1. ## Integration

Could any one possibly help me these questions. I have written the question and then my attempt at the solution next.

Find each of the following indefinite integrals, identifying any general rules
of calculus that you use.

[integral]xcos(1/3x)dx

Solving the equation by integration by parts. Let f(x)=x and g'(x)=cos⁡(1/3x); then f'(x)=1 and g(x)=3 sin⁡(1/3x) Substituting into the equation for integration by parts f(x)g(x)-∫f ' (x)g(x)dx. we obtain ∫xcos(1/3x)dx=3xsin(1/3x)-∫1(3 sin⁡(1/3)x)=3xsin(1/3x)-3∫sin⁡(1/3)x=3xsin(1/3 x)+9cos⁡(1/3 x)+C

x/√1-x^4

Take u=1-x^4; thendu/dx=-4x^3. Hence∫x/√(1-x^4) dx=x/√u (-4x^3 )dx=x/√u du= -1/2 u^(-3/2). Plugging u back into the equation1/2 4x^3 ^(-3/2)

Explain why the graph of the function f(x) = (x2 + 1) ln(1/2x) lies
below the x-axis for 1 < x < 2 and above the x-axis for 2 < x < 3.

The graph of the function f(x)=(x^2+1) ln⁡(1/2x) is positive above the x-axis and negative below the x-axis.

Use this fact to find the area enclosed by the graph and the x-axis
between x = 1 and x = 3, giving your answer to five decimal
places.

I have no idea how to answer this.

Find the volume of the solid of revolution obtained when the graph of
f(x)= sec x + tan x, from x=−[pie]/3 to x = [pie]/4

The volume is [pie]∫(-[pie]/3) ([pie]/4) secx+tanx=[pie](ln⁡(tan⁡(x)+sec⁡(x) )+ln⁡(sec⁡(x)))=[pie]([(ln⁡(tan⁡(-[pie]/3)+sec⁡(-[pie]/3)+ln⁡(sec⁡(-[pie]/3) ]-[(ln⁡(tan⁡([pie]/4)+sec⁡([pie]/4)+ln⁡(sec⁡([pie]/4)]=[pie [ln⁡(-√3+2)+ln⁡(2)-[(ln⁡(1+√2)+ln(√2)]=[pie][-0.63-1.2)=-1.83[pie]

2. ## Re: Integration

Hello, thomasthetankengine!

Thank you for showing your work!

$\int x\cos(\tfrac{x}{3})\,dx$

Solving the equation by integration by parts.

. . $\begin{Bmatrix}u &=& x && dv &=& \cos(\frac{x}{3}) \\ du &=&dx && v &=& 3\sin(\frac{x}{3})\end{Bmatrix}$

Then: . $\int x\cos(\tfrac{x}{3})\,dx \;=\;3x\sin(\tfrac{x}{3}) - 3\int\sin(\tfrac{x}{3})\,dx \;=\;3x\sin(\tfrac{x}{3}) + 9\cos(\tfrac{x}{3}) + C$

Excellent!
Good work!

$\int\frac{x\,dx}{\sqrt{1-x^4}}$

Let $u = x^2 \quad\Rightarrow\quad du = 2x\,dx \quad\Rightarrow\quad x\,dx \,=\,\tfrac{1}{2}du$

Substitute: . $\int \frac{\frac{1}{2}\,du}{\sqrt{1-u^2}} \;=\;\tfrac{1}{2}\int\frac{du}{\sqrt{1-u^2}} \;=\;\tfrac{1}{2}\arcsin u + C$

Back-substitute: . $\tfrac{1}{2}\arcsin(x^2) + C$

3. ## Re: Integration

Thanks. I can see where I have went wrong now. Is there any chance that you can help me with finding the volume of the solid of revolution obtained when the graph of secx+tanx, from x=-pie/3 to x=pie/4, is rotated about the x-axis. I understand that I need to compute the integral then set the limits of integration and then mulitply that by pie? The only problem that I'm having is computing the integral of secx+tanx. I have got secx+tanx=ln(tan(x)+sec(x)+ln(sec). I'm not sure if this is correct, I don't think it is. I'm not sure if the the integral of secx is ln(tan(x)+sec(x)) or ln(sec(x)+tan(x). Also the integral of tan(x) i assume that the integral of tan(x)=-lncos(x). I'm really not sure what i'm doing wrong. Anyhelp would be great. Thanks.

4. ## Re: Integration

Hello, thomasthetankengine!

You forgot to square the function.

