Calculus Help: Find Answers to Questions on Volume & Integration

[Bazzdc89]

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
<br /> \int \frac{1}{(x-1)\sqrt{x^2-2x}} \, dx<br />

Homework Equations

The Attempt at a Solution

so I started with

<br /> u=x^2-2x\text<br />
<br /> {du}=2(x-1)\text{dx}<br />
<br /> \frac{\text{du}}{2}=(x-1)\text{dx}<br />

then i letu=x-1
so u+1=x

that got me
<br /> \frac{1}{2}\int \frac{1}{(u+1)\sqrt{u}} \, du<br />

I then changed that too
<br /> \frac{1}{2}\int \frac{1}{u^{\frac{3}{2}}+\sqrt{u}} \, du<br />

then I let
<br /> v=\sqrt{u}<br />
<br /> v^2=u<br />

<br /> 2\text{vdv}=\text{du}<br />

<br /> \frac{2}{2}\int \frac{v}{\left(v^2\right)^{\frac{3}{2}}+\sqrt{v^2}}<br />
Simplifying I got
<br /> \int \frac{1}{v^2+1} \, dv<br />
Then from there it was easy
<br /> \text{ArcTan}[v]<br />
<br /> {ArcTan}\left[\sqrt{u}\right]<br />
<br /> {ArcTan}\left[\sqrt{x^2-2x}\right]<br />

so that was my final answer
Here is my second question with deals with volume of a solid under revolutionFind 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
<br /> \int \text{Cos}[x]^2-\text{Sin}[x]^2dx<br />
Minus

a= pi/4 , b= pi/2
<br /> \int \text{Sin}[x]^2-\text{Cos}[x]^2dx<br />which resulted my in getting
\pi

My teachers answers to the two questions
For the first one
<br /> \sec ^{-1}(x-1)<br />
For the second one
<br /> \left(\frac{\pi }{2}-1\right)\pi <br />

 

[snipez90]

Your attempt at the first problem is wrong because you have u = x^2 - 2x and u = x-1. You can't define u to be two different expressions and then substitute u back in.

So I completely forgot how to calculate volumes, but in the washer method don't you subtract two integrals that have the same limits of integration?

 

[Bazzdc89]

I noticed that I did use two U substitutions, but maple and wolfram are giving me the same answer so I'm kind of confused on that first step of substitution.

And for washer you do the integration of from a to b, "Big radius"- "small radius" , squaring each of the radius,and for my problem i had to do the washer method twice cause the graph of cos(x)>sin(x),from 0-->Pi/4 ; and the graph of sin(x) > cos(x) from Pi/4 --> Pi/2

 

[snipez90]

Well the issue is that you're basically treating those two expressions to be the same. Not only that, but technically you made a mistake in substituting u since you put u = x-1, so you should end up with u^(3/2) in the denominator (temporarily allowing the substitution). Thus I think it was by coincidence that whatever calculations you made resulted in the correct antiderivative. The point is you can't just let u be two different things. The solution given by wolfram alpha is completely different, and does not let u be two different expressions.

 

[Bazzdc89]

Alright so I figured out what my problem was for question #1, It was coincidence how I got the right answer, and I figured out the proper way to solve the integration.

Any idea if I am solving question #2 right?