Make the internet a better place to learn

3

Answer:

#int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx = x sinh^(-1) x - sqrt(x^2+1) + C#

#color(white)(int ln(x+sqrt(x^2+1)) dx) = x ln(x+sqrt(x^2+1)) - sqrt(x^2+1) + C#

Explanation:

Let us try a hyperbolic substitution.

Let #x = sinh theta#

Then:

#int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx = int color(white)(.)ln(sinh theta + sqrt(sinh^2 theta + 1)) dx/(d theta) d theta#

#color(white)(int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx) = int color(white)(.)ln(sinh theta + cosh theta) cosh theta color(white)(.)d theta#

#color(white)(int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx) = int color(white)(.)ln(e^theta) cosh theta color(white)(.)d theta#

#color(white)(int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx) = int color(white)(.)theta cosh theta color(white)(.)d theta#

#color(white)(int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx) = theta sinh theta - cosh theta + C#

#color(white)(int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)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 = 1/2(e^y-e^(-y))#

Hence:

#e^y-2x-e^(-y) = 0#

So:

#(e^y)^2-(2x)(e^y)-1 = 0#

So using the quadratic formula:

#e^y = (2x+-sqrt((2x)^2+4))/2#

#color(white)(e^y) = x+-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(x+sqrt(x^2+1))#

That is:

#sinh^(-1) x = ln(x+sqrt(x^2+1))#

So:

#int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx = x ln(x+sqrt(x^2+1)) - sqrt(x^2+1) + C#

1

Step #1#: Determine the first derivative

#f'(x) = 6x^2 - 24x + 18#

Step #2#: Determine the critical numbers

These will occur when the derivative equals #0#.

#0 = 6x^2 - 24x + 18#

#0 = 6(x^2 - 4x + 3)#

#0 = (x - 3)(x - 1)#

#x = 3 or 1#

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

We select test points.

Test point 1: #x = 0#

#f'(0) = 6(0)^2 - 24(0) + 18 = 18#

Since this is positive, the function is uniformly increasing on #(-oo, 1)#.

Test point 2: #x = 2#

#f'(2) = 6(2)^2 - 24(2) + 18 = -6#

Since this is negative, the function is decreasing on #(1, 3)#.

I won't select a test point for #(3, oo)# 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''(x) = 12x - 24#

Step #5#: Determine the points of inflection

These will occur when #f''(x) = 0#.

#0 = 12x- 24#

#0 = 12(x - 2)#

#x = 2#

Step #6#: Determine the intervals of concavity

Once again, we select test points.

Test point #1#: #x = 1#

#f''(1) = 12(1) - 24 = -12#

This means that #f(x)# concave down (Since it's negative) on#(-oo, 2)#.

This also means that #f(x)# is concave up on #(2, oo)#.

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

#f(x)# 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(0) = 2(0)^3 - 12(0)^2 + 18x - 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

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.!

1

Answer:

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 = 2lh + 2lw + 2hw#

In this case, it becomes

#600 = 2lh+ 2l^2 + 2lh#

#600 = 4lh + 2l^2#

We want to solve for one of the variables now.

#600 - 2l^2 = 4lh#

#(600 -2l^2)/(4l) =h#

#(300 - l^2)/(2l) = h#

The volume of a rectangular prism is given by:

#V =l xx w xx h#

Which in our case becomes:

#V = l^2 xx h#

Now:

#V = l^2(300 - l^2)/(2l)#

#V = 1/2l(300 - l^2)#

#V = 1/2(300l - l^3)#

#V = 150l - 1/2l^3#

Now find the derivative of this with respect to volume.

#V' = 150 - 3/2l^2#

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

#0 = 150 - 3/2l^2#

#3/2l^2 = 150#

#l^2 = 100#

#l = +- 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.

enter image source here

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 * h#

#1000 = (10)^2 * h#

#1000 = 100h#

#h = 10#

Hopefully this helps!