$\text{Find the volume of the solid of revolution obtained when the graph}$
$\text{of }\,y \:=\:\sec x+\tan x\,\text{ from }x=\text{-}\tfrac{\pi}{3}\text{ to }x=\tfrac{\pi}{4}\,\text{ is rotated about the x-axis.}$

I'll omit the limits for now . . .

$V \;=\;\pi\int (\sec x + \tan x)^2\,dx \;=\;\pi \int\left(\sec^2\!x + 2\sec x\tan x + \tan^2\!x\right)\,dx$

. . $=\;\pi\left[\int\sec^2\!x\,dx + 2\int\sec x\tan x\,dx +\int\tan^2\!x\,dx\right]$

The first two integrals have standard formulas:

. . $\int \sec^2\!x\,dx \;=\;\tan x + C$

. . $\int\sec x\tan x\,dx \;=\;\sec x + C$

The third can handled like this:

. . $\int\tan^2\!x\,dx \;=\;\int(\sec^2\!x - 1)\,dx \;=\;\tan x - x+C$

5. ## Re: Integration

Originally Posted by thomasthetankengine
Thanks. I can see where I have went wrong now. Is there any chance that you can help me with finding the volume of the solid of revolution obtained when the graph of secx+tanx, from x=-pie/3 to x=pie/4, is rotated about the x-axis. I understand that I need to compute the integral then set the limits of integration and then mulitply that by pie? The only problem that I'm having is computing the integral of secx+tanx. I have got secx+tanx=ln(tan(x)+sec(x)+ln(sec). I'm not sure if this is correct, I don't think it is. I'm not sure if the the integral of secx is ln(tan(x)+sec(x)) or ln(sec(x)+tan(x). Also the integral of tan(x) i assume that the integral of tan(x)=-lncos(x). I'm really not sure what i'm doing wrong. Anyhelp would be great. Thanks.
Because I don't like using a lot of integration formulas, I tend to prefer converting things to sines and cosines as they're usually easy to integrate. I usually only remember \displaystyle \begin{align*} \int{\sin{(x)}\,\mathrm{d}x} = -\cos{(x)} + C, \int{\cos{(x)}\,\mathrm{d}x} = \sin{(x)} + C, \int{\sec^2{(x)}\,\mathrm{d}x} = \tan{(x)} + C \end{align*} and a few important identities (Pythagorean, Double Angle and Compound Angle).

Your volume will be calculated as:

\displaystyle \begin{align*} V &= \int_{-\frac{\pi}{3}}^{\frac{\pi}{4}}{ \pi \left[ \sec{(x)} + \tan{(x)} \right] ^2 \, \mathrm{d}x } \\ &= \pi \int_{-\frac{\pi}{3}}^{\frac{\pi}{4}}{ \sec^2{(x)} + 2\sec{(x)}\tan{(x)} + \tan^2{(x)}\,\mathrm{d}x } \\ &= \pi \int_{-\frac{\pi}{3}}^{\frac{\pi}{4}}{ \sec^2{(x)} + 2 \left[ \frac{1}{\cos{(x)}} \right] \left[ \frac{\sin{(x)}}{\cos{(x)}} \right] + \sec^2{(x)} - 1 \,\mathrm{d}x } \\ &= \pi \int_{-\frac{\pi}{3}}^{\frac{\pi}{4}}{ 2\sec^2{(x)} + \frac{2\sin{(x)}}{\cos^2{(x)}} - 1 \, \mathrm{d}x } \\ &= \pi \left\{ 2\int_{-\frac{\pi}{3}}^{\frac{\pi}{4}}{ \sec^2{(x)}\,\mathrm{d}x } - 2\int_{-\frac{\pi}{3}}^{\frac{\pi}{4}}{ \cos^{-2}{(x)} \left[ -\sin{(x)} \right]\,\mathrm{d}x } - \int_{-\frac{\pi}{3}}^{\frac{\pi}{4}}{1\,\mathrm{d}x} \right\} \\ &= \pi \left\{ 2\left[ \tan{(x)} \right]_{-\frac{\pi}{3}}^{\frac{\pi}{4}} - 2 \int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}}{u^{-2}\,\mathrm{d}u} - \left[ x \right]_{-\frac{\pi}{3}}^{\frac{\pi}{4}} \right\} \textrm{ after making the substitution } u = \cos{(x)} \implies \mathrm{d}u = -\sin{(x)}\,\mathrm{d}x \\ &= \pi \left\{ \left[ \tan{ \left( \frac{\pi}{4} \right) } - \tan{ \left( -\frac{\pi}{3} \right) } \right] - 2 \left[ -u^{-1} \right]_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} - \left[ \frac{\pi}{4} - \left( -\frac{\pi}{3} \right) \right] \right\} \\ &= \pi \left[ 1 + \sqrt{3} - 2 \left( -2 + \sqrt{2} \right) - \frac{7\pi}{12} \right] \\ &= \pi \left( 1 + \sqrt{3} + 4 - 2\sqrt{2} - \frac{7\pi}{12} \right) \\ &= \pi \left( 5 - 2\sqrt{2} + \sqrt{3} - \frac{7\pi}{12} \right) \end{align*}

