 2$

Step $6$: Determine the intervals of concavity

Once again, we select test points.

Test point $1$: $x = 1$

$f ' ' \left(1\right) = 12 \left(1\right) - 24 = - 12$

This means that $f \left(x\right)$ concave down (Since it's negative) on$\left(- \infty , 2\right)$.

This also means that $f \left(x\right)$ is concave up on $\left(2 , \infty\right)$.

Step $7$: Determine the x/y- intercept

$f \left(x\right)$ doesn't have any rational factors, so we'll forget about the x-intercepts (as a result, you would need Newton's Method or a similar method to find the x-intercepts).

The y-intercept is

$f \left(0\right) = 2 {\left(0\right)}^{3} - 12 {\left(0\right)}^{2} + 18 x - 1 = - 1$

If you can't connect the graph, you could always make a table of values. In the end, you should get a graph similar to the following. graph{2x^3- 12x^2 + 18x - 1 [-10, 10, -5, 5]}

Hopefully this helps!

2

## If u_n = int (sin nx)/sinx dx, >= 2, prove that u_n = (2sin(n-1)x)/(n-1)+u_(n-2) Hence evaluate: int_0^(pi/2) (sin5x)/ sinx dx?

Ratnaker Mehta
Featured 1 month ago

For the ${2}^{n d}$ Proof, refer to the Explanation.

#### Explanation:

Here is a Second Method to prove the Result for

$n \ge 2 , n \in \mathbb{N} .$

${u}_{n} = \int \sin \frac{n x}{\sin} x \mathrm{dx} .$

Now, $\sin \frac{n x}{\sin} x = \frac{1}{\sin} x \left\{\sin \left(\left(n x - 2 x\right) + 2 x\right)\right\}$

Knowing that, $\sin \left(A + B\right) = \sin A \cos B + \cos A \sin B ,$ we get,

$\sin \frac{n x}{\sin} x = \frac{1}{\sin} x \left\{\sin \left(n x - 2 x\right) \cos 2 x + \cos \left(n x - 2 x\right) \sin 2 x\right\}$

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

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

$= \sin \frac{\left(n - 2\right) x}{\sin} x - 2 \sin x \sin \left(\left(n - 2\right) x\right) + 2 \cos x \cos \left(\left(n - 2\right) x\right)$

$= \sin \frac{\left(n - 2\right) x}{\sin} x + 2 \left\{\cos x \cos \left(\left(n - 2\right) x\right) - \sin x \sin \left(\left(n - 2\right) x\right)\right\}$

$= \sin \frac{\left(n - 2\right) x}{\sin} x + 2 \left\{\cos \left(\left(n - 2\right) x + x\right)\right\}$

$\therefore \sin \frac{n x}{\sin} x = \sin \frac{\left(n - 2\right) x}{\sin} x + 2 \cos \left(\left(n - 1\right) x\right) .$

$\Rightarrow {u}_{n} = \int \sin \frac{n x}{\sin} x \mathrm{dx} = \int \left\{\sin \frac{\left(n - 2\right) x}{\sin} x + 2 \cos \left(\left(n - 1\right) x\right)\right\} \mathrm{dx}$

$= \int \sin \frac{\left(n - 2\right) x}{\sin} x \mathrm{dx} + 2 \int \cos \left(\left(n - 1\right) x\right) \mathrm{dx} .$

"Hence, "u_n=u_(n-2)+(2sin((n-1)x))/(n-1)+C; n in NN, n>=2.

Rest of the Soln. is the same as in the First Method.

Enjoy Maths.!

1

## A closed rectangular box is to have a width equal to the length and a surface area of 600 in3. What is the height of the box that maximizes the volume?

HSBC244
Featured 1 month ago

A height of $10$ inches would maximize the volume. See below for details.

#### Explanation:

Let $l$ be the length, $w$ be the width and $h$ be the height. However, since $l = w$, we can narrow ourselves down to two variables, let them be $l$ and $h$.

The formula for surface area of a rectangular prism is generally:

$S . A = 2 l h + 2 l w + 2 h w$

In this case, it becomes

$600 = 2 l h + 2 {l}^{2} + 2 l h$

$600 = 4 l h + 2 {l}^{2}$

We want to solve for one of the variables now.

$600 - 2 {l}^{2} = 4 l h$

$\frac{600 - 2 {l}^{2}}{4 l} = h$

$\frac{300 - {l}^{2}}{2 l} = h$

The volume of a rectangular prism is given by:

$V = l \times w \times h$

Which in our case becomes:

$V = {l}^{2} \times h$

Now:

$V = {l}^{2} \frac{300 - {l}^{2}}{2 l}$

$V = \frac{1}{2} l \left(300 - {l}^{2}\right)$

$V = \frac{1}{2} \left(300 l - {l}^{3}\right)$

$V = 150 l - \frac{1}{2} {l}^{3}$

Now find the derivative of this with respect to volume.

