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

Apr 9, 2018

$\int \frac{1}{{x}^{2} \sqrt{{x}^{2} - 16}} \mathrm{dx} = \frac{\sqrt{{x}^{2} - 16}}{16 x} + C$

#### Explanation:

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$\int \frac{1}{{x}^{2} \sqrt{{x}^{2} - 16}} \mathrm{dx}$

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 $\alpha$:

$\sin \alpha = \frac{4}{x}$

$\cos \alpha = \frac{\sqrt{{x}^{2} - 16}}{x}$

$\tan \alpha = \frac{4}{\sqrt{{x}^{2} - 16}}$

$x = \frac{4}{\sin} \alpha$

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

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

Let's take the derivative of $\sin \alpha$:

$\cos \alpha \mathrm{da} l p h a = - \frac{4}{x} ^ 2 \mathrm{dx}$

$\mathrm{dx} = \frac{- {x}^{2} \cos \alpha \mathrm{da} l p h a}{4}$

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

$\mathrm{dx} = \frac{- \frac{16}{\sin} ^ 2 \alpha \cos \alpha \mathrm{da} l p h a}{4}$

$\mathrm{dx} = \frac{- 4 \cos \alpha}{\sin} ^ 2 \alpha \mathrm{da} l p h a$

$\frac{1}{\sqrt{{x}^{2} - 16}} = \tan \frac{\alpha}{4} = \sin \frac{\alpha}{4 \cos \alpha}$

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

$\int \frac{1}{{x}^{2} \sqrt{{x}^{2} - 16}} \mathrm{dx} = \int \frac{1}{x} ^ 2 \cdot \frac{1}{\sqrt{{x}^{2} - 16}} \cdot \mathrm{dx} =$

$\int {\sin}^{2} \frac{\alpha}{16} \cdot \sin \frac{\alpha}{4 \cos \alpha} \cdot \left(\frac{- 4 \cos \alpha}{\sin} ^ 2 \alpha\right) \mathrm{da} l p h a =$

$\int \frac{\cancel{\textcolor{red}{{\sin}^{2} \alpha}}}{16} \cdot \sin \frac{\alpha}{\cancel{\textcolor{red}{4 \cos \alpha}}} \cdot \left(\frac{- \cancel{\textcolor{red}{4 \cos \alpha}}}{\cancel{\textcolor{red}{{\sin}^{2} \alpha}}}\right) \mathrm{da} l p h a =$

$- \frac{1}{16} \int \sin \alpha \mathrm{da} l p h a = \frac{1}{16} \cos \alpha + C$

Now, we can substitute back:

$\int \frac{1}{{x}^{2} \sqrt{{x}^{2} - 16}} \mathrm{dx} = \frac{\sqrt{{x}^{2} - 16}}{16 x} + C$

Apr 11, 2018

$V = {\pi}^{2} / 4$

#### Explanation:

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Find the volume of solid generated from revolving a region bounded by $y = \sin x$ and $y = 0$ between $x = \frac{\pi}{2}$ 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 $y = \sin x$, the $x$-axis , $x = \frac{\pi}{2}$, 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 $\mathrm{dx}$.

The circles have a radius $r$ that is equal to $y = \sin x$ 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 $A = \pi {r}^{2} ,$, we can calculate the area of each disc as:

$A = \pi {\sin}^{2} x$

Now if we take the integral of this function and evaluate it between $\frac{\pi}{2}$ 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.

$V = {\int}_{\frac{\pi}{2}}^{\pi} \pi {\sin}^{2} x \mathrm{dx} = \pi {\int}_{\frac{\pi}{2}}^{\pi} \frac{1 - \cos 2 x}{2} \mathrm{dx}$

$V = \frac{\pi}{2} {\int}_{\frac{\pi}{2}}^{\pi} \mathrm{dx} - \frac{\pi}{2} {\int}_{\frac{\pi}{2}}^{\pi} \cos 2 x \mathrm{dx}$

$V = \frac{\pi}{2} x - \frac{\pi}{2} I$

$I = {\int}_{\frac{\pi}{2}}^{\pi} \cos 2 x \mathrm{dx}$

Let $u = 2 x$

$\mathrm{du} = 2 \mathrm{dx}$

$\mathrm{dx} = \frac{\mathrm{du}}{2}$

Let's substitute:

$I = \int \cos u \frac{\mathrm{du}}{2} = \frac{1}{2} \int \cos u \mathrm{du} = \frac{1}{2} \sin u = \frac{1}{2} \sin 2 x$

Let's plug this in:

$V = \frac{\pi}{2} x - \frac{\pi}{2} \cdot \frac{1}{2} \sin 2 x = {\left(\frac{\pi}{4} \left(2 x - \sin 2 x\right)\right)}_{\frac{\pi}{2}}^{\pi}$

$V = \frac{\pi}{4} \left(2 \pi - \sin 2 \pi - \pi + \sin \pi\right)$

$V = \frac{\pi}{4} \left(\pi - 0 + 0\right)$

$V = {\pi}^{2} / 4$