HSC 2016 MX2 Integration Marathon (archive) (2 Viewers)

leehuan · Nov 25, 2015

Re: MX2 2016 Integration Marathon

dan964 said:
what is IBP, is it integration by parts?

Yes

Drsoccerball · Nov 25, 2015

Re: MX2 2016 Integration Marathon

Ekman said:
$\frac{d}{dx} \left( \frac{e^x \cdot \sin{x}}{1-\cos{x}} \right) = \frac{e^x(\sin{x} - 1)}{1-\cos{x}}$

$\therefore \int \left (\frac{1 - \sin x}{1 - \cos x} \right ) e^x \, dx = - \left( \frac{e^x \cdot \sin{x}}{1-\cos{x}} \right) + c$

(This question reminds me of the last question of the 2014 hsc, where doing IBP, via inspection can solve it in a couple of steps)

How did you guess the integral ?

Paradoxica · Nov 25, 2015

Re: MX2 2016 Integration Marathon

Drsoccerball said:
How did you guess the integral ?

Inspection.

Drsoccerball · Nov 25, 2015

Re: MX2 2016 Integration Marathon

Paradoxica said:
Inspection.

AKA Wolfram alpha...

Paradoxica · Nov 25, 2015

Re: MX2 2016 Integration Marathon

Drsoccerball said:
AKA Wolfram alpha...

$$Note that $\frac{\sin{x}}{1-\cos{x}} \equiv \cot{\frac{x}{2}}$

$$An application or two of IBP should suffice. Then the expression should revert back into arguments of $ x.$

Ekman · Nov 26, 2015

Re: MX2 2016 Integration Marathon

Drsoccerball said:
How did you guess the integral ?

Paradoxica said:
Inspection.

Drsoccerball said:
AKA Wolfram alpha...

Paradoxica said:
$$Note that $\frac{\sin{x}}{1-\cos{x}} \equiv \cot{\frac{x}{2}}$

$$An application or two of IBP should suffice. Then the expression should revert back into arguments of $ x.$

I didn't use wolfram alpha to guess it, its something that comes with practice. As I said, it's similar to the last question of the 2014 HSC where you can guess it and solve the integral in a couple of steps.

Guessing a term and differentiating it to make it look like the given integral is my way of doing IBP, and hence the reason why I did it that way.

InteGrand · Nov 26, 2015

Re: MX2 2016 Integration Marathon

$Essentially, here is a way to do it without just completely guessing the answer.$

$Write $\mathcal{I} = \int \frac{1-\sin x }{1-\cos x}\cdot \mathrm{e}^x \text{ d}x$ as $\mathcal{I} = \int \frac{\mathrm{e}^x}{1-\cos x}\text{ d}x - \int \frac{\sin x}{1- \cos x}\cdot \mathrm{e}^x \text{ d}x$.$

$We will use IBP on the second integral in this last expression for $\mathcal{I}$. We will integrate the $\mathrm{e}^x$ and differentiate the $\frac{\sin x}{1-\cos x}$. Clearly, $\int \mathrm{e}^x \text{ d}x = \mathrm{e}^x$. For the differentiation part of the IBP, we have$

$\begin{align*}\frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{\sin x}{1-\cos x} \right)&= \frac{\cos x \left(1 - \cos x \right) - \sin^2 x}{\left(1 - \cos x \right)^2} \quad (\text{quotient rule}) \\ &= \frac{\cos x - \cos^2 x - \sin^2 x}{\left(1 - \cos x \right)^2} \\ &= \frac{\cos x - 1}{\left(1 - \cos x \right)^2} \quad (\text{Pythagorean trig. identity}) \\ & = -\frac{1}{1-\cos x}. \end{align*}$

$So using IBP, we have$

$\begin{align*}\mathcal{I}&= \int \frac{\mathrm{e}^x}{1-\cos x}\text{ d}x - \int \frac{\sin x}{1- \cos x}\cdot \mathrm{e}^x \text{ d}x \\ &= \int \frac{\mathrm{e}^x}{1-\cos x}\text{ d}x - \left[\frac{\sin x}{1-\cos x}\cdot \mathrm{e}^x - \int \frac{-1}{1-\cos x}\cdot \mathrm{e}^x \text{ d}x \right]\\ &= -\frac{\sin x}{1-\cos x}\cdot \mathrm{e}^x +\mathcal{C} \quad (\text{since the two integrals cancel})\\ & = \frac{\sin x}{\cos x - 1}\cdot \mathrm{e}^x + \mathcal{C}.\end{align*}$

omegadot · Nov 26, 2015

Re: MX2 2016 Integration Marathon

Ekman said:
I didn't use wolfram alpha to guess it, its something that comes with practice. As I said, it's similar to the last question of the 2014 HSC where you can guess it and solve the integral in a couple of steps.