$V ' = 150 - \frac{3}{2} {l}^{2}$

Find the critical values. These will occur when the derivative equals $0$.

$0 = 150 - \frac{3}{2} {l}^{2}$

$\frac{3}{2} {l}^{2} = 150$

${l}^{2} = 100$

$l = \pm 10$

A negative side length doesn't make sense, so we don't accept it as a solution. We now verify graphically to insure this is a maximum.

And it is!

We realize the y-coordinate of the maximum is $y = 1000$, which is the maximum volume possible. All that is left for us to do is find the height that produces the maximum volume.

$V = {l}^{2} \cdot h$

$1000 = {\left(10\right)}^{2} \cdot h$

$1000 = 100 h$

$h = 10$

Hopefully this helps!

3

## How do you find the equations for the tangent plane to the surface z=x^2-2xy+y^2 through (1,2,1)?

Steve
Featured 3 weeks ago

$- 2 x + 2 y - z = 1$

#### Explanation:

First we rearrange the equation of the surface into the form $f \left(x , y , z\right) = 0$

$z = {x}^{2} - 2 x y + {y}^{2}$
$\therefore {x}^{2} - 2 x y + {y}^{2} - z = 0$

And so we define our surface function, $f$, by:

$f \left(x , y , z\right) = {x}^{2} - 2 x y + {y}^{2} - z$

In order to find the normal at any particular point in vector space we use the Del, or gradient operator:

$\nabla f \left(x , y , z\right) = \frac{\partial f}{\partial x} \hat{i} + \frac{\partial f}{\partial y} \hat{j} + \frac{\partial f}{\partial z} \hat{k}$

remember when partially differentiating that we differentiate wrt the variable in question whilst treating the other variables as constant. And so:

$\nabla f = \left(\frac{\partial}{\partial x} \left({x}^{2} - 2 x y + {y}^{2} - z\right)\right) \hat{i} +$
$\text{ } \left(\frac{\partial}{\partial y} \left({x}^{2} - 2 x y + {y}^{2} - z\right)\right) \hat{j} +$
$\text{ } \left(\frac{\partial}{\partial z} \left({x}^{2} - 2 x y + {y}^{2} - z\right)\right) \hat{k}$
$\text{ } = \left(2 x - 2 y\right) \hat{i} + \left(- 2 x + 2 y\right) \hat{j} + \left(- 1\right) \hat{k}$

So for the particular point $\left(1 , 2 , 1\right)$ the normal vector to the surface is given by:

$\nabla f \left(1 , 2 , 1\right) = \left(2 - 4\right) \hat{i} + \left(- 2 + 4\right) \hat{j} - \hat{k}$
$\text{ } = - 2 \hat{i} + 2 \hat{j} - \hat{k}$

So the tangent plane to the surface $z = {x}^{2} - 2 x y + {y}^{2}$ has this normal vector and it also passes though the point $\left(1 , 2 , 1\right)$. It will therefore have a vector equation of the form:

$\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$

Where $\vec{r} = \left(\begin{matrix}x \\ y \\ z\end{matrix}\right)$; $\vec{n} = \left(\begin{matrix}- 2 \\ 2 \\ - 1\end{matrix}\right)$, is the normal vector and $a$ is any point in the plane

Hence, the tangent plane equation is:

$\left(\begin{matrix}x \\ y \\ z\end{matrix}\right) \cdot \left(\begin{matrix}- 2 \\ 2 \\ - 1\end{matrix}\right) = \left(\begin{matrix}1 \\ 2 \\ 1\end{matrix}\right) \cdot \left(\begin{matrix}- 2 \\ 2 \\ - 1\end{matrix}\right)$
$\therefore \left(x\right) \left(- 2\right) + \left(y\right) \left(2\right) + \left(z\right) \left(- 1\right) = \left(1\right) \left(- 2\right) + \left(2\right) \left(2\right) + \left(1\right) \left(- 1\right)$
$\therefore - 2 x + 2 y - z = - 2 + 4 - 1$
$\therefore - 2 x + 2 y - z = 1$

We can confirm this graphically: Here is the surface with the normal vector:

and here is the surface with the tangent plane and the normal vector:

1

## How do you find the equations for the tangent plane to the surface z=x^2-2xy+y^2 through (1,2,1)?

Eddie
Featured 3 weeks ago

As an alternative approach, we can parameterise the surface in terms of $\alpha$ and $\beta$ as:

