HSC 2015 MX2 Integration Marathon (archive) (1 Viewer)

Ekman · May 21, 2015

Re: MX2 2015 Integration Marathon

Next Question:

$\int_{0}^{2} \frac{ln(x+2)}{4+x^2} dx$

VBN2470 · May 21, 2015

Re: MX2 2015 Integration Marathon

$Substitute $x=2\tan{\theta}$ and use the property that for some $a\in\mathbb{R}, \\ \int_{0}^{a} f(\theta)\text{ d}\theta=\int_{0}^{a} f(a-\theta)\text{ d}\theta.$ \\ Answer is $\frac{3\pi}{16}\cdot\ln2$$

Ekman · May 21, 2015

Re: MX2 2015 Integration Marathon

Next Question:

$\int \sin^{-1}\left(\sqrt{\frac{x}{x+a}}\right) dx$

InteGrand · May 21, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Next Question:

$\int \sin^{-1}\left(\sqrt{\frac{x}{x+a}}\right) dx$

$Let $I=\int \sin^{-1}\left(\sqrt{\frac{x}{x+a}}\right) \text{ d}x$. Note that:$

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \sin^{-1} \left(\sqrt{\frac{x}{x+a}}\right)\right )=\frac{1}{\sqrt{1-\frac{x}{x+a}}}\times \frac{\mathrm{d} }{\mathrm{d} x}\left(\sqrt{\frac{x}{x+a}} \right )$

$= \frac{1}{2\sqrt{\frac{x}{x+a}}}\times\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{x}{x+a} \right )\times \frac{1}{\sqrt{\frac{x+a-x}{x+a}}}$

$=\frac{\sqrt{x+a}}{2\sqrt{x}}\times \frac{x+a-x}{(x+a)^2}\times \frac{\sqrt{x+a}}{\sqrt{a}}$

$=\frac{x+a}{2\sqrt{x}(x+a)^2}\times \sqrt{a}$

$=\frac{\sqrt{a}}{2\sqrt{x}(x+a)}$

$$\Rightarrow I = x\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\right) - \int \frac{\sqrt{a}x}{2\sqrt{x}(x+a)} \text{ d}x$ (using integration by parts)$

$= x\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\right) - \frac{\sqrt{a}}{2}\int \frac{\sqrt{x}}{(x+a)} \text{ d}x$

$$ = x\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\right) - \frac{\sqrt{a}}{2}J$, where $J=\int \frac{\sqrt{x}}{(x+a)} \text{ d}x$.$

$To find $J$, let $x=a\tan^2 \theta,\mathrm{d}x=2a\tan\theta \sec^2 \theta \text{ d}\theta\Rightarrow \tan \theta = \sqrt{\frac{x}{a}},\theta = \tan ^{-1}\sqrt{\frac{x}{a}} , x+a = a(1+\tan^2 \theta)=a\sec^2 \theta$.$

$Then $J=\int \frac{\sqrt{a}\tan \theta}{a\sec^2 \theta}\times 2a\tan\theta \sec^2 \theta \text{ d}\theta $$

$=2\sqrt{a}\int \tan ^2 \theta \text{ d}\theta$

$=2\sqrt{a}\int (\sec^2 \theta -1) \text{ d}\theta$

$=2\sqrt{a}(\tan \theta -\theta)+c$

$$=2\sqrt{a}\left(\sqrt{\frac{x}{a}} -\tan ^{-1}\sqrt{\frac{x}{a}}\right)+c$.$

$Hence $I=x\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\right) -\frac{\sqrt{a}}{2}\times \left[ 2\sqrt{a}\left(\sqrt{\frac{x}{a}} -\tan ^{-1}\sqrt{\frac{x}{a}}\right)\right]+c$$

$$=x\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\right)-a\left(\sqrt{\frac{x}{a}} -\tan ^{-1}\sqrt{\frac{x}{a}}\right)+c$.$

Drsoccerball · May 21, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
$Let $I=\int \sin^{-1}\left(\sqrt{\frac{x}{x+a}}\right) \text{ d}x$. Note that:$

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \sin^{-1} \left(\sqrt{\frac{x}{x+a}}\right)\right )=\frac{1}{\sqrt{1-\frac{x}{x+a}}}\times \frac{\mathrm{d} }{\mathrm{d} x}\left(\sqrt{\frac{x}{x+a}} \right )$

$= \frac{1}{2\sqrt{\frac{x}{x+a}}}\times\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{x}{x+a} \right )\times \frac{1}{\sqrt{\frac{x+a-x}{x+a}}}$

$=\frac{\sqrt{x+a}}{2\sqrt{x}}\times \frac{x+a-x}{(x+a)^2}\times \frac{\sqrt{x+a}}{\sqrt{a}}$

$=\frac{x+a}{2\sqrt{x}(x+a)^2}\times \sqrt{a}$

$=\frac{\sqrt{a}}{2\sqrt{x}(x+a)}$

$$\Rightarrow I = x\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\right) - \int \frac{\sqrt{a}x}{2\sqrt{x}(x+a)} \text{ d}x$ (using integration by parts)$