Guessing a term and differentiating it to make it look like the given integral is my way of doing IBP, and hence the reason why I did it that way.

$$\noindent For those after a more systematic approach, this is what I used. Observe that $\\\begin{align*}\frac{d}{dx} \left (f(x) e^x \right ) = \left (f(x) + f'(x) \right ) e^x\\\Rightarrow \int \left (f(x) + f'(x) \right ) e^x \, dx = f(x) e^x + {\cal{C}}\end{align*}$

$$\noindent So$\\\begin{align*}\int \left (\frac{1 - \sin x}{1 - \cos x} \right ) e^x \, dx &= \int \left (\frac{1}{1 - \cos x} - \frac{\sin x}{1 - \cos x} \right ) e^x \, dx\\&=\int \left (\frac{1 - \cos x}{(1 - \cos x)^2} - \frac{\sin x}{1 - \cos x} \right ) e^x \, dx\\&= -\int \left (\frac{\sin x}{1 - \cos x} + \frac{-(1-\cos x)}{(1 - \cos x)^2} \right ) e^x \, dx\\&=-\int \left (\frac{e^x \sin x}{1 - \cos x} \right )' \, dx\\&= -\frac{e^x \sin x}{1 - \cos x} + {\cal{C}}\end{align*}$

$$\noindent \textbf{Alternatively}, as Paradoxica has suggested, one can just thrash the integral out using IBP. To do this we begin by noting the following two trig identities:$\\\frac{\sin x}{1 - \cos x} = \cot \frac{x}{2} \quad \mbox{and} \quad \sin^2 \frac{x}{2} = \frac{1 - \cos x}{2}.$

$\begin{align*}\int \left (\frac{1 - \sin x}{1 - \cos x} \right ) e^x \, dx &= \int \frac{e^x}{1 - \cos x} \, dx - \int \frac{e^x \sin x}{1 - \cos x} \, dx\\&= \int \frac{e^x}{1 - \cos x} \, dx - \int e^x \cot \frac{x}{2} \, dx\\&= \int \frac{e^x}{1 - \cos x} \, dx - e^x \cot \frac{x}{2} - \frac{1}{2} \int e^x \mbox{cosec}^2 \,\frac{x}{2} \, dx \quad \mbox{(by parts)}\\&= \int \frac{e^x}{1 - \cos x} \, dx -e^x \cot \frac{x}{2} -\frac{1}{2} \int e^x \cdot \frac{1}{\sin^2 \frac{x}{2}} \, dx\\&= -e^x \cot \frac{x}{2} + \int \frac{e^x}{1 - \cos x} \, dx - \int \frac{e^x}{1 - \cos x} \, dx\\&= - e^x \cot \frac{x}{2} + {\cal{C}}\\&= - \frac{e^x \sin x}{1 - \cos x} + {\cal{C}},\end{align*}$

$\noindent as before.$

omegadot · Nov 26, 2015

Re: MX2 2016 Integration Marathon

InteGrand said:
$Essentially, here is a way to do it without just completely guessing the answer.$

$Write $\mathcal{I} = \int \frac{1-\sin x }{1-\cos x}\cdot \mathrm{e}^x \text{ d}x$ as $\mathcal{I} = \int \frac{\mathrm{e}^x}{1-\cos x}\text{ d}x - \int \frac{\sin x}{1- \cos x}\cdot \mathrm{e}^x \text{ d}x$.$

$We will use IBP on the second integral in this last expression for $\mathcal{I}$. We will integrate the $\mathrm{e}^x$ and differentiate the $\frac{\sin x}{1-\cos x}$. Clearly, $\int \mathrm{e}^x \text{ d}x = \mathrm{e}^x$. For the differentiation part of the IBP, we have$

$\begin{align*}\frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{\sin x}{1-\cos x} \right)&= \frac{\cos x \left(1 - \cos x \right) - \sin^2 x}{\left(1 - \cos x \right)^2} \quad (\text{quotient rule}) \\ &= \frac{\cos x - \cos^2 x - \sin^2 x}{\left(1 - \cos x \right)^2} \\ &= \frac{\cos x - 1}{\left(1 - \cos x \right)^2} \quad (\text{Pythagorean trig. identity}) \\ & = -\frac{1}{1-\cos x}. \end{align*}$

$So using IBP, we have$

$\begin{align*}\mathcal{I}&= \int \frac{\mathrm{e}^x}{1-\cos x}\text{ d}x - \int \frac{\sin x}{1- \cos x}\cdot \mathrm{e}^x \text{ d}x \\ &= \int \frac{\mathrm{e}^x}{1-\cos x}\text{ d}x - \left[\frac{\sin x}{1-\cos x}\cdot \mathrm{e}^x - \int \frac{-1}{1-\cos x}\cdot \mathrm{e}^x \text{ d}x \right]\\ &= -\frac{\sin x}{1-\cos x}\cdot \mathrm{e}^x +\mathcal{C} \quad (\text{since the two integrals cancel})\\ & = \frac{\sin x}{\cos x - 1}\cdot \mathrm{e}^x + \mathcal{C}.\end{align*}$

I see InteGrand beat me to it.....

InteGrand · Nov 26, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
I see InteGrand beat me to it.....

