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Answer:

For a non-rigorous proof, please see below.

Explanation:

For a positive central angle of #x# radians (#0 < x < pi/2#) (not degrees)
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Source:
commons.wikimedia.org

The geometric idea is that

#"Area of "Delta KOA < "Area of " "Sector KOA" < "Area of "Delta LOA#

#"Area of "Delta KOA = 1/2(1)(sinx) \ \ \ # (#1/2"base"*"height"#)

#"Area of " "Sector KOA" = 1/2 (1)^2 x \ \ \ # (#x# is in radians)

#"Area of "Delta LOA = 1/2tanx \ \ \ # (#AL = tanx#)

So we have:

#sinx/2 < x/2 < tanx/2#

For small positive #x#, we have #inx > 0# so we can multiply through by #2/sinx#, to get

#1 < x/sinx < 1/cosx#

So

#cosx < sinx/x < 1# for #0 < x < pi/2#.

#lim_(xrarr0^+) cosx = 1# and #lim_(xrarr0^+) 1= 1#

so #lim_(xrarr0^+) sinx/x = 1#

We also have, for these small #x#, #sin(-x) = -sinx#, so #(-x)/sin(-x) = x/sinx# and #cos(-x) = cosx#, so

#cosx < sinx/x < 1# for #-pi/2 < x < 0#.

#lim_(xrarr0^-) cosx = 1# and #lim_(xrarr0^-) 1= 1#

so #lim_(xrarr0^-) sinx/x = 1#

Since both one sided limits are #1#, the limit is #1#.

Answer:

#int(cos^3x+cos^5x)/(sin^2x+sin^4x)dx=sinx-2cscx-6tan^(-1)(sinx)+C#

Explanation:

Let

#I=int(cos^3x+cos^5x)/(sin^2x+sin^4x)dx#

Factor out a #cosx#:

#I=int((cos^2x+cos^4x)/(sin^2x+sin^4x))*cosxdx#

Rewrite in terms of #sinx#:

#I=int((2-3sin^2x+sin^4x)/(sin^2x+sin^4x))*cosxdx#

For ease of reading, apply the substitution #sinx=u#:

#I=int(2-3u^2+u^4)/(u^2+u^4)du#

Apply long division:

#I=int(1+(2-4u^2)/(u^2(1+u^2)))du#

Apply partial fraction decomposition:

#I=int(1+2/u^2-6/(1+u^2))du#

Integrate term by term:

#I=u-2/u-6tan^(-1)(u)+C#

Reverse the substitution:

#I=sinx-2cscx-6tan^(-1)(sinx)+C#

Answer:

The slope is: #(pi^2+4pi+8sqrt2+64)/(pi^2+12pi+16sqrt2+32)approx0.9564910#

Explanation:

The slope will be given by #dy/dx# of the curve evaluated at #theta=pi/4#.

To get the equation in terms of #x# and #y#, we should use the polar-to-rectangular identities: #{(x=rcostheta),(y=rsintheta):}#

So here, where #r=f(theta)#, we have:

#x=costheta(theta^2-theta-cos^3theta+tan^2theta)#

#color(white)x=theta^2costheta-thetacostheta-cos^4theta+tanthetasintheta#

#y=sintheta(theta^2-theta-cos^3theta+tan^2theta)#

#color(white)y=theta^2sintheta-thetasintheta-cos^3thetasintheta+sinthetatan^2theta#

To find #dy/dx#, we first need to find #dx/(d theta)# and #dy/(d theta)#. This will take a fair amount of product and chain rule, so be careful.

#dx/(d theta)=2thetacostheta-theta^2sintheta-costheta+thetasintheta+4cos^3thetasintheta+sec^2thetasintheta+tanthetacostheta#

#color(white)(dx/(d theta))=2thetacostheta-theta^2sintheta-costheta+thetasintheta+4cos^3thetasintheta+tanthetasectheta+sintheta#

#dy/(d theta)=2thetasintheta+theta^2costheta-sintheta-thetacostheta+3cos^2thetasin^2theta-cos^4theta+costhetatan^2theta+2sinthetatanthetasec^2theta#

#color(white)(dy/(d theta))=2thetasintheta+theta^2costheta-sintheta-thetacostheta+3cos^2thetasin^2theta-cos^4theta+sinthetatantheta+2tan^2thetasectheta#

Now we can find #dy/dx# using #dy/dx=(dy//d theta)/(dx//d theta)#.

