Question

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

Answer 1

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

Explanation:

.

`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`

Answer 2

`V=pi^2/4`

Explanation:

.

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`