1. ## Integration By Parts

I'm having a lot of difficulty figuring out:

csc5(x) dx

My instructor said something about "Looping the Integral" if that helps at all.

The instructor also gave us an example of:

csc3(x) dx

But i'm not sure how to go about the first one I gave.

2. ## Re: Integration By Parts

Hello, Preston019!

I'm having a lot of difficulty figuring out: .$\displaystyle \int\csc^5\!x\,dx$

My instructor said something about "Looping the Integral" if that helps at all.

The instructor also gave us an example of: .$\displaystyle \int \csc^3\!x\,dx$

But I'm not sure how to go about the first one I gave.

Did you understand your instructor's example?
If not, no wonder you can't solve this one.
Maybe I'd better baby-step through it . . .

We have: .$\displaystyle I \;=\;\int\csc^3\!x\,dx \;=\;\int \csc x(\csc^2\!x\,dx)$

Integrate by parts: .$\displaystyle \begin{Bmatrix}u &=& \csc x && dv &=& \csc^2\!x\,dx \\ du &=& \text{-}\csc x\cot x\,dx && v &=& \text{-}\cot x \end{Bmatrix}$

We have: .$\displaystyle I \;=\;\text{-}\csc x\cot x - \int\csc x\cot^2\!x\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc x\cot x - \int\csc x(\csc^2\!x-1)\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc x\cot x - \int(\csc^3\!x - \csc x)\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc x\cot x - \underbrace{\int\csc^3\!x\,dx}_{\text{This is }I} + \int\csc x\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc x\cot x - I + \int\csc x\,dx$

. . . . . . . $\displaystyle 2I \;=\;\text{-}\csc x\cot x + \int\csc x\,dx$

. . . . . . . $\displaystyle 2I \;=\;\text{-}\csc x\cot x + \ln|\csc x - \cot x| + C$

. . . . . . . . $\displaystyle I \;=\;\tfrac{1}{2}\big(\text{-}\csc x\cot x + \ln|\csc x - \cot x|\big) + C$

Therefore: .$\displaystyle \int\csc^3x\,dx \;=\;\tfrac{1}{2}\big(\text{-}\csc x\cot x + \ln|\csc x - \cot x|\big) + C$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We have: .$\displaystyle I \;=\; \int\!\csc^5\!x\,dx \;=\;\int\!\csc^3\!x(\csc^2\!x\,dx)$

Integrate by parts: .$\displaystyle \begin{Bmatrix}u &=& \csc^3\!x && dv &=& \csc^2\!x\,dx \\ du &=& \text{-}3\csc^3\!x\cot x\,dx && v &=& \text{-}\cot x \end{Bmatrix}$

We have: .$\displaystyle I \;=\;\text{-}\csc^3\!x\cot x - 3\!\int\!\csc^3\!x\cot^2\!x\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc^3\!x\cot x - 3\!\int\!\csc^3\!x(\csc^2\!x-1)\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc^3\!x\cot x - 3\!\int(\csc^5\!x - \csc^3\!x)\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc^3\!x\cot x - 3\underbrace{\int\!\csc^5\!x\,dx}_{\text{This is }I} + 3\!\int\!\csc^3\!x\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc^3\!x\cot x - 3I + 3\!\int\!\csc^3\!x\,dx$

. . . . . . .$\displaystyle 4I \;=\;\text{-}\csc^3\!x\cot x + 3\!\int\!\csc^3\!x\,dx$

Since we know the value of $\displaystyle \int\!\csc^3\!x\,dx$, we can finish the problem . . .

3. ## Re: Integration By Parts

Originally Posted by Preston019
I'm having a lot of difficulty figuring out:

csc5(x) dx

My instructor said something about "Looping the Integral" if that helps at all.

