# Problems with some integration

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• Jul 21st 2006, 10:54 PM
scorpion007
Problems with some integration
Hi, im new here, I have some problems with the following integrations:
I've tried various things with all of them, but cant figure them out. I'd love some pointers, hopefully with explanations :)

1. $\displaystyle \int \frac {\sin^3(x)} {\cos^2(x)}dx$
2. $\displaystyle \int^{\frac{\pi}{2}}_{0} \frac {\sin(3x)} {\cos(3x)} dx$ with a suitable substitution is? Apparently the answer is: $\displaystyle \int^{1}_{0} \frac {1}{3u} du$, but im not sure what i should pull put as 'u'. Any ideas?
3. The area of the region bounded by the axis, the line x=1 and the curve: $\displaystyle y = \frac {1}{\sqrt{4-3x^2}}$, so i presume we need to get the definite integral from 0-1 of that curve, but once again i cant figure out the answer.

I'll add to this list as more problems arise.

Thanks for the help :)
• Jul 21st 2006, 11:29 PM
CaptainBlack
Quote:

Originally Posted by scorpion007

1. $\displaystyle \int \frac {\sin^3(x)} {\cos^2(x)}dx$

$\displaystyle \int \frac {\sin^3(x)} {\cos^2(x)}dx =\int (\sin(x))^2\ \frac{sin}{(cos(x))^2}dx$.

Then integrate by parts with:

$\displaystyle u=(\sin(x))^2$, and $\displaystyle dv=\frac{sin}{(cos(x))^2}$,

($\displaystyle dv$ is the derivative of $\displaystyle 1/\cos(x)$)

RonL
• Jul 21st 2006, 11:46 PM
CaptainBlack
Quote:

Originally Posted by scorpion007
2. $\displaystyle \int^{\frac{\pi}{2}}_{0} \frac {\sin(3x)} {\cos(3x)} dx$ with a suitable substitution is? Apparently the answer is: $\displaystyle \int^{1}_{0} \frac {1}{3u} du$, but im not sure what i should pull put as 'u'. Any ideas?)

$\displaystyle u=\cos(3x)$

look plausible.

RonL
• Jul 21st 2006, 11:52 PM
CaptainBlack
Quote:

Originally Posted by scorpion007
3. The area of the region bounded by the axis, the line x=1 and the curve: $\displaystyle y = \frac {1}{\sqrt{4-3x^2}}$, so i presume we need to get the definite integral from 0-1 of that curve, but once again i cant figure out the answer.

By the look of it you want:

$\displaystyle \int_0^1\ \frac {1}{\sqrt{4-3x^2}}\ dx=\int_0^1\ \frac {1}{2\ \sqrt{1-(3/4)x^2}}\ dx$.

Try the change of variable $\displaystyle \sqrt{3/4}\ x=\sin(u)$

RonL
• Jul 22nd 2006, 12:45 AM
scorpion007
Hey, thanks for all the help :)

Could i get a little more help as to how to differentiate $\displaystyle \sin^2(x)$? I've forgotten how to do it, and i need to be able to do it for the first question to solve by parts.

EDIT: nevermind, sorry, i remembered how to differentiate sin^2x. and i got the correct answer :)
Time for the other Q's.
• Jul 22nd 2006, 01:46 AM
scorpion007
ok, got the first 2 questions sorted.

But for number 3, im getting the incorrect answer, this is what i did briefly:
$\displaystyle \frac {1}{2} \int^{1}_{0} \frac{1}{\sqrt{1- \frac{3}{4}x^2}} dx = \frac {1}{2} \Bigl[ \arcsin (\frac {\sqrt{3}}{2}x) \Bigr]^{1}_{0}$
Hence:
$\displaystyle \frac{1}{2} \Bigl[ \frac{\pi}{3} - 0 \Bigr] = \frac{\pi}{6}$
right?
This is apparently incorrect, as the answer is suppose to be $\displaystyle \frac {\sqrt{3}\pi}{9}$

Did i screw up somewhere?

EDIT: Nevermind, i missed a division by sqrt(3)/2, which caused it to be wrong..
• Jul 22nd 2006, 03:45 AM
scorpion007
ok, now i have another problem, different question:

The area of the region bounded by the axes, the line x = a, where 0 < a < 1, and the curve with equation $\displaystyle y = \arccos (x)$ is determined by:

Apparently this is the answer, but I don't understand how they derived it:
$\displaystyle a\arccos(a) + \int^{\frac{\pi}{2}}_{\arccos(a)} \cos(x) dx$

Any ideas?
• Jul 22nd 2006, 04:34 AM
CaptainBlack
Quote:

Originally Posted by scorpion007
ok, now i have another problem, different question:

The area of the region bounded by the axes, the line x = a, where 0 < a < 1, and the curve with equation $\displaystyle y = \arccos (x)$ is determined by:

Apparently this is the answer, but I don't understand how they derived it:
$\displaystyle a\arccos(a) + \int^{\frac{\pi}{2}}_{\arccos(a)} \cos(x) dx$

Any ideas?

Looks like making the substitution $\displaystyle \cos(u)=x$, and then
integrating by parts.

RonL
• Jul 22nd 2006, 04:48 AM
Soroban
Hello, scorpion007!

Quote:

The area of the region bounded by the axes, the line $\displaystyle x = a,\;0 < a < 1$,
and the curve with equation $\displaystyle y = \arccos (x)$ is determined by:

Apparently this is the answer, but I don't understand how they derived it:
$\displaystyle a\arccos(a) + \int^{\frac{\pi}{2}}_{\arccos(a)} \cos(x) dx$

Any ideas? . . . yes

They were very inconsiderate . . .