$m a t h b f r \left(\alpha , \beta\right) = \left\langle \alpha , \beta , {\alpha}^{2} - 2 \alpha \beta + {\beta}^{2}\right\rangle$

Noting that to first order :

• $d m a t h b f {r}_{\alpha} = \frac{\partial m a t h b f r}{\partial \alpha} d \alpha = \left\langle 1 , 0 , 2 \left(\alpha - \beta\right)\right\rangle d \alpha$

• $d m a t h b f {r}_{\beta} = \frac{\partial m a t h b f r}{\partial \beta} d \beta = \left\langle 0 , 1 , - 2 \left(\alpha - \beta\right)\right\rangle d \beta$

....we have this:

The shaded area is a vector, $d m a t h b f A$, whose direction is the normal to the surface at that point, and given by the vector cross product :

$d m a t h b f A = \mathrm{dA} \setminus \textcolor{red}{m a t h b f {e}_{n}} = d m a t h b f {r}_{\alpha} \times d m a t h b f {r}_{\beta}$

$= \det \left(\begin{matrix}m a t h b f {e}_{x} & m a t h b f {e}_{y} & m a t h b f {e}_{z} \\ 1 & 0 & 2 \left(\alpha - \beta\right) \\ 0 & 1 & - 2 \left(\alpha - \beta\right)\end{matrix}\right) \mathrm{da} l p h a \mathrm{db} \eta$

$= \left\langle - 2 \left(\alpha - \beta\right) , 2 \left(\alpha - \beta\right) , 1\right\rangle \setminus \mathrm{da} l p h a \setminus \mathrm{db} \eta$

We are not really bothered by the magnitide of $d m a t h b f A$, its the direction (of any normal vector $m a t h b f n$, ie we don't even need to normalise to $m a t h b f {e}_{n}$) that matters, so we can forget about $\setminus \mathrm{da} l p h a \setminus \mathrm{db} \eta$ and evaluate this as one of many normal vectors:

$m a t h b f n = {\left\langle - 2 \left(\alpha - \beta\right) , 2 \left(\alpha - \beta\right) , 1\right\rangle}_{\left(\alpha = 1 , \beta = 2\right)} = \left\langle 2 , - 2 , 1\right\rangle$

Now we know from the scalar dot product that, for a plane surface:

$\left(m a t h b f r - m a t h b f {r}_{o}\right) \cdot m a t h b f n = 0$

$\implies m a t h b f r \cdot m a t h b f n = m a t h b f {r}_{o} \cdot m a t h b f n$

$\implies \left\langle x , y , z\right\rangle \cdot \left\langle 2 , - 2 , 1\right\rangle = \left\langle {x}_{o} , {y}_{o} , {z}_{o}\right\rangle \cdot \left\langle 2 , - 2 , 1\right\rangle$

So we plug in $m a t h b f {r}_{0} = \left\langle 1 , 2 , 1\right\rangle$ and we get:

$2 x - 2 y + z = - 1$

3

## How do you evaluate \int \frac { ( x - 2) } { x ( x ^ { 2} - 4x + 5) ^ { 2} } d x?

Steve
Featured 1 week ago

$\int \setminus \frac{x - 2}{x {\left({x}^{2} - 4 x + 5\right)}^{2}} \setminus \mathrm{dx} = - \frac{2}{25} \ln | x | + \frac{1}{25} \ln | {x}^{2} - 4 x + 5 | - \frac{3}{50} \arctan \left(x - 2\right) + \frac{x - 4}{10 \left({x}^{2} - 4 x + 5\right)} + c$

#### Explanation:

Let us denote the required integral by $I$

$I = \int \setminus \frac{x - 2}{x {\left({x}^{2} - 4 x + 5\right)}^{2}} \setminus \mathrm{dx}$

Partial Fraction Decomposition

We can decompose the integrand using partial fractions, the partial fraction decomposition will be of the form:

$\frac{x - 2}{x {\left({x}^{2} - 4 x + 5\right)}^{2}} = \frac{A}{x} + \frac{B x + C}{{x}^{2} - 4 x + 5} + \frac{D x + E}{{x}^{2} - 4 x + 5} ^ 2$
$\text{ } = \frac{A {\left({x}^{2} - 4 x + 5\right)}^{2} + \left(B x + C\right) x \left({x}^{2} - 4 x + 5\right) + \left(D x + E\right) x}{x {\left({x}^{2} - 4 x + 5\right)}^{2}}$

$x - 2 = A {\left({x}^{2} - 4 x + 5\right)}^{2} + \left(B x + C\right) x \left({x}^{2} - 4 x + 5\right) + \left(D x + E\right) x$

