Diferantion Integration: int cos^2(6x-9)dx, int (8dx)/(...

[tazmania9]

Determine the following integrals:
1. Integration cos^2(6x-9)dx
2. integration 8dx divided by 9x^2 + 8
3. integration dx divided by x^2-4x-11 (note that x^2-4x-11 can be written as (x-2)^2 -15


Diferantion: x^2 dy/dx - 3xy = x^6sin5x with the given boundary condition that when x= ?/5, y=0

Can u guys pls help me with the above ? Thenkx a lot =)

 

[Deleted member 4993]

Re: Diferantion Integration

Determine the following integrals:
1. Integration cos^2(6x-9)dx

\(\displaystyle \int \cos^2(6x-9) dx\)

hint: substitute

u = 6x - 9

2. integration 8dx divided by 9x^2 + 8

\(\displaystyle \int \frac {8 dx}{9x^2+8} dx\)

hint: substitute

\(\displaystyle x \, = \, \frac {\sqrt{8}}{3}tan(u)\)

3. integration dx divided by x^2-4x-11 (note that x^2-4x-11 can be written as (x-2)^2 -15

\(\displaystyle \int \frac{dx}{(x-2)^2 - 15}\)

use:

\(\displaystyle a^2 \, - \, b^2 \, = \, (a+b)\cdot (a-b)\)

Diferantion: x^2 dy/dx - 3xy = x^6sin5x with the given boundary condition that when x= ?/5, y=0

x^2 dy/dx - 3xy = x^6sin5x

y' - (3/x) y = (x^4) sin(5x)

Above is a standard form of first order ODE.

Can u guys pls help me with the above ? Thenkx a lot =)

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.

 

[soroban]

Re: Diferantion Integration

Hello, tazmania9!

The last one is a killer . . .

\(\displaystyle \text{Differential Equation: }x^2\,\frac{dy}{dx} - 3xy \:=\ x^6\sin5x\;\text{ where }y\left(\tfrac{\pi}{5}\right) = 0\)

\(\displaystyle \text{Divide by }x^2\!:\quad \frac{dy}{dx} - \frac{3}{x}\,y \:=\:x^4\sin5x\)

\(\displaystyle \text{Integrating factor: }\:I \;=\;e^{\int\left(\text{-}\frac{3}{x}\right)dx} \;=\;e^{\text{-}3\ln x} \;=\;e^{\ln(x^{\text{-}3})} \;=\;x^{\text{-}3} \;=\;\frac{1}{x^3}\)

\(\displaystyle \text{Multiply by }I\!:\quad \frac{1}{x^3}\,\frac{dy}{dx} - \frac{3}{x^4}\,y \;=\;x\sin5x\)

\(\displaystyle \text{And we have: }\;\frac{d}{dx}\left(\frac{1}{x^3}\,y\right) \;=\;x\sin5x\)

\(\displaystyle \text{Integrate: }\;\frac{1}{x^3}\,y \;=\;\int x\sin5x\,dx \quad\Leftarrow\;\text{(Integrate by parts)}\)

\(\displaystyle \text{And we have: }\;\frac{y}{x^3} \;=\;-\tfrac{1}{5}x\cos5x + \tfrac{1}{25}\sin5x + C\)


Evaluate \(\displaystyle C.\)

\(\displaystyle y(\tfrac{\pi}{5}) \,=\,0\!:\;\;\frac{0}{(\frac{\pi}{5})^3} \;=\;-\tfrac{1}{5}\left(\tfrac{\pi}{5}\right)\cos\pi + \tfrac{1}{25}\sin\pi + C \quad\Rightarrow\quad 0 \;=\;\text{-}\tfrac{\pi}{25}(\text{-}1) + \tfrac{1}{25}(0) + C\)

. . \(\displaystyle \text{Hence: }\:C \:=\:\text{-}\tfrac{\pi}{25}\)


\(\displaystyle \text{We have: }\;\frac{y}{x^3} \;=\;\text{-}\tfrac{1}{5}x\cos5x + \tfrac{1}{25}\sin5x - \tfrac{\pi}{25}\)


\(\displaystyle \text{Multiply by }x^3\!\!:\quad y \;=\;\text{-}\tfrac{1}{5}x^4\cos5x + \tfrac{1}{25}x^3\sin5x - \tfrac{\pi}{25}x^3\)

 

[tazmania9]

For question number 2 this is the work i did:
let u=9x^2+8
du/dx= 18x

so

1/9x^2=8 * 18x/18x * 8
8/9x^2+8
9/18x

is it correct? how do i continue?

For the question number 3: I dont understand how i remove the number 15 from my equation ? And how to continue?
Can u help me pls?

Thanks in advanced.

 

[Deleted member 4993]

Determine the following integrals:

3. integration dx divided by x^2-4x-11 (note that x^2-4x-11 can be written as (x-2)^2 -15

\(\displaystyle \int\frac{dx}{(x-2)^2-15}\)

\(\displaystyle = \, \int\frac{dx}{ [(x-2) \, -\, \sqrt{15}]\cdot [(x-2) \, + \, \sqrt{15}]}\)

\(\displaystyle = \,\frac{1}{2\cdot \sqrt{15}} [\int\frac{dx}{ [(x-2) \, -\, \sqrt{15}]} \, - \, \int\frac{dx}{[(x-2) \, + \, \sqrt{15}]}]\)

\(\displaystyle = \,\frac{1}{2\cdot \sqrt{15}} \, ln\frac{ (x-2) \, -\, \sqrt{15}}{(x-2) \, + \, \sqrt{15}} \, + \, C\)

 

[Deleted member 4993]

For question number 2 this is the work i did:
let u=9x^2+8
du/dx= 18x

so

1/9x^2=8 * 18x/18x * 8
8/9x^2+8
9/18x

is it correct? No...

Follow the hint given above..
how do i continue?

For the question number 3: I dont understand how i remove the number 15 from my equation ? And how to continue?
Can u help me pls?

Thanks in advanced.