If they had written: .$\displaystyle a\arccos(a) + \int^{\frac{\pi}{2}}_{\arccos(a)} \cos (y)\,dy$

. . we might have had a chance of figuring it out.

The integration limits made me suspicious . . .
Code:

        |     π/2* *         |::::::*  (a, arccos(a))         + - - - - o         |:::::::::: *         |::::::::::  *       - + - - - - + - -*- -         0        a

They integrated with respect to $\displaystyle y$ ... and didn't tell us!

We have: .$\displaystyle x = \cos(y)$

The area of the rectangle is: .$\displaystyle a\cdot\arccos(a)$

The area of the upper segment is: .$\displaystyle \int^{\frac{\pi}{2}}_{arccos(a)} \cos y\,dy$

. . Diabolical!
• Jul 22nd 2006, 04:32 PM
scorpion007
thanks for the help guys.

Soroban, your reasoning seems very logical to me, and looks like that's what it *should* be. Unfortunately they say the answer is cos(x) dx, instead of your cos(y) dy. Which is obviously not correct according to your diagram. So im very confused.

CaptainBlack, do you have any further advice on this perhaps? I'm not sure where to go from your substitution...
• Jul 22nd 2006, 04:52 PM
Soroban
Hello, scorpion007!

I hope you agree that the area is: .$\displaystyle a\cdot\arccos(a) + \int^{\frac{\pi}{2}}_{arccos(a)} \cos y\,dy$

Recall that, with a definite integral, the variable itself doesn't matter.

For example: .$\displaystyle \int^b_b f(\theta)\,d\theta \;= \;\int^b_a f(t)\,dt \;=\;\int^b_af(x)\,dx \;=$ $\displaystyle \int^b_af(y)\,dy$

So my answer is equivalent to theirs . . . and mine is clearer.

• Jul 22nd 2006, 05:00 PM
scorpion007
Here's also another question i couldnt solve. Its multiple choice. See if you can get it.

$\displaystyle f$ is a continuous, differentiable function with $\displaystyle f(x) = f(-x)$ such that $\displaystyle f(x) \ge 0$ for $\displaystyle x \in [0, a]$ and $\displaystyle f(x) < 0$ for $\displaystyle x \in (a, b]$. It is also known that $\displaystyle \int^{b}_{0} f(x) dx = 0$ and $\displaystyle \int^{a}_{0} f(x) dx = A$. Which of the following is false?

A: $\displaystyle \int^{b}_{-b} f(x) dx = 0$
B: $\displaystyle -\int^{-a}_{0} f(x) dx = A$
C: $\displaystyle -\int^{b}_{a} f(x) dx + \int^{a}_{0} f(x) dx - \int^{-a}_{0} f(x) dx = 3A$
D: $\displaystyle \Bigl\vert \int^{b}_{a} f(x) dx \Bigr\vert + \Bigl\vert \int^{a}_{0} f(x) dx \Bigr\vert = 2A$
E: $\displaystyle \int^{-a}_{-b} f(x) dx = 0$

I got confused just by the sheer amount of info in this question. :(
• Jul 22nd 2006, 05:07 PM
scorpion007
Soroban, ah ok, i see what you mean. EDIT: nevermind, im an idiot :P, figured it out
• Jul 22nd 2006, 08:48 PM
CaptainBlack
Quote:

Originally Posted by scorpion007
Here's also another question i couldnt solve. Its multiple choice. See if you can get it.

$\displaystyle f$ is a continuous, differentiable function with $\displaystyle f(x) = f(-x)$ such that $\displaystyle f(x) \ge 0$ for $\displaystyle x \in [0, a]$ and $\displaystyle f(x) < 0$ for $\displaystyle x \in (a, b]$. It is also known that $\displaystyle \int^{b}_{0} f(x) dx = 0$ and $\displaystyle \int^{a}_{0} f(x) dx = A$. Which of the following is false?

A: $\displaystyle \int^{b}_{-b} f(x) dx = 0$(

Is true, as:

$\displaystyle \int^{b}_{-b} f(x) dx=\int^{b}_{0} f(x) dx+\int^{0}_{-b} f(x) dx$

$\displaystyle =\int^{b}_{0} f(x) dx-\int^{0}_{b} f(-u) du$
by change of variable $\displaystyle u=-x$ in the second integral

$\displaystyle =\int^{b}_{0} f(x) dx-\int^{0}_{b} f(u) du$
by symetry of $\displaystyle f(u)$ about $\displaystyle u=0$, ie $\displaystyle f(u)=f(-u)$

$\displaystyle =\int^{b}_{0} f(x) dx+\int^{b}_{a} f(u) du=2\int^{b}_{0} f(x) dx=0$
by interchanging the upper and lower limits of integration.

RonL
• Jul 22nd 2006, 08:54 PM
CaptainBlack
Quote:

Originally Posted by scorpion007
Here's also another question i couldnt solve. Its multiple choice. See if you can get it.

$\displaystyle f$ is a continuous, differentiable function with $\displaystyle f(x) = f(-x)$ such that $\displaystyle f(x) \ge 0$ for $\displaystyle x \in [0, a]$ and $\displaystyle f(x) < 0$ for $\displaystyle x \in (a, b]$. It is also known that $\displaystyle \int^{b}_{0} f(x) dx = 0$ and $\displaystyle \int^{a}_{0} f(x) dx = A$. Which of the following is false?

B: $\displaystyle -\int^{-a}_{0} f(x) dx = A$

Is true:

$\displaystyle -\int^{-a}_{0} f(x) dx=\int^{a}_{0} f(-u) du$

by change of variable $\displaystyle u=-x$

$\displaystyle =\int^{a}_{0} f(u) du$
by symetry of $\displaystyle f(x)$
$\displaystyle =A$

as this is given in the question.

RonL
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