$\therefore x - 2 = A \left({x}^{4} - 8 {x}^{3} + 26 {x}^{2} - 40 x + 25\right) + \left(B x + C\right) \left({x}^{3} - 4 {x}^{2} + 5 x\right) + \left(D x + E\right) x$

We can use various methods to find our unknown constants:

$\text{Put } x = 0 \implies - 2 = A \cdot {5}^{2}$
$\therefore A = - \frac{2}{25}$

Compare coefficients:

$\text{Coeff} \left({x}^{4}\right) \implies 0 = A + B$
$\therefore B = \frac{2}{25}$

$\text{Coeff} \left({x}^{3}\right) \implies 0 = - 8 A + C - 4 B$
$\therefore 0 = \frac{16}{25} + C - \frac{8}{25} \implies C = - \frac{8}{25}$

$\text{Coeff} \left({x}^{2}\right) \implies 0 = 26 A + 5 B - 4 C + D$
$\therefore 0 = - \frac{52}{25} + \frac{10}{25} + \frac{32}{25} + D \implies D = \frac{10}{25} = \frac{2}{5}$

$\text{Coeff} \left({x}^{1}\right) \implies 1 = - 40 A + 5 C + E$
$\therefore 1 = \frac{80}{25} - \frac{40}{25} + E \implies E = - \frac{3}{5}$

Hence we can the integral as:

$I = \int \setminus \frac{- \frac{2}{25}}{x} + \frac{\frac{2}{25} x - \frac{8}{25}}{{x}^{2} - 4 x + 5} + \frac{\frac{2}{5} x - \frac{3}{5}}{{x}^{2} - 4 x + 5} ^ 2 \setminus \mathrm{dx}$

$\setminus \setminus = - \frac{2}{35} \setminus \int \setminus \frac{1}{x} \mathrm{dx} + \frac{2}{25} \int \frac{x - 4}{{x}^{2} - 4 x + 5} \setminus \mathrm{dx} + \frac{1}{5} \int \frac{2 x - 3}{{x}^{2} - 4 x + 5} ^ 2 \setminus \mathrm{dx}$

$\setminus \setminus = - \frac{2}{25} \setminus {I}_{1} + \frac{2}{25} \setminus {I}_{2} + \frac{1}{5} \setminus {I}_{3}$

Where:

${I}_{1} = \int \setminus \frac{1}{x} \mathrm{dx} + \frac{2}{25} \setminus \mathrm{dx}$
${I}_{2} = \int \frac{x - 4}{{x}^{2} - 4 x + 5} \setminus \mathrm{dx}$
${I}_{3} = \int \frac{2 x - 3}{{x}^{2} - 4 x + 5} ^ 2 \setminus \mathrm{dx}$

Now let us take each of these three separate integrals in turn:

Integral 1: ${I}_{1}$

The first integral, ${I}_{1}$, we can just evaluate directly:

${I}_{1} = \int \setminus \frac{1}{x} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \ln | x |$

Integral 2: ${I}_{2}$

The second integral, ${I}_{2}$, we manipulate by completing the square:

${I}_{2} = \int \setminus \frac{x - 4}{{x}^{2} - 4 x + 5} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \int \setminus \frac{x - 4}{{\left(x - 2\right)}^{2} - {2}^{2} + 5} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \int \setminus \frac{x - 4}{{\left(x - 2\right)}^{2} + 1} \setminus \mathrm{dx}$

Let $u = x - 2 \implies \frac{\mathrm{du}}{\mathrm{dx}} = 1$, and $x - 4 = u - 2$, then substituting gives

${I}_{2} = \int \setminus \frac{u - 2}{{u}^{2} + 1} \setminus \mathrm{du}$
$\setminus \setminus \setminus = \int \setminus \frac{u}{{u}^{2} + 1} - \frac{2}{{u}^{2} + 1} \setminus \mathrm{du}$
$\setminus \setminus \setminus = \frac{1}{2} \int \setminus \frac{2 u}{{u}^{2} + 1} \setminus \mathrm{du} - 2 \int \setminus \frac{1}{{u}^{2} + 1} \setminus \mathrm{du}$
$\setminus \setminus \setminus = \frac{1}{2} \ln | {u}^{2} + 1 | - 2 \arctan u$

Restoring the substitution we get:

${I}_{2} = \frac{1}{2} \ln | {x}^{2} - 4 x + 5 | - 2 \arctan \left(x - 2\right)$

Integral 3: ${I}_{3}$

The third integral, ${I}_{3}$ we continue as earlier by completing the square to get:

${I}_{3} = \int \setminus \frac{2 x - 3}{{x}^{2} - 4 x + 5} ^ 2 \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \int \frac{2 x - 3}{{\left(x - 2\right)}^{2} + 1} ^ 2 \setminus \mathrm{dx}$

