# Solving an integral by substitution (two questions)

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• December 3rd 2007, 09:22 PM
ebonyscythe
Solving an integral by substitution (two questions)
1. Evaluate $\int_{e}^{e^4} \frac{1}{x\sqrt{ln(x)}} dx$.

2. Evaluate $\int_{0}^{2\pi} |cos(x)^2sin(x)| dx$
--------
I know I need to use substitutions in both cases but I'm not sure what to substitute so it works out nice... I've tried a few things like $u=\sqrt{ln(x)}$ for the first one but I couldn't get $du$ to work out...
• December 3rd 2007, 09:47 PM
Jhevon
Quote:

Originally Posted by ebonyscythe
1. Evaluate $\int_{e}^{e^4} \frac{1}{x\sqrt{ln(x)}} dx$.

u = ln x

Quote:

2. Evaluate $\int_{0}^{2\pi} |cos(x)^2sin(x)| dx$
u = cos(x)
• December 3rd 2007, 09:58 PM
DivideBy0
What is the effect of the absolute value signs in the second question? (Thinking)
• December 3rd 2007, 10:23 PM
ebonyscythe
Thanks much... I got the first question just fine after that, but when I try to use u=cos(x), then wouldn't du=-sin(x)dx? I'm having a tough time with the absolute value.

My train of thought for the second one... where am I going wrong?

$-\int_{0}^{2\pi}cos(x)^2*-|sin(x)| dx
=-\int_{0}^{1} cos(x) du
=-(sin(1)-sin(0))
=-sin(1)
$

Which I know is wrong because I got 4/3 from Derive...
• December 3rd 2007, 10:34 PM
Jhevon
Quote:

Originally Posted by DivideBy0
What is the effect of the absolute value signs in the second question? (Thinking)

it affects the limit. you may have to split up the integral

it will be: $\int_0^{\pi}\cos^2 x \sin x~dx + \int_{\pi}^{2 \pi} -\cos^2 x \sin x~dx$
• December 3rd 2007, 10:50 PM
ebonyscythe
How would I know to split the interval at pi? Maybe it's because it's two in the morning but this problem is giving me some real trouble... so if I split the interval and substitute... would I get

$\int_{0}^{-1}cos(x)u*du - \int_{-1}^{1}cos(x)u*du$?
• December 3rd 2007, 10:53 PM
Jhevon
Quote:

Originally Posted by ebonyscythe
How would I know to split the interval at pi? Maybe it's because it's two in the morning but this problem is giving me some real trouble... so if I split the interval and substitute... would I get

$\int_{0}^{-1}cos(x)u*du - \int_{-1}^{1}cos(x)u*du$?

the absolute values turns the function positive when it is negative. thus you must find the interval on which the function is negative, and integrate the negative of the function (which will be positive) over that interval. graphing is the easiest way to see where it is negative, and it is negative on pi to 2pi in that interval
• December 3rd 2007, 11:00 PM
ebonyscythe
Ah, okay... I was thinking for some reason that I wanted the interval of which cos(x) is negative instead of the whole function cos(x)^2sin(x)... dumb mistake made, thanks for explaining.

Anyways, not sure if you're getting to it but am I heading in the right direction for the substitution?
• December 3rd 2007, 11:01 PM
Jhevon
Quote:

Originally Posted by ebonyscythe
Ah, okay... I was thinking for some reason that I wanted the interval of which cos(x) is negative instead of the whole function cos(x)^2sin(x)... dumb mistake made, thanks for explaining.

Anyways, not sure if you're getting to it but am I heading in the right direction for the substitution?

...um, no actually. write out all your steps, what you wrote so far makes little sense
• December 3rd 2007, 11:13 PM
ebonyscythe
$\int_{0}^{\pi} cos(x)^2 sin(x) dx + \int_{\pi}^{2\pi} -cos(x)^2 sin(x) dx$

Let u = cos(x) so du = -sin(x) dx
*need a -1 in the first integrand to substitute, so multiply by a factor of one
$(-1)\int_{0}^{\pi} cos(x)^2 sin(x) (-1)dx + \int_{\pi}^{2\pi} -cos(x)^2 sin(x) dx$

substituting, I get...
$-\int_{0}^{-1} cos(x)*u*du + \int_{-1}^{1} cos(x)*u*du$

I'm trying to make sense of what I've got in my notes, so I've no idea where to go next or what I'm doing wrong here...
• December 3rd 2007, 11:25 PM
Jhevon
Quote:

Originally Posted by ebonyscythe
$\int_{0}^{\pi} cos(x)^2 sin(x) dx + \int_{\pi}^{2\pi} -cos(x)^2 sin(x) dx$

Let u = cos(x) so du = -sin(x) dx
*need a -1 in the first integrand to substitute, so multiply by a factor of one
$(-1)\int_{0}^{\pi} cos(x)^2 sin(x) (-1)dx + \int_{\pi}^{2\pi} -cos(x)^2 sin(x) dx$

substituting, I get...
$-\int_{0}^{-1} cos(x)*u*du + \int_{-1}^{1} cos(x)*u*du$

I'm trying to make sense of what I've got in my notes, so I've no idea where to go next or what I'm doing wrong here...

you said cos(x) is u, i don't understand why you still have cos(x) in your integrand
• December 3rd 2007, 11:30 PM
ebonyscythe
Ah crud, it's u^2, not cos(x)*u... right? That would make more sense... So...

$\int_{0}^{\pi} cos(x)^2 sin(x) dx + \int_{\pi}^{2\pi} -cos(x)^2 sin(x) dx$

Let u = cos(x) so du = -sin(x) dx
$(-1)\int_{0}^{\pi} cos(x)^2 sin(x) (-1)dx + \int_{\pi}^{2\pi} -cos(x)^2 sin(x) dx$

$-\int_{0}^{-1} u^2*du + \int_{-1}^{1} u^2*du$
• December 4th 2007, 12:01 AM
angel.white
Also, a friendly warning.

$cos(x)^{2}$ looks like $cos(x^{2})$

which is not the same as $cos^{2}(x)$

:)

$y=ln\left(\frac{4+\sqrt{2}x^{3}}{4+2x^{3}} \right)^{3/5}$

And I interpreted the exponent 3/5 as being applied to the entire natural log, rather than to the inside of the natural log. Cost me 2.5% of my total test grade -.-
• December 4th 2007, 11:04 AM
Jhevon
Quote:

Originally Posted by angel.white
Also, a friendly warning.

$cos(x)^{2}$ looks like $cos(x^{2})$

which is not the same as $cos^{2}(x)$

:)

$y=ln\left(\frac{4+\sqrt{2}x^{3}}{4+2x^{3}} \right)^{3/5}$

You are correct. I was going to make a similar comment last night...don't know why i didn't. i'm getting too old.

Quote:

And I interpreted the exponent 3/5 as being applied to the entire natural log, rather than to the inside of the natural log. Cost me 2.5% of my total test grade -.-
so you only got 97.5%? :D
• December 4th 2007, 11:22 AM
angel.white
Quote:

Originally Posted by Jhevon
You are correct. I was going to make a similar comment last night...don't know why i didn't. i'm getting too old.

so you only got 97.5%? :D

:) I got 74/80 = 92.5%

Also lost a point for incorrectly transcribing a correct answer, explaining in a paragraph why one of the limit questions must be equal to zero rather than manipulating it to be equal to zero, and 2 points for forgetting the very last link in a chain rule differentiation.

But then my instructor curved it, so the grade on the books is out of 74 ;)
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