The instructor also gave us an example of:

csc3(x) dx

But i'm not sure how to go about the first one I gave.
\displaystyle \displaystyle \begin{align*} \int{\csc^5{x}\,dx} &= \int{\frac{1}{\sin^5{x}}\,dx} \\ &= \int{\frac{\sin{x}}{\sin^6{x}}\,dx} \\ &= -\int{\frac{-\sin{x}}{\left(1 - \cos^2{x} \right)^3}\,dx} \\ &= -\int{\frac{1}{\left(1 - u^2\right)^3}\,du} \textrm{ after making the substitution }u = \cos{x} \\ &= -\int{\frac{1}{(1 - u)^3(1 + u)^3}\,du} \end{align*}

You should be able to solve this using Partial Fractions.

4. ## Re: Integration By Parts

Originally Posted by Soroban
Hello, Preston019!

Did you understand your instructor's example?
If not, no wonder you can't solve this one.
Maybe I'd better baby-step through it . . .

We have: .$\displaystyle I \;=\;\int\csc^3\!x\,dx \;=\;\int \csc x(\csc^2\!x\,dx)$

Integrate by parts: .$\displaystyle \begin{Bmatrix}u &=& \csc x && dv &=& \csc^2\!x\,dx \\ du &=& \text{-}\csc x\cot x\,dx && v &=& \text{-}\cot x \end{Bmatrix}$

We have: .$\displaystyle I \;=\;\text{-}\csc x\cot x - \int\csc x\cot^2\!x\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc x\cot x - \int\csc x(\csc^2\!x-1)\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc x\cot x - \int(\csc^3\!x - \csc x)\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc x\cot x - \underbrace{\int\csc^3\!x\,dx}_{\text{This is }I} + \int\csc x\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc x\cot x - I + \int\csc x\,dx$

. . . . . . . $\displaystyle 2I \;=\;\text{-}\csc x\cot x + \int\csc x\,dx$

. . . . . . . $\displaystyle 2I \;=\;\text{-}\csc x\cot x + \ln|\csc x - \cot x| + C$

. . . . . . . . $\displaystyle I \;=\;\tfrac{1}{2}\big(\text{-}\csc x\cot x + \ln|\csc x - \cot x|\big) + C$

Therefore: .$\displaystyle \int\csc^3x\,dx \;=\;\tfrac{1}{2}\big(\text{-}\csc x\cot x + \ln|\csc x - \cot x|\big) + C$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We have: .$\displaystyle I \;=\; \int\!\csc^5\!x\,dx \;=\;\int\!\csc^3\!x(\csc^2\!x\,dx)$

Integrate by parts: .$\displaystyle \begin{Bmatrix}u &=& \csc^3\!x && dv &=& \csc^2\!x\,dx \\ du &=& \text{-}3\csc^3\!x\cot x\,dx && v &=& \text{-}\cot x \end{Bmatrix}$

We have: .$\displaystyle I \;=\;\text{-}\csc^3\!x\cot x - 3\!\int\!\csc^3\!x\cot^2\!x\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc^3\!x\cot x - 3\!\int\!\csc^3\!x(\csc^2\!x-1)\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc^3\!x\cot x - 3\!\int(\csc^5\!x - \csc^3\!x)\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc^3\!x\cot x - 3\underbrace{\int\!\csc^5\!x\,dx}_{\text{This is }I} + 3\!\int\!\csc^3\!x\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc^3\!x\cot x - 3I + 3\!\int\!\csc^3\!x\,dx$

. . . . . . .$\displaystyle 4I \;=\;\text{-}\csc^3\!x\cot x + 3\!\int\!\csc^3\!x\,dx$

Since we know the value of $\displaystyle \int\!\csc^3\!x\,dx$, we can finish the problem . . .
That helped a lot! Thank you! But, I think you messed up with distributing your minus sign here:

Originally Posted by Soroban

We have: .$\displaystyle I \;=\;\text{-}\csc^3\!x\cot x - 3\!\int\!\csc^3\!x\cot^2\!x\,dx$

[/size]
I think it should be a plus and not a minus.

