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Answer:

# 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:

# s/(s^4+4a^4) = s/( (s^2+2a^2)^2 - 4a^2s^2) #
# " " = s/( (s^2+2a^2)^2 - (2as)^2) #

We now have the difference of two squares, so:

# s/(s^4+4a^4) = s/( (s^2+2a^2 + 2as)(s^2+2a^2 - 2as) ) #

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

# s/(s^4+4a^4) = A/(s^2+2a^2 + 2as) + B/(s^2+2a^2 - 2as) #
# " " = (A(s^2+2a^2 - 2as) + B(s^2+2a^2 + 2as))/ ((s^2+2a^2 + 2as)(s^2+2a^2 - 2as)) #

# :. s = A(s^2+2a^2 - 2as) + B(s^2+2a^2 + 2as) #

Equating Coefficients we get:

# Coeff(s) \ \ => 1 = -2aA+2aB \ \ \ \ \ ..... [1]#
# Coeff(s^2) => 0 = A+B \ \ \ \ \ ..... [2]#

Now we solve these simultaneous equations:

# Eq[1] + (2a)Eq[2] => 1=4aB #
# :. A=-1/(4a), B= 1/(4a) #

Se we can write the original expression as:

# s/(s^4+4a^4) = 1/(4a) { 1/(s^2+2a^2 - 2as) - 1/(s^2+2a^2 + 2as) } #
# " " = 1/(4a) { 1/((s-a)^2-a^2+2a^2) - 1/((s+a)^2-a^2+2a^2) } #
# " " = 1/(4a) { 1/((s-a)^2+a^2) - 1/((s+a)^2+a^2) } #

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} #
# " " = 1/(4a^2)sin(at){e^(at) - e^(-at)} #

# " " = 1/(4a^2)sin(at)2sinh(at) #

# " " = (sin(at)sinh(at))/(2a^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 = (s^2+1/2)^2 +3/4#

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

# s/(s^4+s^2+1) = s/{ (s^4+2s^2+1} -s^2 } #
# " " = s/{ (s^2+1)^2 -s^2 } #
# " " = s/{ (s^2+1-s)(s^2+1+s) } #

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

# s/(s^4+s^2+1) = A/(s^2+1-s) + B/(s^2+1+s) #
# " " = (A(s^2+1+s) + B(s^2+1-s)) /{ (s^2+1-s)(s^2+1+s) } #

# :. s = A(s^2+1+s) + B(s^2+1-s) #

Equating Coefficients we get:

# Coeff(s) \ \ => 1 = A-B \ \ \ \ \ ..... [1]#
# Coeff(s^2) => 0 = A+B \ \ \ \ \ ..... [2]#

Now we solve these simultaneous equations:

# Eq[1] + Eq[2] => 1=2A #
# :. A=1/2, B= -1/2 #

Se we can write the original expression as:

# s/(s^4+s^2+1) = 1/2{1/(s^2+1-s) - 1/(s^2+1+s)} #
# " " = 1/2{1/((s-1/2)^2-(1/2)^2+1) - 1/((s+1/2)^2-(1/2)^2+1)} #
# " " = 1/2{1/((s-1/2)^2-3/4) - 1/((s+1/2)^2-3/4) } #
# " " = 1/2{1/((s-1/2)^2-(sqrt(3)/2)^2) - 1/((s+1/2)^2-(sqrt(3)/2)^2) } #

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))} #
# " " = 1/2*2/sqrt(3)sin(sqrt(3)/2){e^(1/2t) - e^(-1/2t)} #
# " " = 1/sqrt(3)sin(sqrt(3)/2)2sinh(1/2t) #
# " " = 2/sqrt(3)sin(sqrt(3)/2)sinh(1/2t) #

Answer:

# 2+pi/4 #

Explanation:

Here is the graph of the two curves. The shaded area, #A#, is the area of interest:
enter image source here

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 #Q1# for the point of intrestion by solving the simultaneous equations:

# r=1+cos 2theta #
# r=1 #

However, intuition is faster, and it looks like angle of intersection in #Q1# is #pi/4#, we can verify this by a quick evaluation:

# theta = pi/4 => r=1+cos 2theta =1+cos(pi/2) = 1 #

Confirming our intuition. So now we have the following:

enter image source here

Where the shaded area repents #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(2theta)#
#theta# varies for #0# to #pi/4#

So then:

# 1/4A = int_0^(pi/4) int_1^(1+cos2theta) r \ dr \ d theta #

If we evaluate the inner integral first then we get:

# int_1^(1+cos2theta) r \ dr = [1/2r^2]_1^(1+cos2theta) #
# " " = 1/2( (1+cos2theta)^2 - 1^2) #
# " " = 1/2( 1+2cos2theta+cos^2 2theta - 1) #
# " " = 1/2( 2cos2theta+cos^2 2theta) #
# " " = 1/2( 2cos2theta+1/2(cos4theta+1)) #
# " " = cos2theta+1/4cos4theta+1/4 #

And so our double integral becomes:

# 1/4A \ = int_0^(pi/4) int_1^(1+cos2theta) r \ dr \ d theta #
# " " = int_0^(pi/4) cos2theta+1/4cos4theta+1/4 \ d theta #
# " " = [1/2sin2theta+1/16sin4theta+1/4theta]_0^(pi/4) #
# " " = (1/2sin(pi/2)+1/16sinpi+1/4pi/4) - (0) #
# " " = 1/2+pi/16 #
# :. A = 2+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 \ 1/2r^2 \ d theta# and calculating the two shaded sections separately:
enter image source here

Here this shaded area is given by:

# A_1 = 1/2 \ int_0^(pi/4) \ (1+cos 2theta)^2 \ d theta #
# \ \ \ \ = 1/2 \ int_0^(pi/4) \ 1+2cos 2theta+ cos^2 2theta \ d theta #
# \ \ \ \ = 1/2 \ int_0^(pi/4) \ 1+2cos 2theta+1/2(cos4theta+1) \ d theta#
# \ \ \ \ = 1/2 \ int_0^(pi/4) \ 1+2cos 2theta+1/2cos4theta+1/2 \ d theta#
# \ \ \ \ = 1/2 [3/2theta+sin2theta+1/8sin4theta]_0^(pi/4)#
# \ \ \ \ = 1/2 ((3pi)/8+1+0)#
# \ \ \ \ = (3pi)/16+1/2#

And this shaded area
enter image source here

is given by:

# A_2 = 1/2 \ int_0^(pi/4) \ (1)^2 \ d theta #
# \ \ \ \ = 1/2 [theta]_0^(pi/4)#
# \ \ \ \ = 1/2 (pi/4-0)#
# \ \ \ \ = pi/8#

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

# 1/4 A \ = A_1 - A_2 #
# " " = (3pi)/16+1/2 - pi/8 #
# " " = (pi)/16+1/2 #
#:. A = (pi)/4+2 #

Answer:

For the #2^(nd)# Proof, refer to the Explanation.

Explanation:

Here is a Second Method to prove the Result for

#n>=2, n in NN.#

#u_n=intsin(nx)/sinxdx.#

Now, #sin(nx)/sinx=1/sinx{sin((nx-2x)+2x)}#

Knowing that, #sin(A+B)=sinAcosB+cosAsinB,# we get,

#sin(nx)/sinx=1/sinx{sin(nx-2x)cos2x+cos(nx-2x)sin2x}#

#=1/sinx{(sin((n-2)x))(1-2sin^2x)+(cos((n-2)x))(2sinxcosx)}#

#=1/sinx{sin((n-2)x)-2sin^2xsin((n-2)x)+2sinxcosxcos((n-2)x)}#

#=sin((n-2)x)/sinx-2sinxsin((n-2)x)+2cosxcos((n-2)x)#

#=sin((n-2)x)/sinx+2{cosxcos((n-2)x)-sinxsin((n-2)x)}#

#=sin((n-2)x)/sinx+2{cos((n-2)x+x)}#

#:. sin(nx)/sinx=sin((n-2)x)/sinx+2cos((n-1)x).#

#rArr u_n=intsin(nx)/sinxdx=int{sin((n-2)x)/sinx+2cos((n-1)x)}dx#

#=intsin((n-2)x)/sinxdx+2intcos((n-1)x)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.!

Answer:

Use #lim_(theta rarr0)sintheta/theta = 1#

Explanation:

#lim_(xrarr0)(1−cos(8x))/(1−cos(3x))# has initial form #0/0#

Recall (from trigonometry class) that

#(1-costheta)(1-costheta) = 1-cos^2theta = sin^2theta#

#(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_(xrarr0)(1+cos(3x))/(1+cos(8x)) = 2/2=1#,

so we will focus on

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

We'll use the facts that

#lim_(xrarr0)sin^2(8x)/(8x)^2 = 1# and #lim_(xrarr0)(3x)^2/sin^2(3x) = 1#

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

# = lim_(xrarr0)64/9 (sin(8x)/(8x))^2/(sin(3x)/(3x))^2#

# = 64/9 * 1/1 = 64/9#

Answer:

The smallest area is #864 \ cm^2# which occurs when the dimensions are #36 \ cm xx 24 \ cm #

Explanation:

Let us set up the following variables:

# {(x, "Width of poster (cm)"), (y, "Height of poster (cm)"), (A, "Area of poster ("cm^3")") :} #

Then the dimensions of the printed matter are:

