Integral #int dx/(x^2sqrt(x^2-16))#?

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#int1/(x^2sqrt(x^2-16))dx#

We will use trigonometric substitution to solve this. Let's set up our triangle:

Now, let's write the basic three trigonometric functions for angle #alpha#:

#sinalpha=4/x#

#cosalpha=sqrt(x^2-16)/x#

#tanalpha=4/sqrt(x^2-16)#

#x=4/sinalpha#

#x^2=16/sin^2alpha#

#1/x^2=sin^2alpha/16#

Let's take the derivative of #sinalpha#:

#cosalphadalpha=-4/x^2dx#

#dx=(-x^2cosalphadalpha)/4#

Let's plug in for #x^2#:

#dx=(-16/sin^2alphacosalphadalpha)/4#

#dx=(-4cosalpha)/sin^2alphadalpha#

#1/sqrt(x^2-16)=tanalpha/4=sinalpha/(4cosalpha)#

Let's plug in all the pieces to convert our integral into a trigonometric integral:

#int1/(x^2sqrt(x^2-16))dx=int1/x^2*1/sqrt(x^2-16)*dx=#

#intsin^2alpha/16*sinalpha/(4cosalpha)*((-4cosalpha)/sin^2alpha)dalpha=#

#intcancelcolor(red)(sin^2alpha)/16*sinalpha/(cancelcolor(red)(4cosalpha))*((-cancelcolor(red)(4cosalpha))/cancelcolor(red)(sin^2alpha))dalpha=#

#-1/16intsinalphadalpha=1/16cosalpha+C#

Now, we can substitute back:

#int1/(x^2sqrt(x^2-16))dx=sqrt(x^2-16)/(16x)+C#

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Find the volume of solid generated from revolving a region bounded by #y=sinx# and #y=0# between #x=pi/2# and #x=pi# around the #x#-axis.

#y=0# is the #x#-axis. As such, we want to revolve the area between the curve of #y=sinx#, the #x#-axis , #x=pi/2#, and #x=pi# around the #x#-axis and calculate the volume of the solid generated.

The graph below shows this area:

If we revolve this area around the #x#-axis we will get the solid shown below:

If you can imagine this solid being divided into vertical slices parallel to the #y#-axis with a extremely small thicknesses each one would look like a thin disc with the surface area of a circle and thickness of #dx#.

The circles have a radius #r# that is equal to #y=sinx# and vary in size depending on where on the #x#-axis you perform the slice.

So, because the formula for the area of a circle is #A=pir^2,#, we can calculate the area of each disc as:

#A=pisin^2x#

Now if we take the integral of this function and evaluate it between #pi/2# and #pi# we will have the volume of the solid.

This is because the integral adds the areas of infinite number of discs between the two limits together.

#V=int_(pi/2)^pipisin^2xdx=piint_(pi/2)^pi(1-cos2x)/2dx#

#V=pi/2int_(pi/2)^pidx-pi/2int_(pi/2)^picos2xdx#

#V=pi/2x-pi/2I#

#I=int_(pi/2)^picos2xdx#

Let #u=2x#

#du=2dx#

#dx=(du)/2#

Let's substitute:

#I=intcosu(du)/2=1/2intcosudu=1/2sinu=1/2sin2x#

Let's plug this in:

#V=pi/2x-pi/2*1/2sin2x=(pi/4(2x-sin2x))_(pi/2)^pi#

#V=pi/4(2pi-sin2pi-pi+sinpi)#

#V=pi/4(pi-0+0)#

#V=pi^2/4#