- #1

Bazzdc89

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Hello everyone, first time poster here. Well here is my problem , I recently had a test and right after the test my teacher handed our class the solutions to the test ,there were two problems that my answers conflicted with his. So basically my questions are why are my answers different from his? Am I doing my problems correctly?. I would greatly appreciate it if you could review my answers

Hello everyone, first time poster here. Well here is my problem , I recently had a test and right after the test my teacher handed our class the solutions to the test ,there were two problems that my answers conflicted with his. So basically my questions are why are my answers different from his? Am I doing my problems correctly?. I would greatly appreciate it if you could review my answers

[tex]

\int \frac{1}{(x-1)\sqrt{x^2-2x}} \, dx

[/tex]

## Homework Equations

## The Attempt at a Solution

so I started with

[tex]

u=x^2-2x\text

[/tex]

[tex]

{du}=2(x-1)\text{dx}

[/tex]

[tex]

\frac{\text{du}}{2}=(x-1)\text{dx}

[/tex]

then i let[tex]u=x-1[/tex]

so [tex]u+1=x[/tex]

that got me

[tex]

\frac{1}{2}\int \frac{1}{(u+1)\sqrt{u}} \, du

[/tex]

I then changed that too

[tex]

\frac{1}{2}\int \frac{1}{u^{\frac{3}{2}}+\sqrt{u}} \, du

[/tex]

then I let

[tex]

v=\sqrt{u}

[/tex]

[tex]

v^2=u

[/tex]

[tex]

2\text{vdv}=\text{du}

[/tex]

[tex]

\frac{2}{2}\int \frac{v}{\left(v^2\right)^{\frac{3}{2}}+\sqrt{v^2}}

[/tex]

Simplifying I got

[tex]

\int \frac{1}{v^2+1} \, dv

[/tex]

Then from there it was easy

[tex]

\text{ArcTan}[v]

[/tex]

[tex]

{ArcTan}\left[\sqrt{u}\right]

[/tex]

[tex]

{ArcTan}\left[\sqrt{x^2-2x}\right]

[/tex]

so that was my final answer

**Here is my second question with deals with volume of a solid under revolution**Find the volume of the solid generated by revolving the region bounded by the graphs of

y=sin(x) y=cos(x) x= Pi/2 ( we could use which ever method we preferred, I used washer)

**My attempt Using Washer method**

a=0 , b= pi/4

[tex]

\int \text{Cos}[x]^2-\text{Sin}[x]^2dx

[/tex]

Minus

a= pi/4 , b= pi/2

[tex]

\int \text{Sin}[x]^2-\text{Cos}[x]^2dx

[/tex]which resulted my in getting

[tex]\pi [/tex]

**My teachers answers to the two questions**

For the first one

[tex]

\sec ^{-1}(x-1)

[/tex]

For the second one

[tex]

\left(\frac{\pi }{2}-1\right)\pi

[/tex]

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