5. ## Re: Integration By Parts

Never mind! ^^

6. ## Re: Integration By Parts

There is an error in the integral shown for csc^3(x)dx
The result should be:
-1/2(csc(x)cot(x) + ln|csc(x) + cot(x)|) + C

7. ## Re: Integration By Parts

It looks to me like Soroban's derivation and answer are correct. Where do you think he made a mistake?

- Hollywood

8. ## Re: Integration By Parts

Originally Posted by hollywood
It looks to me like Soroban's derivation and answer are correct. Where do you think he made a mistake?

- Hollywood
The error occurs when first showing the integration of: intg csc^3(x) dx

We get to:
2I = -csc{x}cot{x} + intg csc(x) dx

But then the next line should be:
2I = -csc(x)cot(x) - ln|csc(x) + cot(x)| + C

Sorry I couldn't get the Latex format right, so I'm just using pain text that I hope is okay. It would help to learn to Latex formatting if a person could preview their posts before actually submitting them.

9. ## Re: Integration By Parts

No, the integral of $\displaystyle \csc{x}$ is $\displaystyle \ln{|\csc{x}-\cot{x}|}+C$. Here's the derivation:

$\displaystyle \int \csc{x} \,dx$

Multiply numerator and denominator by $\displaystyle \csc{x} - \cot{x}$:

$\displaystyle \int \frac{\csc^2{x}-\csc{x}\cot{x}}{\csc{x}-\cot{x}} \,dx$

and since $\displaystyle \frac{d}{dx}\csc{x} = -\csc{x}\cot{x}$ and $\displaystyle \frac{d}{dx}\cot{x} = -\csc^2{x}$, the numerator is the derivative of the denominator, which means we can make the substitution and get:

$\displaystyle \int \csc{x} \,dx = \ln{|\csc{x}-\cot{x}|}+C$

- Hollywood

10. ## Re: Integration By Parts

Originally Posted by AMathGuy
Sorry I couldn't get the Latex format right, so I'm just using pain text that I hope is okay. It would help to learn to Latex formatting if a person could preview their posts before actually submitting them.
When you are typing a post you will see three buttons on the lower right hand side. They are "Post Quick Reply," "Go Advanced," and "Cancel." Click on "Go Advanced" and you will see copy of your message up top and the edit panel below it. You can check your post at any time after that by hitting the "Preview Post" button on the lower right hand side.

We also have a LaTeX Help forum where you can test out your coding and ask questions about it.

-Dan

11. ## Re: Integration By Parts

Originally Posted by hollywood
No, the integral of $\displaystyle \csc{x}$ is $\displaystyle \ln{|\csc{x}-\cot{x}|}+C$. Here's the derivation:

$\displaystyle \int \csc{x} \,dx$

Multiply numerator and denominator by $\displaystyle \csc{x} - \cot{x}$:

$\displaystyle \int \frac{\csc^2{x}-\csc{x}\cot{x}}{\csc{x}-\cot{x}} \,dx$

and since $\displaystyle \frac{d}{dx}\csc{x} = -\csc{x}\cot{x}$ and $\displaystyle \frac{d}{dx}\cot{x} = -\csc^2{x}$, the numerator is the derivative of the denominator, which means we can make the substitution and get:

$\displaystyle \int \csc{x} \,dx = \ln{|\csc{x}-\cot{x}|}+C$

- Hollywood
Here are a couple links to other math sites showing the answer I used. This first link shows the basic technique you used showing the other answer:

Integral csc(x)

I did further investigation and found an explanation showing that BOTH answers are equivalent!
calculus - Integral of $\csc(x)$ - Mathematics Stack Exchange

Sorry I jumped the gun on this!! My apologizes!

12. ## Re: Integration By Parts

I didn't see that. So we're both right. For those who missed it, since

$\displaystyle |\csc{x}+\cot{x}||\csc{x}-\cot{x}|=|\csc^2{x}-\cot^2{x}|=1$,

$\displaystyle \ln|\csc{x}-\cot{x}|=\ln\frac{1}{|\csc{x}+\cot{x}|}=-\ln|\csc{x}+\cot{x}|$.

- Hollywood

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