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Featured 4 days ago

For a non-rigorous proof, please see below.

For a positive central angle of

Source:

The geometric idea is that

So we have:

For small positive

So

so

We also have, for these small

so

Since both one sided limits are

Featured 2 weeks ago

Let

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

Factor out a

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

Rewrite in terms of

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

For ease of reading, apply the substitution

#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#

Featured 1 week ago

The slope is:

The slope will be given by

To get the equation in terms of

So here, where

#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

#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

Just so we don't have to write out a giant fraction, let's evaluate

#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

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

Featured 6 days ago

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:

In other "words":

i) Necessary condition:

Using polynomial long division (taking some time to find

Now using the pq-formula for quadratic equations:

ii) Necessary and sufficient condition:

iii) y - values and points

Provided that

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

I hope it helps.

Featured 4 days ago

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

By definition of an integral, then

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

represents the area under the curve

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

# 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

# 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

# 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

# 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 #

Featured 4 days ago

We are given;

The first states that

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

In order to show that

With the information we have, only one term is potentially negative,

Of course, we already made the assumption that

Lastly,

The numerator

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