Let $u = x - 2 \implies \frac{\mathrm{du}}{\mathrm{dx}} = 1$, and $2 x - 3 = 2 u + 1$, then substituting gives

${I}_{3} = \int \frac{2 u + 1}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du}$
$\setminus \setminus \setminus = \int \frac{2 u}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du} + \int \frac{1}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du}$

Consider the integral:

$\int \frac{2 u}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du}$

Let $w = {u}^{2} + 1 \implies \frac{\mathrm{dw}}{\mathrm{du}} = 2 u$, so we can substitute to get:

$\int \frac{2 u}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du} = \int \setminus \frac{1}{w} ^ 2 \setminus \mathrm{dw}$
$\text{ } = - \frac{1}{w}$

Restoring the substitutions we get:

$\int \frac{2 u}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du} = - \frac{1}{{u}^{2} + 1}$
$\text{ } = - \frac{1}{{\left(x - 2\right)}^{2} + 1}$
$\text{ } = - \frac{1}{{x}^{2} - 4 x + 5}$

And now we consider:

$\int \frac{1}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du}$

Let $u = \tan \theta \implies \frac{\mathrm{du}}{d \theta} = {\sec}^{2} \theta$, then substituting gives

$\int \frac{1}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du} = \int \setminus \frac{1}{{\tan}^{2} \theta + 1} ^ 2 \setminus {\sec}^{2} \theta \setminus d \theta$
$\text{ } = \int \setminus \frac{1}{{\sec}^{2} \theta} ^ 2 \setminus {\sec}^{2} \theta \setminus d \theta$
$\text{ } = \int \setminus \frac{1}{{\sec}^{2} \theta} \setminus d \theta$
$\text{ } = \int \setminus {\cos}^{2} \theta \setminus d \theta$
$\text{ } = \int \setminus \frac{1}{2} \left(1 + \cos 2 \theta\right) \setminus d \theta$
$\text{ } = \frac{1}{2} \setminus \int \setminus 1 + \cos 2 \theta \setminus d \theta$
$\text{ } = \frac{1}{2} \left(\theta + \frac{1}{2} \sin 2 \theta\right)$
$\text{ } = \frac{1}{2} \theta + \frac{1}{2} \sin \theta \cos \theta$

Now, $u = \tan \theta \implies u = \sin \frac{\theta}{\cos} \theta$

$\therefore \sin \theta \cos \theta = {\cos}^{2} \theta \cdot u$
$\text{ } = \frac{1}{\sec} ^ 2 \theta \cdot u$
$\text{ } = \frac{1}{1 + {\tan}^{2} \theta} \cdot u$
$\text{ } = \frac{u}{{u}^{2} + 1}$

Restoring the substitutions we get:

$\int \frac{1}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du} = \frac{1}{2} \arctan u + \frac{1}{2} \frac{u}{{u}^{2} + 1}$
$\text{ } = \frac{1}{2} \arctan \left(x - 2\right) + \frac{1}{2} \frac{x - 2}{{\left(x - 2\right)}^{2} + 1}$
$\text{ } = \frac{1}{2} \arctan \left(x - 2\right) + \frac{1}{2} \frac{x - 2}{{x}^{2} - 4 x + 5}$

Hence, we can write the third integral as:

${I}_{3} = \int \frac{2 u}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du} + \int \frac{1}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du}$
$\setminus \setminus \setminus = - \frac{1}{{x}^{2} - 4 x + 5} + \frac{1}{2} \arctan \left(x - 2\right) + \frac{1}{2} \frac{x - 2}{{x}^{2} - 4 x + 5}$

Combine Results

Now we return back to our original partial fraction result from earlier, and replace the three integrals with the above results:

$I = - \frac{2}{25} \setminus {I}_{1} + \frac{2}{25} \setminus {I}_{2} + \frac{1}{5} \setminus {I}_{3}$

$\setminus \setminus = - \frac{2}{25} \left\{\ln | x |\right\} + \frac{2}{25} \left\{\frac{1}{2} \ln | {x}^{2} - 4 x + 5 | - 2 \arctan \left(x - 2\right)\right\} + \frac{1}{5} \left\{- \frac{1}{{x}^{2} - 4 x + 5} + \frac{1}{2} \arctan \left(x - 2\right) + \frac{1}{2} \frac{x - 2}{{x}^{2} - 4 x + 5}\right\} + c$

$\setminus \setminus = - \frac{2}{25} \ln | x | + \frac{1}{25} \ln | {x}^{2} - 4 x + 5 | - \frac{4}{25} \arctan \left(x - 2\right) - \frac{1}{5} \frac{1}{{x}^{2} - 4 x + 5} + \frac{1}{10} \arctan \left(x - 2\right) + \frac{1}{10} \frac{x - 2}{{x}^{2} - 4 x + 5} + c$

