10
Active contributors today

## Can u guys helpe please? Ive been trying to solve two questions and i simply have no luck. They are: Inverse laplace of #s/(s^4+4a^4)# And Inverse laplace of #s/(s^4+s^2+1)# If you guys could gelp me I wpuld be grateful

Steve
Featured 2 months ago

# ℒ^-1 {s/(s^4+4a^4)} \ \ \ \ \ = (sin(at)sinh(at))/(2a^2) #
# ℒ^-1 {s/(s^4+s^2+1)} = 2/sqrt(3)sin(sqrt(3)/2)sinh(1/2t) #

#### Explanation:

Part (A)

First we can complete the square:

$\frac{s}{{s}^{4} + 4 {a}^{4}} = \frac{s}{{\left({s}^{2} + 2 {a}^{2}\right)}^{2} - 4 {a}^{2} {s}^{2}}$
$\text{ } = \frac{s}{{\left({s}^{2} + 2 {a}^{2}\right)}^{2} - {\left(2 a s\right)}^{2}}$

We now have the difference of two squares, so:

$\frac{s}{{s}^{4} + 4 {a}^{4}} = \frac{s}{\left({s}^{2} + 2 {a}^{2} + 2 a s\right) \left({s}^{2} + 2 {a}^{2} - 2 a s\right)}$

Now let us form the partial fraction decomposition for the expression, which will be of the form:

$\frac{s}{{s}^{4} + 4 {a}^{4}} = \frac{A}{{s}^{2} + 2 {a}^{2} + 2 a s} + \frac{B}{{s}^{2} + 2 {a}^{2} - 2 a s}$
$\text{ } = \frac{A \left({s}^{2} + 2 {a}^{2} - 2 a s\right) + B \left({s}^{2} + 2 {a}^{2} + 2 a s\right)}{\left({s}^{2} + 2 {a}^{2} + 2 a s\right) \left({s}^{2} + 2 {a}^{2} - 2 a s\right)}$

$\therefore s = A \left({s}^{2} + 2 {a}^{2} - 2 a s\right) + B \left({s}^{2} + 2 {a}^{2} + 2 a s\right)$

Equating Coefficients we get:

$C o e f f \left(s\right) \setminus \setminus \implies 1 = - 2 a A + 2 a B \setminus \setminus \setminus \setminus \setminus \ldots . . \left[1\right]$
$C o e f f \left({s}^{2}\right) \implies 0 = A + B \setminus \setminus \setminus \setminus \setminus \ldots . . \left[2\right]$

Now we solve these simultaneous equations:

$E q \left[1\right] + \left(2 a\right) E q \left[2\right] \implies 1 = 4 a B$
$\therefore A = - \frac{1}{4 a} , B = \frac{1}{4 a}$

Se we can write the original expression as:

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

And we can now take inverse Laplace Transforms using standard results to get:

# ℒ^-1 {s/(s^4+4a^4)} = 1/(4a){e^(at)sin(at)/a - e^(-at)sin(at)/a} #
$\text{ } = \frac{1}{4 {a}^{2}} \sin \left(a t\right) \left\{{e}^{a t} - {e}^{- a t}\right\}$

$\text{ } = \frac{1}{4 {a}^{2}} \sin \left(a t\right) 2 \sinh \left(a t\right)$

$\text{ } = \frac{\sin \left(a t\right) \sinh \left(a t\right)}{2 {a}^{2}}$

Part (B)
In a similar way to the first solution we want to make the denominator the difference of two squares so that we have two factors leading to two partial fractions. If we complete the square on this denominator the method fails as we get

${s}^{4} + {s}^{2} + 1 = {\left({s}^{2} + \frac{1}{2}\right)}^{2} + \frac{3}{4}$

Which is not the difference of two squares; so let's try a little manipulation:

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

Again we find the partial fraction decomposition; which will be of the form;:

$\frac{s}{{s}^{4} + {s}^{2} + 1} = \frac{A}{{s}^{2} + 1 - s} + \frac{B}{{s}^{2} + 1 + s}$
$\text{ } = \frac{A \left({s}^{2} + 1 + s\right) + B \left({s}^{2} + 1 - s\right)}{\left({s}^{2} + 1 - s\right) \left({s}^{2} + 1 + s\right)}$

$\therefore s = A \left({s}^{2} + 1 + s\right) + B \left({s}^{2} + 1 - s\right)$

Equating Coefficients we get:

