# how to show that....

• Jan 8th 2010, 09:02 AM
how to show that....
1) Using:

$\int (f(x))^n f'(x) dx = \frac{1}{n + 1}(f(x))^{n+1} + c$

how can i show...

$\int \frac{sin x}{(2 - cos x)^3} dx = - \frac{1}{2(2 - cos x)^2} + c$

where c is an arbitrary constant.

2) Now in implicit form, find the solution

$\frac{dy}{dx} = \frac{6y^{2/3}sin x}{(2- cos x)^3}$

3) what is the corresponding answer, implicit that satisifies y = 1 when x = 0

4) Whats the explicit form of the answer?

5) whats y's value, given by the solution when $x = 1/2 \pi$

Thanks :)
• Jan 8th 2010, 09:29 AM
Shanks
$\int \frac{sin x}{(2 - cos x)^3} dx = \int \frac{1}{(2 - cos x)^3} d(2-cos x) = - \frac{1}{2(2 - cos x)^2} + c$

$\frac{dy}{dx} = \frac{6y^{2/3}sin x}{(2- cos x)^3}$
implies
$\frac{dy}{6y^{2/3}}=\frac{sin x dx}{(2- cos x)^3}$
Integrate both sides to get the explicit form.

• Jan 8th 2010, 09:29 AM
Henryt999
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• Jan 8th 2010, 11:23 AM
That helps with 1), any help with the others please? :)
• Jan 8th 2010, 12:26 PM
Amer
2) $\frac{dy}{dx} = \frac{6y^{\frac{2}{3}} \sin x}{(2-\cos x)^3 }$

$\frac{dy}{6y^{\frac{2}{3}}} = \frac{\sin x\cdot dx}{(2-\cos x)^3}$

$\int \frac{dy}{6y^{\frac{2}{3}}} = \int \frac{\sin x\cdot dx}{(2-\cos x)^3}$

the right side we find it in the first question

$\frac{y^{\frac{-2}{3} +1}}{6\left(\frac{-2}{3}+1\right)}=- \frac{1}{2(2 - cos x)^2} + c
$

$\frac{y^{\frac{1}{3}}}{2} = - \frac{1}{2(2 - cos x)^2} + c
$

3)when y=1 and x=0

$\frac{1}{2} = \frac{-1}{2(2-1)^2}+c$

$\frac{1}{2} = \frac{-1}{2} + c \Rightarrow c = 1$ so

$\frac{y^{\frac{1}{3}}}{2} = \frac{-1}{2(2-\cos x)^2} + 1$

$y^{\frac{1}{3}} = \frac{-1}{(2-\cos x)^2} + 2$

4)

$y = \left( \frac{-1}{(2-\cos x)^2} +2 \right)^3$

5)when $x= \frac{1}{2\pi}$

$y = \left( \frac{-1}{(2-\cos (1/2\pi))^2} +2 \right)^3$
• Jan 9th 2010, 01:45 AM
I thought i understood the first bit properly, but i don't.

So we have:

$\int (f(x))^n f'(x) dx = \frac{1}{n + 1}(f(x))^{n+1} + c$

and i need to show that:

$\int \frac{sin x}{(2 - cos x)^3} dx = - \frac{1}{2(2 - cos x)^2} + c$

Where c is an arbitray constant.

____________________________

I Answered it to show that:

$\int f^n f ' = \frac{1}{n + 1} f^{n+1}$

$\int (2 - cos x)^{-3} sin x dx$

$\frac{1}{-3+1} (2 . cos x)^{-3+1} + c$

Is this correct?
• Jan 9th 2010, 02:10 AM
Shanks
a little error, pay attention to the operation between 2 and cos x.
• Jan 9th 2010, 02:21 AM
ok but what am i missing because i feel the question doesnt finish there.

after

$\frac{1}{-3+1} (2 . cos x)^{-3+1} + c$

what do i need to do next, this is important as it effects all the subsquent questions?
• Jan 9th 2010, 02:24 AM
Prove It
The substitution rule is easier if written as

$\int{f\left(u(x)\right)\,\frac{du}{dx}\,dx} = \int{f(u)\,du}$.

So in your case

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

Let $u = 2 - \cos{x}$ so that $\frac{du}{dx} = \sin{x}$.

The integral becomes

$\int{u^{-3}\,\frac{du}{dx}\,dx} = \int{u^{-3}\,du}$

$= -\frac{1}{2}u^{-2} + C$

$= -\frac{1}{2}(2 - \cos{x})^{-2} + C$

$= -\frac{1}{2(2 - \cos{x})^2} + C$.
• Jan 9th 2010, 02:27 AM
Ok great,

so now with your solution has that changed the subsquent questions...

2) Now in implicit form, find the solution

$\frac{dy}{dx} = \frac{6y^{2/3}sin x}{(2- cos x)^3}$
• Jan 9th 2010, 02:35 AM
Prove It
Quote:

Originally Posted by ADY
Ok great,

so now with your solution has that changed the subsquent questions...