P.S. I absolutely love your screen name

6. ## Re: Integration

May I ask why the function must be squared?

7. ## Re: Integration

I think I understand where I was going wrong now. I didn't square the function. The formula of the 'Volume of revolution'=[pie]int(f(x))^2dx.

8. ## Re: Integration

Just an aside: \displaystyle \begin{align*} \pi \end{align*} is the Greek letter "pi", not "pie". Interestingly, they use that symbol because it is their letter p, and this number is DEFINED as the Perimeter of a circle with a diameter of 1 unit.

9. ## Re: Integration

Here's what I got so far but i'm really not sure what next to do, or even if it's right.

The volume is π∫▒〖〖(sec⁡〖x+tan⁡〖x)〗 〗〗^2 dx=〗 π∫▒〖(sec^2⁡〖x+2 sec⁡〖x tan⁡〖x+tan^2⁡〖x)〗 〗 〗 〗=π(〗 ∫▒sec^2⁡〖x dx+2∫▒sec⁡〖x tan⁡〖x dx+∫▒tan^2⁡〖x dx=π(tan⁡〖x+sec⁡〖x+〗 〗 〗 〗 〗 〗 tan⁡〖x+c=〗 〖〖(tan〗⁡〖x+sec⁡〖x+tan⁡〖x-x〗)〗 〗〗_(-π/3)^(π/4)=π[〖tan π/4〗⁡〖+sec⁡〖π/4+tan⁡〖π/4-π/4〗]-〖[tan〗⁡〖-π/3+sec⁡〖-π/3+tan⁡〖-π/3∓π/3〗]=π[1+√2+1-π/4]-[+2〗 〗 〗 〗

10. ## Re: Integration

Originally Posted by thomasthetankengine
Here's what I got so far but i'm really not sure what next to do, or even if it's right.

The volume is π∫▒〖〖(sec⁡〖x+tan⁡〖x)〗 〗〗^2 dx=〗 π∫▒〖(sec^2⁡〖x+2 sec⁡〖x tan⁡〖x+tan^2⁡〖x)〗 〗 〗 〗=π(〗 ∫▒sec^2⁡〖x dx+2∫▒sec⁡〖x tan⁡〖x dx+∫▒tan^2⁡〖x dx=π(tan⁡〖x+sec⁡〖x+〗 〗 〗 〗 〗 〗 tan⁡〖x+c=〗 〖〖(tan〗⁡〖x+sec⁡〖x+tan⁡〖x-x〗)〗 〗〗_(-π/3)^(π/4)=π[〖tan π/4〗⁡〖+sec⁡〖π/4+tan⁡〖π/4-π/4〗]-〖[tan〗⁡〖-π/3+sec⁡〖-π/3+tan⁡〖-π/3∓π/3〗]=π[1+√2+1-π/4]-[+2〗 〗 〗 〗
Can I beg you to use Latex (see our help forum) or some other sort of math compiler? Or at least put it on several lines so it isn't bunched up like this. For example, I just tried to translate this out so it is more legible, but your errors in the parenthesis made this astoundingly painful until I scrolled back and found Prove It's post.

Here's the last line as best I can figure:
$= \pi \left [ tan \left ( \frac{\pi}{4} \right ) + sec \left ( \frac{\pi}{4} \right ) + tan \left ( \frac{\pi}{4} - \frac{\pi}{4} \right ) - \left ( \frac{\pi}{3} \right) + sec \left ( -\frac{\pi}{3} \right ) + tan \left ( -\frac{\pi}{3} \mp \frac{\pi}{3} \right ) \right ]$

$= \pi \left [ 1 + \sqrt{2} + 1 - \frac{\pi}{4} \right ] - (2)$

11. ## Re: Integration

Did you read my post? I put the complete solution...