# Math Help - Integration By Parts

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: . $\int\csc^5\!x\,dx$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Since we know the value of $\int\!\csc^3\!x\,dx$, 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 \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: . $I \;=\;\int\csc^3\!x\,dx \;=\;\int \csc x(\csc^2\!x\,dx)$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Since we know the value of $\int\!\csc^3\!x\,dx$, 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: . $I \;=\;\text{-}\csc^3\!x\cot x - 3\!\int\!\csc^3\!x\cot^2\!x\,dx$

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I think it should be a plus and not a minus.

5. ## Re: Integration By Parts

Never mind! ^^