3

Answer:

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

Explanation:

First we rearrange the equation of the surface into the form # f(x,y,z)=0#

# z=x^2-2xy+y^2 #
# :. x^2-2xy+y^2-z = 0 #

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

# f(x,y,z) = x^2-2xy+y^2-z #

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

# grad f(x,y,z) = (partial f)/(partial x) hat(i) + (partial f)/(partial y) hat(j) + (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:

# grad f = ((partial)/(partial x) (x^2-2xy+y^2-z))hat(i) + #
# " " ((partial)/(partial y) (x^2-2xy+y^2-z))hat(j) + #
# " " ((partial)/(partial z) (x^2-2xy+y^2-z))hat(k) #
# " "= (2x-2y)hat(i) + (-2x+2y)hat(j) + (-1)hat(k) #

So for the particular point #(1,2,1)# the normal vector to the surface is given by:

# grad f(1,2,1) = (2-4)hat(i) +(-2+4)hat(j) -hat(k) #
# " " = -2hat(i) +2hat(j) -hat(k) #

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

# vec r * vec n = vec a * vec n #

Where #vec r=((x),(y),(z))#; #vec n=( (-2), (2), (-1) )#, is the normal vector and #a# is any point in the plane

Hence, the tangent plane equation is:

# ((x),(y),(z)) * ( (-2), (2),(-1) ) = ((1),(2),(1)) * ( (-2), (2),(-1) ) #
# :. (x)(-2) + (y)(2) + (z)(-1) = (1)(-2) + (2)(2) + (1)(-1) #
# :. -2x+2y-z = -2+4-1 #
# :. -2x+2y-z = 1 #

We can confirm this graphically: Here is the surface with the normal vector:
enter image source here
and here is the surface with the tangent plane and the normal vector:
enter image source here

1

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

#mathbf r(alpha, beta) = langle alpha, beta, alpha^2 - 2 alpha beta + beta^2 rangle#

Noting that to first order :

  • #d mathbf r_alpha = (partial mathbf r)/(partial alpha) d alpha = langle 1,0,2 (alpha - beta) rangle d alpha#

  • #d mathbf r_beta = (partial mathbf r)/(partial beta) d beta = langle 0,1,- 2 (alpha - beta) rangle d beta #

....we have this:

enter image source here

The shaded area is a vector, #d mathbf A #, whose direction is the normal to the surface at that point, and given by the vector cross product :

#d mathbf A = dA \ color(red)( mathbf e_n ) =d mathbf r_alpha xx d mathbf r_beta#

# = det ((mathbf e_x, mathbf e_y, mathbf e_z),(1,0,2 (alpha - beta)),(0,1,- 2 (alpha - beta))) dalpha dbeta#

#= langle - 2(alpha - beta), 2 (alpha - beta), 1 rangle \ dalpha \ dbeta#

We are not really bothered by the magnitide of #d mathbf A#, its the direction (of any normal vector #mathbf n#, ie we don't even need to normalise to #mathbf e_n#) that matters, so we can forget about # \ dalpha \ dbeta# and evaluate this as one of many normal vectors:

#mathbf n = langle - 2(alpha - beta), 2 (alpha - beta), 1 rangle_{(alpha = 1, beta = 2)} = langle 2, -2, 1 rangle#

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

#(mathbf r - mathbf r_o) cdot mathbf n = 0#

#implies mathbf r cdot mathbf n = mathbf r_o cdot mathbf n#

#implies langle x, y, z rangle cdot langle 2, -2, 1 rangle = langle x_o, y_o, z_o rangle cdot langle 2, -2, 1 rangle#

So we plug in #mathbf r_0 = langle 1, 2, 1 rangle# and we get:

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

3

Answer:

# int \ (x-2)/(x(x^2-4x+5)^2) \ dx = -2/25 ln|x| + 1/25 ln|x^2-4x+5| - 3/50 arctan(x-2) + (x-4)/(10(x^2-4x+5)) + c #

Explanation:

Let us denote the required integral by #I#

# I = int \ (x-2)/(x(x^2-4x+5)^2) \ dx #

Partial Fraction Decomposition

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

# (x-2)/(x(x^2-4x+5)^2) = A/x + (Bx+C)/(x^2-4x+5) + (Dx+E)/(x^2-4x+5)^2 #
# " " = {A(x^2-4x+5)^2 + (Bx+C)x(x^2-4x+5) + (Dx+E)x}/{x(x^2-4x+5)^2} #

Leading to:

# x-2 = A(x^2-4x+5)^2 + (Bx+C)x(x^2-4x+5) + (Dx+E)x #

# :. x-2 = A(x^4-8x^3+26x^2-40x+25) + (Bx+C)(x^3-4x^2+5x) + (Dx+E)x #

We can use various methods to find our unknown constants:

# "Put " x=0 => -2=A*5^2 #
# :. A = -2/25 #

Compare coefficients:

# "Coeff"(x^4) => 0 = A +B #
# :. B = 2/25 #

# "Coeff"(x^3) => 0 = -8A+C-4B#
# :. 0=16/25+C-8/25 => C=-8/25#

# "Coeff"(x^2) => 0 = 26A+5B-4C+D#
# :. 0=-52/25+10/25+32/25+D => D=10/25=2/5#

# "Coeff"(x^1) => 1 = -40A+5C+E#
# :. 1=80/25-40/25+E => E=-3/5#

Hence we can the integral as:

# I = int \ (-2/25)/x + (2/25x-8/25)/(x^2-4x+5) + (2/5x-3/5)/(x^2-4x+5)^2 \ dx#

# \ \ = -2/35 \ int \ 1/x dx + 2/25 int (x-4)/(x^2-4x+5) \ dx+ 1/5 int (2x-3)/(x^2-4x+5)^2 \ dx#

# \ \ = -2/25 \ I_1 + 2/25 \ I_2 + 1/5 \ I_3#

Where:

# I_1 = int \ 1/x dx + 2/25 \ dx#
# I_2 = int (x-4)/(x^2-4x+5) \ dx #
# I_3 = int (2x-3)/(x^2-4x+5)^2 \ 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 \ 1/x \ dx #
# \ \ \ = ln|x| #

Integral 2: #I_2#

The second integral, #I_2#, we manipulate by completing the square:

# I_2 = int \ (x-4)/(x^2-4x+5) \ dx #
# \ \ \ = int \ (x-4)/((x-2)^2-2^2+5) \ dx#
# \ \ \ = int \ (x-4)/((x-2)^2+1) \ dx#

Let #u=x-2 => (du)/dx = 1 #, and #x-4=u-2#, then substituting gives

# I_2 = int \ (u-2)/(u^2+1) \ du#
# \ \ \ = int \ u/(u^2+1) - 2/(u^2+1)\ du#
# \ \ \ = 1/2int \ (2u)/(u^2+1) \ du - 2 int \ 1/(u^2+1)\ du#
# \ \ \ = 1/2ln|u^2+1| - 2 arctanu#

Restoring the substitution we get:

# I_2 = 1/2ln|x^2-4x+5| - 2 arctan(x-2)#

Integral 3: #I_3#

The third integral, #I_3# we continue as earlier by completing the square to get:

# I_3 = int \ (2x-3)/(x^2-4x+5)^2 \ dx #
# \ \ \ = int (2x-3)/((x-2)^2+1)^2 \ dx #

Let #u=x-2 => (du)/dx = 1 #, and #2x-3=2u+1#, then substituting gives

# I_3 = int (2u+1)/(u^2+1)^2 \ du #
# \ \ \ = int (2u)/(u^2+1)^2 \ du+ int 1/(u^2+1)^2 \ du #

Consider the integral:

# int (2u)/(u^2+1)^2 \ du #

Let #w=u^2+1 => (dw)/(du)=2u#, so we can substitute to get:

# int (2u)/(u^2+1)^2 \ du = int \ 1/w^2 \ dw #
# " " = -1/w #

Restoring the substitutions we get:

# int (2u)/(u^2+1)^2 \ du = -1/(u^2+1) #
# " " = -1/((x-2)^2+1) #
# " " = -1/(x^2-4x+5) #

And now we consider:

# int 1/(u^2+1)^2 \ du #

Let #u=tan theta => (du)/(d theta) = sec^2 theta #, then substituting gives

# int 1/(u^2+1)^2 \ du = int \ 1/(tan^2theta+1)^2 \ sec^2 theta \ d theta #
# " " = int \ 1/(sec^2theta)^2 \ sec^2 theta \ d theta #
# " " = int \ 1/(sec^2theta) \ d theta #
# " " = int \ cos^2theta \ d theta #
# " " = int \ 1/2(1+cos2theta) \ d theta #
# " " = 1/2 \ int \ 1+cos2theta \ d theta #
# " " = 1/2 (theta + 1/2 sin2theta) #
# " " = 1/2 theta + 1/2 sintheta costheta#

