Integration of 1/2x & sinxcosx & x+1

[ankur29]

Hello I was wondering why in this differential equation they did not use double angle identity, I understand how to get the other result also

So

1)Dif eqn:

Dy/dx +ycotx= cosx


Integrating factor IF= e^ (integrand of cotx) = sinx

Solution
yIF= integrand of QIF where Q is cotx from original question)

y sinx= integral of cosxsinx

Now I recognised this as ½ sin2x and the integral would be

ysinx= ½(-cos2x/2+ c)

therefore y= 1/sinx (-cos2x/4+ c/2)

however this could be done by substitution

y sinx= integral of cosxsinx let u=sinx to give
ysinx= integral of cosx./u du/cosx
ysinx= sin^2x +c
y=1/sinx (sin^2x +c)

why are they not the same ?? or are they really the same

I mean if we differentiate either sin^2x or -cos2x/2 we would get back to cosxsinx

2)
(x-2)dy/dx –y = (x-2)^3
dy/dx – y/(x-2)= (x-2)^2
IF= 1/(x-2)
yIF= intergral of QIF

y/(x-2)= integral of (x-2)^2 .1/(x-2)
y/(x-2)= integral of (x-2)

If I were to integrate (x-2) why is it (x-2)^2/2 +c is it the same as x^2/2-2x+c
(I am confused about this one as from expansion I can see its similar but if I put any value for c into either say 1 I get different answers)



3)Finally integral of 1/2x I can do this by dividing by a half
i.e ½ integral of 1/x =1/2 lnx

but is itthe same as 1/2ln2x as this wway I say what must I do to denominator (2x) to get to numerator and then simply multiply this coefficient to the natural log of the bottom

i.e 1/4x+3 = ½ ln 4x+3

Can I only do that if I cannot factorise out a coeff?

 

[D H]

Two typical reasons why using two integration techniques yield different answers are

• You made a mistake in one or both integrations.
• The answers are the same, to within an arbitrary constant.

You did both here. $\int u du = 1/2u^2 + c$. You forgot the factor of 1/2.

The trig substitution and u substitution techniques yield solutions of

$$\aligned<br /> y\sin x &= -\,\frac 1 4 \cos 2x + c_1 && \text{trig substitution} \\<br /> &= \phantom{-}\frac 1 2 \sin^2 x + c_2 && \text{ u-substitution}<br /> \endaligned$$

With the trig substitution $\cos 2x = 1 - 2\sin^2x$, the first becomes

[tex]y\sin x = -\,\frac 1 4 (1-2\sin^2 x) + c_1 = \frac 1 2 \sin^2 x + c_1 - \frac 1 4[/itex]<br /> <br /> So the two techniques do agree to within an arbitrary constant.[/tex]

 

[ankur29]

so a diff technique will give a diff constant?

 

[D H]

Not always, but in this case, yes.