Just so we don't have to write out a giant fraction, let's evaluate #dx/(d theta)# and #dy/(d theta)# at #theta=pi/4# separately.

#dx/(d theta)=2 pi/4 1/sqrt2+pi^2/16 1/sqrt2-1/sqrt2+pi/4 1/sqrt2+4 1/(2sqrt2)1/sqrt2+2 1/sqrt2+1 1/sqrt2#

#color(white)(dx/(d theta))=pi/(2sqrt2)+pi^2/(16sqrt2)+pi/(4sqrt2)+1+sqrt2#

#color(white)(dx/(d theta))=1/(16sqrt2)(pi^2+12pi+16sqrt2+32)#

#dy/(d theta)=2 pi/4 1/sqrt2+pi^2/16 1/sqrt2-1/sqrt2-pi/4 1/sqrt2+3 1/2 1/2-1/4+1/sqrt2 1+2*1sqrt2#

#color(white)(dy/(d theta))=pi/(2sqrt2)+pi^2/(16sqrt2)-pi/(4sqrt2)+1/2+2sqrt2#

#color(white)(dy/(d theta))=1/(16sqrt2)(pi^2+4pi+8sqrt2+64)#

Thus, at #theta=pi/4#, the slope is:

#dy/dx=(dy//d theta)/(dx//d theta)=(pi^2+4pi+8sqrt2+64)/(pi^2+12pi+16sqrt2+32)approx0.9564910#

To find the critical points we have to examine three items as follows.
Besides we need the first (given) derivative and the second derivative of the function k.

Remarks:

#k^'( x)# means the term of the first derivative of k, #k^('') (x)# is the term of the second derivative of k.

In other "words":

#d/dx k (x ) = k^'( x) # and #d^2/dx^2 k ( x ) = k^('') ( x)#.

i) Necessary condition: #k^'(x) =0#

Using polynomial long division (taking some time to find #x = 2# that fulfils #k^'(x) =0#) we get

# (x^3-3x^2-18x+40):(x-2)=x^2-x-20#
#- (x^3-2x^2)#
# = - x^2 - 18 x#
#- (-x^2 + 2x) #
# = - 20 x + 40#
#- (-20x + 40)#
#= 0#

Now using the pq-formula for quadratic equations:

#x^2-x-20=0#
#=> x = 1/2 +- sqrt(1/4 + 20) = 1/2 +- sqrt(81/4) = 1/2 +- 9/2#
# => x= - 4 vv x = 5#.

ii) Necessary and sufficient condition: #k^'( x) = 0 ^^ k^('') (x) != 0#.
#k^('') (- 4) = 54 > 0#, so minimum at #x=-4#,
#k^('') (2) = - 18 < 0#, so maximum at #x=2#,
#k^('') (5) = 27 > 0#, so minimum at #x=5#.

iii) y - values and points

Provided that #k ( x ) = x^4/4 - x^3 -9x^2 +40 x# we get

#k (- 4) = -176#, so the first mimimum point is #M_1 (- 4 ; - 176) #,

#k (2) = 40#, so maximum point is #M (2 ; 40) #,

#k (5) = 25/4#, so the second mimimum point is #M_2 (5 ; 25/4) #.

For other equations of the function k there are other values and therefore other points.

I hope it helps.

Answer:

# int_0^(pi/2) \ sinx \ dx = 1 #

Explanation:

By definition of an integral, then

# int_a^b \ f(x) \ dx #

represents the area under the curve #y=f(x)# between #x=a# and #x=b#. We can estimate this area under the curve using thin rectangles. The more rectangles we use, the better the approximation gets, and calculus deals with the infinite limit of a finite series of infinitesimally thin rectangles.

That is

# int_a^b \ f(x) \ dx = lim_(n rarr oo) (b-a)/n sum_(i=1)^n \ f(a + i(b-a)/n)#

And we partition the interval #[a,b]# equally spaced using:

# Delta = {a+0((b-a)/n), a+1((b-a)/n), ..., a+n((b-a)/n) } #
# \ \ \ = {a, a+1((b-a)/n),a+2((b-a)/n), ... ,b } #

Here we have #f(x)=sinx# and so we partition the interval #[0,pi/2]# using:

# Delta = {0, 0+1(pi/2)/n, 0+2 (pi/2)/n, 0+3 (pi/2)/n, ..., pi/2 } #

And so:

# I = int_0^(pi/2) \ sinx \ dx #
# \ \ = lim_(n rarr oo) (pi/2)/n sum_(i=1)^n \ f(0+i*(pi/2)/n)#
# \ \ = lim_(n rarr oo) pi/(2n) sum_(i=1)^n \ sin((ipi)/(2n) )# ..... [A}

If we were dealing with polynomials, we would now utilise standard summation formulas for powers, but as we are dealing with a sine summation those results will not help.