2) Now in implicit form, find the solution

$\frac{dy}{dx} = \frac{6y^{2/3}sin x}{(2- cos x)^3}$

Well

$\frac{dy}{dx} = \frac{6y^{\frac{2}{3}}\sin{x}}{(2 - \cos{x})^3}$

$y^{-\frac{2}{3}}\,\frac{dy}{dx} = \frac{6\sin{x}}{(2 - \cos{x})^3}$

Now integrate both sides with respect to $x$...

$\int{y^{-\frac{2}{3}}\,\frac{dy}{dx}\,dx} = \int{\frac{6\sin{x}}{(2 - \cos{x})^3}\,dx}$

$\int{y^{-\frac{2}{3}}\,dy} = 6\int{\frac{\sin{x}}{(2 - \cos{x})^3}\,dx}$.

Can you go from here? Hint: the RHS is the same as Q1...
• Jan 9th 2010, 02:43 AM
Shanks
Amer provide you the correct solution, you can refer it.
If you have any question, don't hesitate to ask for help.
• Jan 9th 2010, 02:45 AM
Quote:

Originally Posted by Prove It
Well

$\frac{dy}{dx} = \frac{6y^{\frac{2}{3}}\sin{x}}{(2 - \cos{x})^3}$

$y^{-\frac{2}{3}}\,\frac{dy}{dx} = \frac{6\sin{x}}{(2 - \cos{x})^3}$

Now integrate both sides with respect to $x$...

$\int{y^{-\frac{2}{3}}\,\frac{dy}{dx}\,dx} = \int{\frac{6\sin{x}}{(2 - \cos{x})^3}\,dx}$

$\int{y^{-\frac{2}{3}}\,dy} = 6\int{\frac{\sin{x}}{(2 - \cos{x})^3}\,dx}$.

Can you go from here? Hint: the RHS is the same as Q1...

so

$\int{y^{-\frac{2}{3}}\,\frac{dy}{dx}\,dx} = \int{\frac{6\sin{x}}{(2 - \cos{x})^3}\,dx}$ = $- \frac{1}{2(2 - cos x)^2} + c$
• Jan 9th 2010, 03:07 AM
This is getting quite a messy thread - only because of my posts, so i'll do a consolidation from post 1.

Using:

$\int{f\left(u(x)\right)\,\frac{du}{dx}\,dx} = \int{f(u)\,du}$

show that:

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

where c is an arbitray constant.

We Let $u = 2 - \cos{x}$ so that $\frac{du}{dx} = \sin{x}$

So The integral becomes

$\int{u^{-3}\,\frac{du}{dx}\,dx} = \int{u^{-3}\,du}$

$= -\frac{1}{2}u^{-2} + C$

$= -\frac{1}{2}(2 - \cos{x})^{-2} + C$

$= -\frac{1}{2(2 - \cos{x})^2} + C$.[/quote]

Now in implicit form, find the solution

$\frac{dy}{dx} = \frac{6y^{2/3}sin x}{(2- cos x)^3}$

$\frac{dy}{dx} = \frac{6y^{\frac{2}{3}} \sin x}{(2-\cos x)^3 }$

$\frac{dy}{6y^{\frac{2}{3}}} = \frac{\sin x\cdot dx}{(2-\cos x)^3}$

$\int \frac{dy}{6y^{\frac{2}{3}}} = \int \frac{\sin x\cdot dx}{(2-\cos x)^3}$

In the right side we find it in the 1st part of the question

$\frac{y^{\frac{-2}{3} +1}}{6\left(\frac{-2}{3}+1\right)}=- \frac{1}{2(2 - cos x)^2} + c
$

$\frac{y^{\frac{1}{3}}}{2} = - \frac{1}{2(2 - cos x)^2} + c
$

We now need to find the corresponding implicit solution that satisifies intial condition y = 1 when x = 0

$\frac{1}{2} = \frac{-1}{2(2-1)^2}+c$

$\frac{1}{2} = \frac{-1}{2} + c \Rightarrow c = 1$ so

$\frac{y^{\frac{1}{3}}}{2} = \frac{-1}{2(2-\cos x)^2} + 1$

$y^{\frac{1}{3}} = \frac{-1}{(2-\cos x)^2} + 2$

So to find the explicit form of the solution:

$y = \left( \frac{-1}{(2-\cos x)^2} +2 \right)^3$

Then the value of y when $x= \frac{1}{2\pi}$

$y = \left( \frac{-1}{(2-\cos (1/2\pi))^2} +2 \right)^3$

= $5.3594$
• Jan 9th 2010, 03:08 AM
Prove It
Quote:

Originally Posted by ADY
so

$\int{y^{-\frac{2}{3}}\,\frac{dy}{dx}\,dx} = \int{\frac{6\sin{x}}{(2 - \cos{x})^3}\,dx}$ = $- \frac{1}{2(2 - cos x)^2} + c$

No.

Perform the integration on the LHS.

Perform the integration on the RHS.