$C o e f f \left(s\right) \setminus \setminus \implies 1 = A - B \setminus \setminus \setminus \setminus \setminus \ldots . . \left[1\right]$
$C o e f f \left({s}^{2}\right) \implies 0 = A + B \setminus \setminus \setminus \setminus \setminus \ldots . . \left[2\right]$

Now we solve these simultaneous equations:

$E q \left[1\right] + E q \left[2\right] \implies 1 = 2 A$
$\therefore A = \frac{1}{2} , B = - \frac{1}{2}$

Se we can write the original expression as:

$\frac{s}{{s}^{4} + {s}^{2} + 1} = \frac{1}{2} \left\{\frac{1}{{s}^{2} + 1 - s} - \frac{1}{{s}^{2} + 1 + s}\right\}$
$\text{ } = \frac{1}{2} \left\{\frac{1}{{\left(s - \frac{1}{2}\right)}^{2} - {\left(\frac{1}{2}\right)}^{2} + 1} - \frac{1}{{\left(s + \frac{1}{2}\right)}^{2} - {\left(\frac{1}{2}\right)}^{2} + 1}\right\}$
$\text{ } = \frac{1}{2} \left\{\frac{1}{{\left(s - \frac{1}{2}\right)}^{2} - \frac{3}{4}} - \frac{1}{{\left(s + \frac{1}{2}\right)}^{2} - \frac{3}{4}}\right\}$
$\text{ } = \frac{1}{2} \left\{\frac{1}{{\left(s - \frac{1}{2}\right)}^{2} - {\left(\frac{\sqrt{3}}{2}\right)}^{2}} - \frac{1}{{\left(s + \frac{1}{2}\right)}^{2} - {\left(\frac{\sqrt{3}}{2}\right)}^{2}}\right\}$

And we can now take inverse Laplace Transforms using standard results to get:

# ℒ^-1 {s/(s^4+s^2+1)} = 1/2{e^(1/2t)sin(sqrt(3)/2)/(sqrt(3)/2)) - e^(-1/2t)sin(sqrt(3)/2)/(sqrt(3)/2))} #
$\text{ } = \frac{1}{2} \cdot \frac{2}{\sqrt{3}} \sin \left(\frac{\sqrt{3}}{2}\right) \left\{{e}^{\frac{1}{2} t} - {e}^{- \frac{1}{2} t}\right\}$
$\text{ } = \frac{1}{\sqrt{3}} \sin \left(\frac{\sqrt{3}}{2}\right) 2 \sinh \left(\frac{1}{2} t\right)$
$\text{ } = \frac{2}{\sqrt{3}} \sin \left(\frac{\sqrt{3}}{2}\right) \sinh \left(\frac{1}{2} t\right)$

## Find the area inside #r=1+cos2theta# and outside circle #r=1# ?

Steve
Featured 2 months ago

$2 + \frac{\pi}{4}$

#### Explanation:

Here is the graph of the two curves. The shaded area, $A$, is the area of interest:

It is a symmetrical problems so we only need find the shaded area of the RHS of Quadrant $1$ and multiply by $4$.

We could find the angle $\theta$ in $Q 1$ for the point of intrestion by solving the simultaneous equations:

$r = 1 + \cos 2 \theta$
$r = 1$

However, intuition is faster, and it looks like angle of intersection in $Q 1$ is $\frac{\pi}{4}$, we can verify this by a quick evaluation:

$\theta = \frac{\pi}{4} \implies r = 1 + \cos 2 \theta = 1 + \cos \left(\frac{\pi}{2}\right) = 1$

Confirming our intuition. So now we have the following:

Where the shaded area repents $\frac{1}{4}$ of the total area sought:

We can now start to set up a double integral to calculate this area

$r$ sweeps out a ray from $1$ to $1 + \cos \left(2 \theta\right)$
$\theta$ varies for $0$ to $\frac{\pi}{4}$

So then:

$\frac{1}{4} A = {\int}_{0}^{\frac{\pi}{4}} {\int}_{1}^{1 + \cos 2 \theta} r \setminus \mathrm{dr} \setminus d \theta$

If we evaluate the inner integral first then we get:

${\int}_{1}^{1 + \cos 2 \theta} r \setminus \mathrm{dr} = {\left[\frac{1}{2} {r}^{2}\right]}_{1}^{1 + \cos 2 \theta}$
$\text{ } = \frac{1}{2} \left({\left(1 + \cos 2 \theta\right)}^{2} - {1}^{2}\right)$
$\text{ } = \frac{1}{2} \left(1 + 2 \cos 2 \theta + {\cos}^{2} 2 \theta - 1\right)$
$\text{ } = \frac{1}{2} \left(2 \cos 2 \theta + {\cos}^{2} 2 \theta\right)$
$\text{ } = \frac{1}{2} \left(2 \cos 2 \theta + \frac{1}{2} \left(\cos 4 \theta + 1\right)\right)$
$\text{ } = \cos 2 \theta + \frac{1}{4} \cos 4 \theta + \frac{1}{4}$

And so our double integral becomes:

$\frac{1}{4} A \setminus = {\int}_{0}^{\frac{\pi}{4}} {\int}_{1}^{1 + \cos 2 \theta} r \setminus \mathrm{dr} \setminus d \theta$
$\text{ } = {\int}_{0}^{\frac{\pi}{4}} \cos 2 \theta + \frac{1}{4} \cos 4 \theta + \frac{1}{4} \setminus d \theta$
$\text{ } = {\left[\frac{1}{2} \sin 2 \theta + \frac{1}{16} \sin 4 \theta + \frac{1}{4} \theta\right]}_{0}^{\frac{\pi}{4}}$
$\text{ } = \left(\frac{1}{2} \sin \left(\frac{\pi}{2}\right) + \frac{1}{16} \sin \pi + \frac{1}{4} \frac{\pi}{4}\right) - \left(0\right)$
$\text{ } = \frac{1}{2} + \frac{\pi}{16}$
$\therefore A = 2 + \frac{\pi}{4}$

# #

METHOD 2

If you are not happy with double integrals, then we can evaluate the area using the polar area formula $A = \int \setminus \frac{1}{2} {r}^{2} \setminus d \theta$ and calculating the two shaded sections separately:

Here this shaded area is given by:

${A}_{1} = \frac{1}{2} \setminus {\int}_{0}^{\frac{\pi}{4}} \setminus {\left(1 + \cos 2 \theta\right)}^{2} \setminus d \theta$
$\setminus \setminus \setminus \setminus = \frac{1}{2} \setminus {\int}_{0}^{\frac{\pi}{4}} \setminus 1 + 2 \cos 2 \theta + {\cos}^{2} 2 \theta \setminus d \theta$
$\setminus \setminus \setminus \setminus = \frac{1}{2} \setminus {\int}_{0}^{\frac{\pi}{4}} \setminus 1 + 2 \cos 2 \theta + \frac{1}{2} \left(\cos 4 \theta + 1\right) \setminus d \theta$
$\setminus \setminus \setminus \setminus = \frac{1}{2} \setminus {\int}_{0}^{\frac{\pi}{4}} \setminus 1 + 2 \cos 2 \theta + \frac{1}{2} \cos 4 \theta + \frac{1}{2} \setminus d \theta$
$\setminus \setminus \setminus \setminus = \frac{1}{2} {\left[\frac{3}{2} \theta + \sin 2 \theta + \frac{1}{8} \sin 4 \theta\right]}_{0}^{\frac{\pi}{4}}$
$\setminus \setminus \setminus \setminus = \frac{1}{2} \left(\frac{3 \pi}{8} + 1 + 0\right)$
$\setminus \setminus \setminus \setminus = \frac{3 \pi}{16} + \frac{1}{2}$

is given by:

${A}_{2} = \frac{1}{2} \setminus {\int}_{0}^{\frac{\pi}{4}} \setminus {\left(1\right)}^{2} \setminus d \theta$
$\setminus \setminus \setminus \setminus = \frac{1}{2} {\left[\theta\right]}_{0}^{\frac{\pi}{4}}$
$\setminus \setminus \setminus \setminus = \frac{1}{2} \left(\frac{\pi}{4} - 0\right)$
$\setminus \setminus \setminus \setminus = \frac{\pi}{8}$

And so the total sought area is $4$ times the difference between these results:

$\frac{1}{4} A \setminus = {A}_{1} - {A}_{2}$
$\text{ } = \frac{3 \pi}{16} + \frac{1}{2} - \frac{\pi}{8}$
$\text{ } = \frac{\pi}{16} + \frac{1}{2}$
$\therefore A = \frac{\pi}{4} + 2$

## 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.!

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

Jim H
Featured 1 month 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 ${\lim}_{x \rightarrow 0} {\left(3 x\right)}^{2} / {\sin}^{2} \left(3 x\right) = 1$

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

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

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

## The top and bottom margins of a poster are 6 cm each and the side margins are 4 cm each. If the area of the printed material is 384 cm^2, what are the dimensions of the poster with the smallest area?