$\setminus \setminus = - \frac{2}{25} \ln | x | + \frac{1}{25} \ln | {x}^{2} - 4 x + 5 | - \frac{3}{50} \arctan \left(x - 2\right) - \frac{1}{5} \frac{1}{{x}^{2} - 4 x + 5} + \frac{1}{10} \frac{x - 2}{{x}^{2} - 4 x + 5} + c$

$\setminus \setminus = - \frac{2}{25} \ln | x | + \frac{1}{25} \ln | {x}^{2} - 4 x + 5 | - \frac{3}{50} \arctan \left(x - 2\right) + \frac{x - 2 - 2}{10 \left({x}^{2} - 4 x + 5\right)} + c$

$\setminus \setminus = - \frac{2}{25} \ln | x | + \frac{1}{25} \ln | {x}^{2} - 4 x + 5 | - \frac{3}{50} \arctan \left(x - 2\right) + \frac{x - 4}{10 \left({x}^{2} - 4 x + 5\right)} + c$

2

## What will be the Fourier series of the periodic function that is obtained by passing the voltage v(t) = Vcos{100(pi)(t)} through a half-wave rectifier?

Eddie
Featured 1 week ago

This is one cycle of what you're trying to analyse:

It's the cosine function, with period $P = \frac{1}{50} \text{ s}$, but without the negative bits. So it's an even function requiring a Fourier cosine series:

$S \left\{f \left(t\right)\right\} = {a}_{o} + {\sum}_{n = 1}^{\infty} {a}_{n} \cos \left(\frac{2 n \pi}{P} t\right)$

With: ${a}_{o} = \frac{1}{P} {\int}_{- \frac{P}{2}}^{\frac{P}{2}} f \left(t\right) \setminus \mathrm{dt}$, we narrow the interval as follows:

${a}_{o} = 50 V {\int}_{- \frac{1}{200}}^{\frac{1}{200}} \cos 100 \pi t \setminus \mathrm{dt}$

$= 50 V \cdot \frac{1}{50 \pi} = \frac{V}{\pi}$

For the other term:

${a}_{n} = \frac{2}{P} {\int}_{- \frac{P}{2}}^{\frac{P}{2}} f \left(t\right) \cos \left(\frac{2 n \pi}{P} t\right) \setminus \mathrm{dt}$

$= 100 V {\int}_{- \frac{1}{200}}^{\frac{1}{200}} \cos \left(100 \pi t\right) \cos \left(100 n \pi t\right) \setminus \mathrm{dt}$

Using a sum-product formula:

$= 50 V {\int}_{- \frac{1}{200}}^{\frac{1}{200}} \cos \left(100 \left(n + 1\right) \pi t\right) + \cos \left(100 \left(n - 1\right) \pi t\right) \setminus \mathrm{dt}$

$= 50 V {\left[\frac{\sin \left(100 \left(n + 1\right) \pi t\right)}{100 \left(n + 1\right) \pi} + \frac{\sin \left(100 \left(n - 1\right) \pi t\right)}{100 \left(n - 1\right) \pi}\right]}_{- \frac{1}{200}}^{\frac{1}{200}}$

$= 100 V {\left[\frac{\sin \left(100 \left(n + 1\right) \pi t\right)}{100 \left(n + 1\right) \pi} + \frac{\sin \left(100 \left(n - 1\right) \pi t\right)}{100 \left(n - 1\right) \pi}\right]}_{0}^{\frac{1}{200}}$

$= \frac{V}{\pi} \left(\frac{\sin \left(\left(n + 1\right) \frac{\pi}{2}\right)}{\left(n + 1\right)} + \frac{\sin \left(\left(n - 1\right) \frac{\pi}{2}\right)}{\left(n - 1\right)}\right)$

Generally, both of these terms are zero for odd $n$. For even $n$, they alternate between $- 1$ and $1$. So if we now count $n = 2 , 4 , 6. . .$ in terms of $k = 1 , 2 , 3 , \ldots$, and sub $n = 2 k$, we have:

${a}_{k} = \frac{V}{\pi} \left(\frac{{\left(- 1\right)}^{k}}{\left(2 k + 1\right)} + \frac{- {\left(- 1\right)}^{k}}{\left(2 k - 1\right)}\right)$

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

However, we also need to consider $n = 1$. At that point, we note ${\lim}_{n \to 1} \frac{\sin \left(\left(n - 1\right) \frac{\pi}{2}\right)}{\left(n - 1\right)} = \frac{\pi}{2} \implies {a}_{1} = \frac{V}{2}$

Therefore:

S{f(t)} = V/pi (1 + pi/2 cos (100 pi t) + 2 sum_(k=1)^(oo) (-1)^(k+1) cdot ( cos (200 k pi t))/((4k^2-1)) )

The first 3 terms looks like this:

2

## How do you find the limit of (1−cos(8x))/(1−cos(3x)) as x approaches 0?

Jim H
Featured 4 weeks ago

Use ${\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta} = 1$

#### Explanation:

lim_(xrarr0)(1−cos(8x))/(1−cos(3x)) has initial form $\frac{0}{0}$

Recall (from trigonometry class) that

$\left(1 - \cos \theta\right) \left(1 - \cos \theta\right) = 1 - {\cos}^{2} \theta = {\sin}^{2} \theta$

(1−cos(8x))/(1−cos(3x)) * ((1+cos(8x))(1+cos(3x)))/((1+cos(8x))(1+cos(3x))) = ((1-cos^2(8x))(1+cos(3x)))/((1-cos^2(3x))(1+cos(8x))

 = (sin^2(8x)(1+cos(3x)))/(sin^2(3x)(1+cos(8x))

We can see that ${\lim}_{x \rightarrow 0} \frac{1 + \cos \left(3 x\right)}{1 + \cos \left(8 x\right)} = \frac{2}{2} = 1$,

so we will focus on

lim_(xrarr0)sin^2(8x)/(sin^2(3x)

We'll use the facts that

${\lim}_{x \rightarrow 0} {\sin}^{2} \frac{8 x}{8 x} ^ 2 = 1$ and ${\lim}_{x \rightarrow 0} {\left(3 x\right)}^{2} / {\sin}^{2} \left(3 x\right) = 1$

${\lim}_{x \rightarrow 0} {\sin}^{2} \frac{8 x}{{\sin}^{2} \left(3 x\right)} = {\lim}_{x \rightarrow 0} {\left(8 x\right)}^{2} / {\left(3 x\right)}^{2} \frac{{\sin}^{2} \frac{8 x}{8 x} ^ 2}{{\sin}^{2} \frac{3 x}{3 x} ^ 2}$

$= {\lim}_{x \rightarrow 0} \frac{64}{9} {\left(\sin \frac{8 x}{8 x}\right)}^{2} / {\left(\sin \frac{3 x}{3 x}\right)}^{2}$

$= \frac{64}{9} \cdot \frac{1}{1} = \frac{64}{9}$

2

## Find out the solution of this integration -int 1/(cos(x)+cos(a))dx???

Andrea S.
Featured 2 days ago

$\int \frac{\mathrm{dx}}{\cos x + \cos a} = \frac{1}{\sin} a \ln \left(\cos \frac{\frac{x - a}{2}}{\cos} \left(\frac{x + a}{2}\right)\right) + C$

#### Explanation:

Use the parametric formula:

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

and substitute:

$t = \tan \left(\frac{x}{2}\right)$

$x = 2 \arctan t$

$\mathrm{dx} = \frac{2 \mathrm{dt}}{1 + {t}^{2}}$

we have:

$\int \frac{\mathrm{dx}}{\cos x + \cos a} = 2 \int \frac{\mathrm{dt}}{1 + {t}^{2}} \frac{1}{\frac{1 - {t}^{2}}{1 + {t}^{2}} + \cos a}$

 int dx/(cosx+cosa) = 2 int dt/((1-t^2) +cosa(1+t^2)

$\int \frac{\mathrm{dx}}{\cos x + \cos a} = 2 \int \frac{\mathrm{dt}}{\left(1 + \cos a\right) - {t}^{2} \left(1 - \cos a\right)}$

Factor the denominator:

 int dx/(cosx+cosa) = 2 int dt/(( sqrt(1+cosa) - tsqrt(1-cosa)) ( sqrt(1+cosa) + tsqrt(1-cosa) )

and use partial fractions decomposition:

$\frac{1}{\left(\sqrt{1 + \cos a} - t \sqrt{1 - \cos a}\right) \left(\sqrt{1 + \cos a} + t \sqrt{1 - \cos a}\right)} = \frac{A}{\sqrt{1 + \cos a} - t \sqrt{1 - \cos a}} + \frac{B}{\sqrt{1 + \cos a} + t \sqrt{1 - \cos a}}$

$A \left(\sqrt{1 + \cos a} + t \sqrt{1 - \cos a}\right) + B \left(\sqrt{1 + \cos a} - t \sqrt{1 - \cos a}\right) = 1$

$\left\{\begin{matrix}\left(A + B\right) \sqrt{1 + \cos a} = 1 \\ \left(A - B\right) \sqrt{1 - \cos a} = 0\end{matrix}\right.$

