Integral $\int \frac{d x}{x^{2} \sqrt{x^{2} - 16}}$ $int dx/(x^2sqrt(x^2-16))$ ?

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Integral $\int \frac{d x}{x^{2} \sqrt{x^{2} - 16}}$ $int dx/(x^2sqrt(x^2-16))$ ?

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Answer 1 · 2018-04-09T21:57:23

$\int \frac{1}{x^{2} \sqrt{x^{2} - 16}} d x = \frac{\sqrt{x^{2} - 16}}{16 x} + C$ $int1/(x^2sqrt(x^2-16))dx=sqrt(x^2-16)/(16x)+C$

Explanation:

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$\int \frac{1}{x^{2} \sqrt{x^{2} - 16}} d x$ $int1/(x^2sqrt(x^2-16))dx$

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

enter image source here

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

$sin α = \frac{4}{x}$ $sinalpha=4/x$

$cos α = \frac{\sqrt{x^{2} - 16}}{x}$ $cosalpha=sqrt(x^2-16)/x$

$tan α = \frac{4}{\sqrt{x^{2} - 16}}$ $tanalpha=4/sqrt(x^2-16)$

$x = \frac{4}{sin α}$ $x=4/sinalpha$

$x^{2} = \frac{16}{{sin}^{2} α}$ $x^2=16/sin^2alpha$

$\frac{1}{x^{2}} = \frac{{sin}^{2} α}{16}$ $1/x^2=sin^2alpha/16$

Let's take the derivative of $sin α$ $sinalpha$ :

$cos α d α = - \frac{4}{x^{2}} d x$ $cosalphadalpha=-4/x^2dx$

$d x = \frac{- x^{2} cos α d α}{4}$ $dx=(-x^2cosalphadalpha)/4$

Let's plug in for $x^{2}$ $x^2$ :

$d x = \frac{- \frac{16}{{sin}^{2} α} cos α d α}{4}$ $dx=(-16/sin^2alphacosalphadalpha)/4$

$d x = \frac{- 4 cos α}{{sin}^{2} α} d α$ $dx=(-4cosalpha)/sin^2alphadalpha$

$\frac{1}{\sqrt{x^{2} - 16}} = \frac{tan α}{4} = \frac{sin α}{4 cos α}$ $1/sqrt(x^2-16)=tanalpha/4=sinalpha/(4cosalpha)$

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

$\int \frac{1}{x^{2} \sqrt{x^{2} - 16}} d x = \int \frac{1}{x^{2}} \cdot \frac{1}{\sqrt{x^{2} - 16}} \cdot d x =$ $int1/(x^2sqrt(x^2-16))dx=int1/x^2*1/sqrt(x^2-16)*dx=$

$\int \frac{{sin}^{2} α}{16} \cdot \frac{sin α}{4 cos α} \cdot (\frac{- 4 cos α}{{sin}^{2} α}) d α =$ $intsin^2alpha/16*sinalpha/(4cosalpha)*((-4cosalpha)/sin^2alpha)dalpha=$

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

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

Now, we can substitute back:

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

Answer 2 · 2018-04-11T07:29:46

$V=pi^2/4$

Explanation:

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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:

enter image source here

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

enter image source here

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:

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

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

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

$V=pi^2/4$

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Integral $\int \frac{d x}{x^{2} \sqrt{x^{2} - 16}}$ $int dx/(x^2sqrt(x^2-16))$ ?

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Integral $\int \frac{d x}{x^{2} \sqrt{x^{2} - 16}}$ $int dx/(x^2sqrt(x^2-16))$ ?