In order to evaluate the sine sum, We consider the following:

# 1 + z + z^2 + ... + z^n = (z^(n+1)-1)/(z-1) \ \ # for z #!= 1#

in combination with Euler's formula by taking

#z = e^(i theta) =cos theta+i sin theta#

Then applying De Moivre's theorem:

# e^(i n theta)=cosntheta+isin n theta #

and equating real and imaginary parts, we eventually find that:

# sum_(j=1)^n sin(jtheta) = (cos(theta/2)-cos(n+1/2)theta) / (2sin(theta/2)#
# " " = (cos(theta/2)-cos(ntheta+theta/2)) / (2sin(theta/2)#
# " " = (cos(theta/2)-(cos ntheta cos(theta/2) - sinnthetasin(theta/2))) / (2sin(theta/2)#

# " " = (cos(theta/2)-cos ntheta cos(theta/2) + sinnthetasin(theta/2)) / (2sin(theta/2))#

So, if we put #theta=(pi)/(2n)# and use this result in the earlier summation [A], we get:

# I = lim_(n rarr oo) pi/(2n) {(cos((pi)/(4n))-cos (pi/2) cos((pi)/(4n)) + sin (pi/2) sin((pi)/(4n))) / (2sin(pi/(4n)))}#

# \ \ = lim_(n rarr oo) pi/(2n) {(cos((pi)/(4n)) + sin((pi)/(4n))) / (2sin(pi/(4n)))}#

# \ \ = pi/4 \ lim_(n rarr oo) 1/(n) (cot((pi)/(4n)) + 1 ) #

# \ \ = pi/4 { lim_(n rarr oo) 1/(n) (1/tan((pi)/(4n))) + lim_(n rarr oo) 1/(n) }#

# \ \ = pi/4 { lim_(n rarr oo) 1/(n) (( pi/(4n))/tan((pi)/(4n))) * (4n)/pi + 0 }#

# \ \ = lim_(n rarr oo) ( pi/(4n))/tan((pi)/(4n)) #

And a standard calculus limit is:

# lim_(x rarr 0) x/tanx =1 #

Utilising this result with #x=pi/(4n)# we get:

# I = 1 #

Using Calculus

If we use Calculus and our knowledge of integration to establish the answer, for comparison, we get:

# int_0^(pi/2) \ sinx \ dx = [ -cos ]_0^(pi/2) #
# " " = -(cos (pi/2)-cos0) #
# " " = -(0-1) #
# " " = 1 #

We are given;

#f(x)<=0#
#g(x) > 0#
#h(x) = f(x)/g(x)#

The first states that #f(x)# is negative or zero, the second that #g(x)# is strictly positive, and the third establishes a new function, #h(x)# which is a ratio of the first two. We are also told that both are differentiable and increasing.

If a function is increasing, then we can infer that its derivative function must be positive. This can be written as;

#f'(x) > 0#
#g'(x) > 0#

In order to show that #h(x)# is increasing, we must show that #h'(x)# is also positive. We can find #h'(x)# using the quotient rule.

#h'(x)= (f'(x)g(x) - f(x)g'(x))/(g^2(x))#

With the information we have, only one term is potentially negative, #f(x)#, and #f(x)# appears only once, in the subtracted term in the numerator. When subtracting a negative, the sign becomes positive, so the numerator is therefore positive. The denominator must also be positive, as #g(x)# is strictly positive, so #h'(x)# is positive. Notice that #f(x)# can also be 0, in which case, #h'(x)# is still positive. #h(x)# is therefore increasing.

Of course, we already made the assumption that #h(x)# is differentiable. Returning to #h'(x)#, the denominator term is strictly positive, so we don't have to worry about any 0's there. Furthermore, we are told that #f(x)# and #g(x)# are differentiable, so they must also be continuous, and #f'(x)# and #g'(x)# exist. Therefore, #h'(x)# exists, so #h(x)# is differentiable.

Lastly, #h(x)# is indeed negative for all points where it is not 0.
The numerator #f(x)# is negative or 0, and the denominator, #g(x)#, is strictly positive, and a ratio of a positive and a negative is always negative.

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