Now, #u=tan theta => u=sintheta/costheta #

# :. sinthetacostheta = cos^2theta * u #
# " " =1/sec^2theta * u #
# " " =1/(1+tan^2theta) * u #
# " " =u/(u^2+1) #

Restoring the substitutions we get:

# int 1/(u^2+1)^2 \ du = 1/2 arctanu+ 1/2 u/(u^2+1)#
# " " = 1/2 arctan(x-2) + 1/2 (x-2)/((x-2)^2+1)#
# " " = 1/2 arctan(x-2) + 1/2 (x-2)/(x^2-4x+5)#

Hence, we can write the third integral as:

# I_3 = int (2u)/(u^2+1)^2 \ du+ int 1/(u^2+1)^2 \ du #
# \ \ \ = -1/(x^2-4x+5)+ 1/2 arctan(x-2) + 1/2 (x-2)/(x^2-4x+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 = -2/25 \ I_1 + 2/25 \ I_2 + 1/5 \ I_3#

# \ \ = -2/25 {ln|x|} + 2/25 {1/2ln|x^2-4x+5| - 2 arctan(x-2)} + 1/5 {-1/(x^2-4x+5)+ 1/2 arctan(x-2) + 1/2 (x-2)/(x^2-4x+5)} + c #

# \ \ = -2/25 ln|x| + 1/25 ln|x^2-4x+5| - 4/25 arctan(x-2) -1/5 1/(x^2-4x+5)+ 1/10 arctan(x-2) + 1/10 (x-2)/(x^2-4x+5) + c #

# \ \ = -2/25 ln|x| + 1/25 ln|x^2-4x+5| - 3/50 arctan(x-2) -1/5 1/(x^2-4x+5)+ 1/10 (x-2)/(x^2-4x+5) + c #

# \ \ = -2/25 ln|x| + 1/25 ln|x^2-4x+5| - 3/50 arctan(x-2) + (x-2-2)/(10(x^2-4x+5)) + c #

# \ \ = -2/25 ln|x| + 1/25 ln|x^2-4x+5| - 3/50 arctan(x-2) + (x-4)/(10(x^2-4x+5)) + c #

2

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

enter image source here

It's the cosine function, with period #P = 1/50 " s"#, but without the negative bits. So it's an even function requiring a Fourier cosine series:

#S{f(t)} = a_o + sum_(n=1)^(oo) a_n cos ((2 n pi)/P t) #

With: #a_o =1/P int_(-P/2)^(P/2) f(t) \ dt#, we narrow the interval as follows:

#a_o =50V int_(-1/200)^(1/200) cos 100 pi t \ dt #

#= 50V * 1/(50pi) = V/pi#

For the other term:

#a_n =2/P int_(-P/2)^(P/2) f(t) cos ((2 n pi)/P t) \ dt#

# =100V int_(-1/200)^(1/200) cos ( 100 pi t) cos ( 100 n pi t) \ dt#

Using a sum-product formula:

# =50V int_(-1/200)^(1/200) cos ( 100 (n+1) pi t) + cos ( 100 (n-1) pi t) \ dt#

# =50 V [(sin( 100 (n+1) pi t))/(100(n+1)pi) + ( sin ( 100 (n-1) pi t))/(100(n-1)pi)]_(-1/200)^(1/200) #

# =100 V [(sin( 100 (n+1) pi t))/(100(n+1)pi) + ( sin ( 100 (n-1) pi t))/(100(n-1)pi)]_(0)^(1/200) #

# =V/pi ((sin( (n+1) pi/2 ))/((n+1)) + ( sin ( (n-1) pi/2 ))/((n-1))) #

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,...#, and sub #n = 2k#, we have:

#a_k = V/pi(( (-1)^(k))/((2k+1)) + ( - (-1)^(k) )/((2k-1)) ) #

#= (2V)/pi (-1)^(k+1) (1)/((4k^2-1) ) #

However, we also need to consider #n = 1#. At that point, we note #lim_(n to 1) ( sin ( (n-1) pi/2 ))/((n-1)) = pi/2 implies a_1 = 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:

enter image source here

2

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#

2

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...