Please help with part b of this integration question (1 Viewer)

Life'sHard · Jul 7, 2021

For i)
Let x = a-u on the LHS then differentiate to solve for du

Life'sHard · Jul 7, 2021

For b)
Whenever there's an x change it to pi-x I think.

CM_Tutor · Jul 7, 2021

Applying the result in part (a) to the question in part (b), you get:

$\begin{align*} \text{Let} \quad I &= \int_0^{\pi} \cfrac{x\sin{x}}{3+\sin^2{x}}\ dx \\ \text{Then} \quad I &= \int_0^{\pi} \cfrac{(\pi - x)\sin{(\pi - x)}}{3+\sin^2{(\pi - x)}}\ dx \qquad \text{using the formula from part (a)} \\ &= \int_0^{\pi} \cfrac{(\pi - x)\sin{x}}{3+\sin^2{x}}\ dx \qquad \text{noting that $\sin{(\pi - x)} = \sin{x}$} \\ &= \int_0^{\pi} \cfrac{\pi\sin{x}}{3+\sin^2{x}} - \cfrac{x\sin{x}}{3+\sin^2{x}}\ dx \\ &= \int_0^{\pi} \cfrac{\pi\sin{x}}{3+\sin^2{x}}\ dx - \int_0^{\pi} \cfrac{x\sin{x}}{3+\sin^2{x}}\ dx \\ &= \int_0^{\pi} \cfrac{\pi\sin{x}}{3 + \sin^2{x}}\ dx - I \end{align*}$

We have re-generated our original integral and can combine them to get:

$\begin{align*} 2I &= \int_0^{\pi} \cfrac{\pi\sin{x}}{3 + 1 - \cos^2{x}}\ dx \\ I &= \cfrac{\pi}{2}\int_0^{\pi} \cfrac{\sin{x}}{4 - \cos^2{x}}\ dx \\ &= \cfrac{\pi}{2}\int_1^{-1} \cfrac{-du}{4 - u^2} \qquad \text{after making the substitution $u = \cos{x}$ and noting that $x = 0 \implies u = 1$ and $x = \pi \implies u = -1$} \\ &= \cfrac{\pi}{2}\int_{-1}^1 \cfrac{du}{(2 + u)(2 - u)} \\ &= \cfrac{\pi}{8}\int_{-1}^1 \cfrac{1}{u + 2} - \cfrac{1}{u - 2}\ du \qquad \text{using partial fractions to find that $\cfrac{1}{4 - u^2} = \cfrac{1}{4}\left(\cfrac{1}{u + 2} - \cfrac{1}{u - 2}\right)$}\\ &= \cfrac{\pi}{8}\bigg[\ln|u + 2| - \ln|u - 2|\bigg]_{-1}^1 \\ &= \cfrac{\pi}{8}\left[\ln{3} - \ln{1} - \left(\ln{1} - \ln{3}\right)\right] \\ &= \cfrac{\pi}{4}\ln{3} \end{align*}$

kimtuluu · Jul 7, 2021

CM_Tutor said:
Applying the result in part (a) to the question in part (b), you get:

$\begin{align*} \text{Let} \quad I &= \int_0^{\pi} \cfrac{x\sin{x}}{3+\sin^2{x}}\ dx \\ \text{Then} \quad I &= \int_0^{\pi} \cfrac{(\pi - x)\sin{(\pi - x)}}{3+\sin^2{(\pi - x)}}\ dx \qquad \text{using the formula from part (a)} \\ &= \int_0^{\pi} \cfrac{(\pi - x)\sin{x}}{3+\sin^2{x}}\ dx \qquad \text{noting that $\sin{(\pi - x)} = \sin{x}$} \\ &= \int_0^{\pi} \cfrac{\pi\sin{x}}{3+\sin^2{x}} - \cfrac{x\sin{x}}{3+\sin^2{x}}\ dx \\ &= \int_0^{\pi} \cfrac{\pi\sin{x}}{3+\sin^2{x}}\ dx - \int_0^{\pi} \cfrac{x\sin{x}}{3+\sin^2{x}}\ dx \\ &= \int_0^{\pi} \cfrac{\pi\sin{x}}{3 + \sin^2{x}}\ dx - I \end{align*}$

We have re-generated our original integral and can combine them to get:

$\begin{align*} 2I &= \int_0^{\pi} \cfrac{\pi\sin{x}}{3 + 1 - \cos^2{x}}\ dx \\ I &= \cfrac{\pi}{2}\int_0^{\pi} \cfrac{\sin{x}}{4 - \cos^2{x}}\ dx \\ &= \cfrac{\pi}{2}\int_1^{-1} \cfrac{-du}{4 - u^2}\ du \qquad \text{after making the substitution $u = \cos{x}$ and noting that $x = 0 \implies u = 1$ and $x = \pi \implies u = -1$} \\ &= \cfrac{\pi}{2}\int_{-1}^1 \cfrac{du}{(2 + u)(2 - u)}\ du \\ &= \cfrac{\pi}{8}\int_{-1}^1 \cfrac{1}{u + 2} - \cfrac{1}{u - 2}\ du \qquad \text{using partial fractions to find that $\cfrac{1}{4 - u^2} = \cfrac{1}{4}\left(\cfrac{1}{u + 2} - \cfrac{1}{u - 2}\right)$}\\ &= \cfrac{\pi}{8}\bigg[\ln|u + 2| - \ln|u - 2|\bigg]_{-1}^1 \\ &= \cfrac{\pi}{8}\left[\ln{3} - \ln{1} - \left(\ln{1} - \ln{3}\right)\right] \\ &= \cfrac{\pi}{4}\ln{3} \end{align*}$

That's perfect. Many thanks for your help. Note: there are couple of lines in the solution that have double du in it.

CM_Tutor · Jul 7, 2021

kimtuluu said:
That's perfect. Many thanks for your help. Note: there are couple of lines in the solution that have double du in it.

Thanks, now fixed

CM_Tutor · Jul 7, 2021

FYI, when I see the proof in part (a), I expect to apply it to an integral that either regenerates the same integral or an equivalent that allows a major simplification of the problem.

kimtuluu · Jul 7, 2021

CM_Tutor said:
FYI, when I see the proof in part (a), I expect to apply it to an integral that either regenerates the same integral or an equivalent that allows a major simplification of the problem.

I thought so too. Just didn't think that it would involve substitution as well as integration by parts where full working out is required for a 4 mark question.

kimtuluu · Jul 8, 2021

This is my answer of part (a). Please let me know if it is acceptable

quickoats · Jul 8, 2021

kimtuluu said:
This is my answer of part (a). Please let me know if it is acceptable
View attachment 31027

No this isn’t the proper way to prove it using a substitution.
You start by saying let u = a - x. Then you find du/dx, find the new bounds, then sub everything back into the original equation to change it from being in terms of x to in terms of u.
This method should be familiar if you are doing MX1/2

stupid_girl · Jul 8, 2021

kimtuluu said:
I thought so too. Just didn't think that it would involve substitution as well as integration by parts where full working out is required for a 4 mark question.

This question does not require integration by parts.

kimtuluu · Jul 8, 2021

stupid_girl said:
This question does not require integration by parts.

My bad. I meant partial fraction not integration by parts

CM_Tutor · Jul 8, 2021

kimtuluu said:
This is my answer of part (a). Please let me know if it is acceptable
View attachment 31027

As @quickoats has noted, this is not the standard way to do this proof. However, that doesn't mean that it is either wrong or unacceptable.

It would be more usual to start your approach with "Let $F(x)$ be the primitive function of $f(x)$ ", allowing you to go straight to

$\int_0^a f(x)\ dx = \bigg[F(x)\bigg]_0^a = F(a) - F(0)$

Having said that, I see no flaw in your method and it is a valid proof. I would give it full marks.

The more usual approach would be:

$\begin{align*} \int_0^a f(x)\ dx &= \int_{x=0}^{x=a} f(a - u)\times -1\ du \qquad \text{on making the substitution $x = a - u \implies dx = -du$} \\ &= -\int_a^0 f(a - u)\ du \qquad \text{as, if $x = 0$ then $u = a$, and if $x = a$ then $u = 0$} \\ &= \int_0^a f(a - u)\ du \qquad \\ &= \int_0^a f(a - x)\ dx \qquad \text{as $u$ is a dummy variable} \end{align*}$

mr.habibbi · Jul 29, 2021

Screen Shot 2021-07-29 at 3.14.39 pm.png

is there an alternative way of proving this without needing the "in a definite integral, the variable is interchangeable" statement (because i lost a mark for not writing this statement in my proof)

idkkdi · Jul 29, 2021

mr.habibbi said:
View attachment 31306
is there an alternative way of proving this without needing the "in a definite integral, the variable is interchangeable" statement (because i lost a mark for not writing this statement in my proof)

look above for the non-substitution method.

mr.habibbi · Jul 29, 2021

u mean kimtuluus answer?

kimtuluu · Jul 29, 2021

mr.habibbi said:
u mean kimtuluus answer?

Yes.

kimtuluu · Jul 29, 2021

mr.habibbi said:
u mean kimtuluus answer?

Yes. The common way is to use substitution however you can use my answer as the alternative and it's acceptable for full mark.

rajeshtheting · Aug 5, 2021

kimtuluu said:
This is my answer of part (a). Please let me know if it is acceptable
View attachment 31027

it is right but depends on the q ... if u got a similar q but it mentioend to use a sub then this would be incorrect however if it just asked to prove it then this would be aceptable

Please help with part b of this integration question (1 Viewer)

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