# Thread: how to show that....

1. ## 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

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

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

3. .

4. That helps with 1), any help with the others please?

5. 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 $x= \frac{1}{2\pi}$

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

6. 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?

7. a little error, pay attention to the operation between 2 and cos x.

8. 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?

9. The substitution rule is easier if written as

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

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

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

The integral becomes

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

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

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

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

10. Ok great,

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

2) Now in implicit form, find the solution

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

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...

12. Amer provide you the correct solution, you can refer it.
If you have any question, don't hesitate to ask for help.

13. 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

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

14. 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 $x= \frac{1}{2\pi}$

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

= $5.3594$

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