Steve
Featured yesterday

The smallest area is $864 \setminus c {m}^{2}$ which occurs when the dimensions are $36 \setminus c m \times 24 \setminus c m$

#### Explanation:

Let us set up the following variables:

$\left\{\begin{matrix}x & \text{Width of poster (cm)" \\ y & "Height of poster (cm)" \\ A & "Area of poster ("cm^3")}\end{matrix}\right.$

Then the dimensions of the printed matter are:

$\left\{\begin{matrix}\text{Width" & =x-6-6 & =x-12 \\ "Height" & =y-4-4 & =y-8 \\ :. " Area} & = 384 & = \left(x - 12\right) \left(y - 8\right)\end{matrix}\right.$

So we have;

$384 = \left(x - 12\right) \left(y - 8\right)$
$\therefore y - 8 = \frac{384}{x - 12}$
$\therefore y = 8 + \frac{384}{x - 12}$
$\text{ } = \frac{8 \left(x - 12\right) + 384}{x - 12}$
$\text{ } = \frac{8 x - 96 + 384}{x - 12}$
$\text{ } = \frac{8 x + 288}{x - 12} \setminus \setminus \setminus \setminus \setminus \ldots . . \left(\star\right)$

And the total area of the poster is given by:

$A = x y$
$\setminus \setminus \setminus = \left(x\right) \left(\frac{8 x + 288}{x - 12}\right) \setminus \setminus \setminus$ (using ($\star$))
$\setminus \setminus \setminus = \frac{8 {x}^{2} + 288 x}{x - 12}$

We want to minimize (hopefully) by finding $\frac{\mathrm{dA}}{\mathrm{dx}}$, which we get by applying the product rule:

$\frac{\mathrm{dA}}{\mathrm{dx}} = \frac{\left(x - 12\right) \left(16 x + 288\right) - \left(1\right) \left(8 {x}^{2} + 288 x\right)}{x - 12} ^ 2$
$\text{ } = \frac{16 {x}^{2} + 288 x - 192 x + 3456 - 8 {x}^{2} - 288 x}{x - 12} ^ 2$
$\text{ } = \frac{8 {x}^{2} - 192 x + 3456}{x - 12} ^ 2$

At a min or max $\frac{\mathrm{dA}}{\mathrm{dx}} = 0$

$\therefore \frac{8 {x}^{2} - 192 x + 3456}{x - 12} ^ 2 = 0$
$\therefore 8 {x}^{2} - 192 x - 3456 = 0$
$\therefore 8 \left({x}^{2} - 24 x - 432\right) = 0$
$\therefore {x}^{2} - 24 x - 432 = 0$
$\therefore \left(x + 12\right) \left(x - 36\right) = 0$

This equation leads to the two solutions:

$x = - 12$, or $x = 36$

Obviously $x > 0$, so we can eliminate $x = - 12$, leaving $x = 36$ as the only valid solution. With this value of $x$ we have:

$y = \frac{288 + 288}{36 - 12} \setminus \setminus \setminus \setminus \setminus$ (using ($\star$))
$\setminus \setminus = \frac{576}{24}$
$\setminus \setminus = 24$

With these dimensions we have:

$A = 36 \cdot 24$
$\setminus \setminus \setminus = 864$

We should check that this value leads to a minimum (rather than a maximum) area, which we confirm graphically:

graph{(8x^2+288x)/(x-12) [-2, 50, -2000, 2000]}

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

Andrea S.
Featured 1 week 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...

##### Questions
• · 29 minutes ago
• · 29 minutes ago · in Chain Rule
• · An hour ago
• · An hour ago
• 2 hours ago · in Quotient Rule
• 4 hours ago
• · 4 hours ago
• · 5 hours ago · in Product Rule
• · 6 hours ago
• · 7 hours ago
• · 7 hours ago
• · 10 hours ago · in Integration by Parts
• · 11 hours ago
• · 11 hours ago
• · 11 hours ago
• · 11 hours ago
• · 12 hours ago
• · 12 hours ago
• · 13 hours ago
• · 14 hours ago
• · 14 hours ago
• 14 hours ago · in Quotient Rule