$A = B = \frac{1}{2 \sqrt{1 + \cos a}}$

$\int \frac{\mathrm{dx}}{\cos x + \cos a} = \frac{1}{\sqrt{1 + \cos a}} \int \frac{\mathrm{dt}}{\sqrt{1 + \cos a} - t \sqrt{1 - \cos a}} + \frac{1}{\sqrt{1 + \cos a}} \int \frac{\mathrm{dt}}{\sqrt{1 + \cos a} + t \sqrt{1 - \cos a}}$

Simplify the expression:

 int dx/(cosx+cosa) = int dt/((1+cosa) - tsqrt((1-cosa)(1+cosa)) )+int dt/( (1+cosa) + tsqrt((1-cosa)(1+cosa))

$\int \frac{\mathrm{dx}}{\cos x + \cos a} = \int \frac{\mathrm{dt}}{\left(1 + \cos a\right) - t \sqrt{1 - {\cos}^{2} a}} + \int \frac{\mathrm{dt}}{\left(1 + \cos a\right) + t \sqrt{1 - {\cos}^{2} a}}$

Now:

$\sqrt{1 - {\cos}^{2} a} = \left\mid \sin \right\mid a$

however because of the symmetry of the expression we can ignore the absolute value, as changing $\sin a$ with $- \sin a$ leaves everything unchanged:

$\int \frac{\mathrm{dx}}{\cos x + \cos a} = \int \frac{\mathrm{dt}}{\left(1 + \cos a\right) - t \sin a} + \int \frac{\mathrm{dt}}{\left(1 + \cos a\right) + t \sin a}$

$\int \frac{\mathrm{dx}}{\cos x + \cos a} = \frac{1}{\sin} a \left(\int \frac{\mathrm{dt}}{\frac{1 + \cos a}{\sin} a - t} + \int \frac{\mathrm{dt}}{\frac{1 + \cos a}{\sin} a + t}\right)$

$\int \frac{\mathrm{dx}}{\cos x + \cos a} = \frac{1}{\sin} a \left(- \ln \left(\frac{1 + \cos a}{\sin} a - t\right) + \ln \left(\frac{1 + \cos a}{\sin} a + t\right)\right) + C$

Using now the properties of logarithms:

$\int \frac{\mathrm{dx}}{\cos x + \cos a} = \frac{1}{\sin} a \ln \left(\frac{\frac{1 + \cos a}{\sin} a + t}{\frac{1 + \cos a}{\sin} a - t}\right) + C$

$\int \frac{\mathrm{dx}}{\cos x + \cos a} = \frac{1}{\sin} a \ln \left(\frac{1 + \cos a + t \sin a}{1 + \cos a - t \sin a}\right) + C$

Undoing the substitution:

$\int \frac{\mathrm{dx}}{\cos x + \cos a} = \frac{1}{\sin} a \ln \left(\frac{1 + \cos a + \tan \left(\frac{x}{2}\right) \sin a}{1 + \cos a - \tan \left(\frac{x}{2}\right) \sin a}\right) + C$

Multiply by $\cos \left(\frac{x}{2}\right)$ numerator and denominator of the argument:

$\int \frac{\mathrm{dx}}{\cos x + \cos a} = \frac{1}{\sin} a \ln \left(\frac{\cos \left(\frac{x}{2}\right) + \cos a \cos \left(\frac{x}{2}\right) + \sin \left(\frac{x}{2}\right) \sin a}{\cos \left(\frac{x}{2}\right) + \cos \left(\frac{x}{2}\right) \cos a - \sin \left(\frac{x}{2}\right) \sin a}\right) + C$

$\int \frac{\mathrm{dx}}{\cos x + \cos a} = \frac{1}{\sin} a \ln \left(\frac{\cos \left(\frac{x}{2}\right) + \cos \left(\frac{x}{2} - a\right)}{\cos \left(\frac{x}{2}\right) + \cos \left(\frac{x}{2} + a\right)}\right) + C$

Use now the identity:

$\cos \alpha + \cos \beta = 2 \cos \left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha - \beta}{2}\right)$

$\int \frac{\mathrm{dx}}{\cos x + \cos a} = \frac{1}{\sin} a \ln \left(\frac{2 \cos \left(\frac{x - a}{2}\right) \cos \left(\frac{a}{2}\right)}{2 \cos \left(\frac{x + a}{2}\right) \cos \left(\frac{a}{2}\right)}\right) + C$

$\int \frac{\mathrm{dx}}{\cos x + \cos a} = \frac{1}{\sin} a \ln \left(\cos \frac{\frac{x - a}{2}}{\cos} \left(\frac{x + a}{2}\right)\right) + C$

Further trigonometric semplification is possible, but the answer is getting too long...