Well my method doesn't require those trig. identities like in your second method, but the idea is similar (cancel out one integral by using the integral from the IBP step).

omegadot · Nov 26, 2015

Re: MX2 2016 Integration Marathon

InteGrand said:
Well my method doesn't require those trig. identities like in your second method, but the idea is similar (cancel out one integral by using the integral from the IBP step).

Yes, I agree, your method is definitely more elegant.

omegadot · Nov 26, 2015

Re: MX2 2016 Integration Marathon

$$\noindent In a similar vein to my previous question, find $\int \left (\frac{1 - \tan x}{1 - \sec x} \right ) \mbox{e}^x \, dx.$

Paradoxica · Nov 26, 2015

Re: MX2 2016 Integration Marathon

$$Observe $\frac{1-\tan{x}}{1-\sec{x}} \equiv 1 + \cot{\frac{x}{2}} - \frac{1}{2}\csc^2{\frac{x}{2}}$

$Rewrite this as $ \left ( 1 + \cot{\frac{x}{2}} \right ) + \frac{d}{dx} \left ( 1 + \cot{\frac{x}{2}} \right ) $ by virtue of reverse chaining$

$$Then use the derive-able integration formula:$

$$ \int (f(x) + f'(x))e^x dx = f(x)e^x + C$

leehuan · Nov 26, 2015

Re: MX2 2016 Integration Marathon

Not the easiest thing to observe. Obviously t-formulae are useful but best give a mention.

InteGrand · Nov 26, 2015

Re: MX2 2016 Integration Marathon

Alternatively, we can again do it by breaking up the integral and using IBP to cancel out some integrals and be left with the answer. This would require a tedious differentiation using the quotient rule, but avoids the need for non-well-known trig. identities.

Paradoxica · Nov 26, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
Not the easiest thing to observe. Obviously t-formulae are useful but best give a mention.

Never used t-formulae. Reduced the fraction purely into the two fundamental ratios and applied double angle formulae.

Drsoccerball · Nov 26, 2015

Re: MX2 2016 Integration Marathon

Here's a question for new 2016'ers I just thought of:

$\int{x^2cot^{-1}x}dx$

omegadot · Nov 26, 2015

Re: MX2 2016 Integration Marathon

porcupinetree said:

$\noindent Alternatively, let $x = u^2, dx = 2u \, du$ so that$

$\begin{align*}\int \frac{9 + 6\sqrt{x} + x}{4\sqrt{x} + x} \, dx &= 2 \int \frac{9 + 6u + u^2}{4 + u} \, du\\&= 2 \int \frac{(u + 4)(u + 2) + 1}{u + 4} \, du\\&= 2 \int \left (u + 2 + \frac{1}{u + 4} \right ) \, du\\&= u^2 + 4u + 2 \ln |u + 4| + {\cal{C}}\\&= x + 4 \sqrt{x} + 2 \ln (\sqrt{x} + 4) + {\cal{C}}.\end{align*}$

$$\noindent \textbf{New Question}$

$$\noindent Find $ \int \frac{\ln (2x)}{x \ln (3x)} \, dx.$

kawaiipotato · Nov 27, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent Find $ \int \frac{\ln (2x)}{x \ln (3x)} \, dx.$

$= \int \frac{ \ln 2 + \ln x }{x( \ln 3 + \ln x)} dx$

$$ let u $ = \ln x$

$du = \frac{dx}{x}$

$= \int \frac{ \ln 2 + u }{ \ln 3 + u} du$

$= \int \frac{ \ln 3 - \ln 3 + \ln 2 + u}{ \ln 3 + u} du$

$= \int 1 du + \ln \frac{2}{3} \int \frac{1}{ \ln 3 + u} du$

$= u + \ln \frac{2}{3} * \ln \left| \ln 3 + u \right| + \mathcal{C}$

$= \ln x + \ln \frac{2}{3} * \ln \left| \ln 3 + \ln x \right| + \mathcal{C}$

______________

$New Question$

$$ Show $ \int_{0}^{a} \frac{x}{\cos x \cos (a-x)} dx = \frac{a}{\sin a }\ln \sec a$

leehuan · Nov 27, 2015

Re: MX2 2016 Integration Marathon

Question skipped:

Drsoccerball said:
Here's a question for new 2016'ers I just thought of:

$\int{x^2cot^{-1}x}dx$

Though, if one does not know the derivative of cotangent inverse one may like to note that arccot(x) = arctan(1/x)
-----------------------------
sinx/(1+cosx) I would've felt happy using double angles. I don't feel too comfortable doing so when sec is hanging there even if it's really 1/cos

HSC 2016 MX2 Integration Marathon (archive) (2 Viewers)

Well-Known Member

Well-Known Member

-insert title here-

Well-Known Member

-insert title here-

Well-Known Member

Well-Known Member

Active Member

Active Member

Well-Known Member

Active Member

Active Member

-insert title here-

Well-Known Member

Well-Known Member

-insert title here-

Well-Known Member

Active Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)