# {("Width", =x-6-6,=x-12), ("Height", =y-4-4,=y-8), ( :. " Area",=384 ,=(x-12)(y-8)) :} #

So we have;

# 384 =(x-12)(y-8) #
# :. y-8 = 384/(x-12)#
# :. y = 8+384/(x-12)#
# " " = (8(x-12)+384)/(x-12)#
# " " = (8x-96+384)/(x-12)#
# " " = (8x+288)/(x-12) \ \ \ \ \ ..... (star)#

And the total area of the poster is given by:

# A = xy #
# \ \ \ = (x)((8x+288)/(x-12)) \ \ \ # (using (#star#))
# \ \ \ = (8x^2+288x)/(x-12) #

We want to minimize (hopefully) by finding #(dA)/dx#, which we get by applying the product rule:

# (dA)/dx = { (x-12)(16x+288) - (1)(8x^2+288x) } / (x-12)^2 #
# " " = { 16x^2+288x-192x+3456-8x^2-288x } / (x-12)^2 #
# " " = { 8x^2-192x+3456 } / (x-12)^2 #

At a min or max #(dA)/dx=0#

# :. { 8x^2-192x+3456 } / (x-12)^2 = 0 #
# :. 8x^2-192x-3456 = 0 #
# :. 8(x^2-24x-432)=0 #
# :. x^2-24x-432 = 0 #
# :. (x+12)(x-36) = 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 = (288+288)/(36-12) \ \ \ \ \ # (using (#star#))
# \ \ = 576/24 #
# \ \ = 24 #

With these dimensions we have:

# A = 36*24 #
# \ \ \ = 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]}

Answer:

# int dx/(cosx+cosa) = 1/sina ln ( cos((x-a)/2) / cos((x+a)/2))+C#

Explanation:

Use the parametric formula:

#cosx = (1-tan^2(x/2))/(1+tan^2(x/2))#

and substitute:

#t= tan(x/2)#

#x = 2 arctan t#

#dx = (2dt)/(1+t^2)#

we have:

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

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

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

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:

#1/(( sqrt(1+cosa) - tsqrt(1-cosa)) ( sqrt(1+cosa) + tsqrt(1-cosa) ) )= A/( sqrt(1+cosa) - tsqrt(1-cosa)) + B/( sqrt(1+cosa) + tsqrt(1-cosa))#

#A( sqrt(1+cosa) + tsqrt(1-cosa)) + B(sqrt(1+cosa) - tsqrt(1-cosa)) = 1#

#{ ( (A+B) sqrt(1+cosa) = 1) , ((A-B) sqrt(1-cosa) = 0):}#

#A=B= 1/(2sqrt(1+cosa))#

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

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 dx/(cosx+cosa) = int dt/((1+cosa) - tsqrt(1-cos^2a) )+int dt/( (1+cosa) + tsqrt(1-cos^2a))#

Now:

#sqrt(1-cos^2a) = abs sin 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 dx/(cosx+cosa) = int dt/((1+cosa) - tsina )+int dt/( (1+cosa) + tsina)#

# int dx/(cosx+cosa) = 1/sina (int dt/((1+cosa)/sina - t )+int dt/( (1+cosa)/sina + t))#

# int dx/(cosx+cosa) = 1/sina (-ln((1+cosa)/sina - t )+ln( (1+cosa)/sina + t)) +C#

Using now the properties of logarithms:

# int dx/(cosx+cosa) = 1/sina ln (( (1+cosa)/sina + t) / ((1+cosa)/sina - t ))+C#

# int dx/(cosx+cosa) = 1/sina ln (( 1+cosa + tsina) / (1+cosa - t sina))+C#

Undoing the substitution:

# int dx/(cosx+cosa) = 1/sina ln (( 1+cosa + tan(x/2)sina) / (1+cosa - tan(x/2) sina))+C#

Multiply by #cos(x/2)# numerator and denominator of the argument:

# int dx/(cosx+cosa) = 1/sina ln (( cos(x/2)+cosacos(x/2) + sin(x/2)sina) / (cos(x/2)+cos(x/2)cosa - sin(x/2) sina))+C#

# int dx/(cosx+cosa) = 1/sina ln (( cos(x/2)+cos(x/2-a) ) / (cos(x/2)+cos(x/2+a)))+C#

Use now the identity:

#cos alpha + cos beta = 2cos((alpha+beta)/2)cos((alpha-beta)/2)#

# int dx/(cosx+cosa) = 1/sina ln (( 2cos((x-a)/2)cos(a/2) ) / (2cos((x+a)/2)cos(a/2)))+C#

# int dx/(cosx+cosa) = 1/sina ln ( cos((x-a)/2) / cos((x+a)/2))+C#

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

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