# Thread: Solving an integral by substitution (two questions)

1. ## Solving an integral by substitution (two questions)

1. Evaluate $\displaystyle \int_{e}^{e^4} \frac{1}{x\sqrt{ln(x)}} dx$.

2. Evaluate $\displaystyle \int_{0}^{2\pi} |cos(x)^2sin(x)| dx$
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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 $\displaystyle u=\sqrt{ln(x)}$ for the first one but I couldn't get $\displaystyle du$ to work out...

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

2. Evaluate $\displaystyle \int_{0}^{2\pi} |cos(x)^2sin(x)| dx$
u = cos(x)

3. What is the effect of the absolute value signs in the second question?

4. 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?

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

Which I know is wrong because I got 4/3 from Derive...

5. Originally Posted by DivideBy0
What is the effect of the absolute value signs in the second question?
it affects the limit. you may have to split up the integral

it will be: $\displaystyle \int_0^{\pi}\cos^2 x \sin x~dx + \int_{\pi}^{2 \pi} -\cos^2 x \sin x~dx$

6. 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

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

7. 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

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

8. 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?

9. 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

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

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
$\displaystyle (-1)\int_{0}^{\pi} cos(x)^2 sin(x) (-1)dx + \int_{\pi}^{2\pi} -cos(x)^2 sin(x) dx$

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

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...

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

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
$\displaystyle (-1)\int_{0}^{\pi} cos(x)^2 sin(x) (-1)dx + \int_{\pi}^{2\pi} -cos(x)^2 sin(x) dx$

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

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

12. Ah crud, it's u^2, not cos(x)*u... right? That would make more sense... So...

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

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

$\displaystyle -\int_{0}^{-1} u^2*du + \int_{-1}^{1} u^2*du$

13. Also, a friendly warning.

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

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

I had a test where I had to differentiate:

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

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 -.-

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

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

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

I had a test where I had to differentiate:

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

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%?

15. 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%?
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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