$= x\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\right) - \frac{\sqrt{a}}{2}\int \frac{\sqrt{x}}{(x+a)} \text{ d}x$

$$ = x\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\right) - \frac{\sqrt{a}}{2}J$, where $J=\int \frac{\sqrt{x}}{(x+a)} \text{ d}x$.$

$To find $J$, let $x=a\tan^2 \theta,\mathrm{d}x=2a\tan\theta \sec^2 \theta \text{ d}\theta\Rightarrow \tan \theta = \sqrt{\frac{x}{a}},\theta = \tan ^{-1}\sqrt{\frac{x}{a}} , x+a = a(1+\tan^2 \theta)=a\sec^2 \theta$.$

$Then $J=\int \frac{\sqrt{a}\tan \theta}{a\sec^2 \theta}\times 2a\tan\theta \sec^2 \theta \text{ d}\theta $$

$=2\sqrt{a}\int \tan ^2 \theta \text{ d}\theta$

$=2\sqrt{a}\int (\sec^2 \theta -1) \text{ d}\theta$

$=2\sqrt{a}(\tan \theta -\theta)+c$

$$=2\sqrt{a}\left(\sqrt{\frac{x}{a}} -\tan ^{-1}\sqrt{\frac{x}{a}}\right)+c$.$

$Hence $I=x\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\right) -\frac{\sqrt{a}}{2}\times \left[ 2\sqrt{a}\left(\sqrt{\frac{x}{a}} -\tan ^{-1}\sqrt{\frac{x}{a}}\right)\right]+c$$

$$=x\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\right)-a\left(\sqrt{\frac{x}{a}} -\tan ^{-1}\sqrt{\frac{x}{a}}\right)+c$.$

lol app theres a simple substitution

Ekman · May 21, 2015

Re: MX2 2015 Integration Marathon

Alt Solution:

$$ Substitute $ x=a\tan^{2}{\theta}$

$\therefore I= 2a\int\theta\tan{\theta}\sec^{2}{\theta} d\theta$

$$Through IBP: $ I=a(\theta\tan^{2}{\theta}-\tan{\theta}+\theta) +c$

$\therefore = x\tan^{-1}{\sqrt{\frac{x}{a}}} - \sqrt{ax} + a\tan^{-1}{\sqrt{\frac{x}{a}}} +c$

------------------------------------------------
Next Question:

$\int \tan^{-1}\left({\frac{1}{1+x(x-1)}\right) dx$

InteGrand · May 21, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
lol app theres a simple substitution

A slightly simpler version is to start with $x=a\tan^2 \theta$ .

Edit: just done above

FrankXie · May 21, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
$\int \tan^{-1}\left({\frac{1}{1+x(x-1)}\right) dx$

$\tan^{-1}\left({\frac{1}{1+x(x-1)}\right)=\tan^{-1}x-\tan^{-1}(x-1)$

Drsoccerball · May 21, 2015

Re: MX2 2015 Integration Marathon

FrankXie said:
$\tan^{-1}\left({\frac{1}{1+x(x-1)}\right)=\tan^{-1}x-\tan^{-1}(x-1)$

Not sure how thes are equal

InteGrand · May 21, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Not sure how thes are equal

There is an identity $\tan ^{-1} A - \tan^{-1} B = \tan^{-1} \left(\frac{A-B}{1+AB} \right)$ .

(One can derive it starting from $\tan (\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1+\tan \alpha \tan \beta}$ .)

Drsoccerball · May 21, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
There is an identity $\tan ^{-1} A - \tan^{-1} B = \tan^{-1} \left(\frac{A-B}{1+AB} \right)$ .

(One can derive it starting from $\tan \alpha - \tan \beta = \frac{\tan \alpha - \tan \beta}{1+\tan \alpha \tan \beta}$ .)

never seen this before...

InteGrand · May 21, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
never seen this before...

I don't think the HSC expects students to know it consciously like for the trig compound angle formulae, and if it was required, they'd probably get you to derive it as an earlier part of a Q, or they'd give it to you.

InteGrand · May 21, 2015

Re: MX2 2015 Integration Marathon

(It's basically, for $\alpha,\beta \in \left(-\frac{\pi}{2},\frac{\pi}{2} \right)$ , let $A=\tan \alpha,B=\tan \beta \Leftrightarrow \alpha = \tan^{-1}A, \beta = \tan^{-1}B$ so $\tan (\alpha - \beta)= \frac{\tan \alpha - \tan \beta}{1+\tan\alpha \tan\beta}\Rightarrow \alpha - \beta = \tan ^{-1} \left(\frac{\tan \alpha - \tan \beta}{1+\tan\alpha \tan\beta} \right)$ , provided that $\alpha - \beta \in \left(-\frac{\pi}{2},\frac{\pi}{2} \right)$ , which implies that $\tan^{-1}A-\tan^{-1}B=\tan ^{-1} \left(\frac{A - B}{1+A B} \right)$ )

leehuan · May 24, 2015

Re: MX2 2015 Integration Marathon

Tell me if this is going a bit too far..

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

InteGrand · May 24, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
Tell me if this is going a bit too far..

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

$We use integration by parts, using $u=\cot ^{-1} x, \mathrm{d}v = \mathrm{d}x\Rightarrow \mathrm{d}u=\frac{-1}{1+x^2}\text{ d}x,v=x$.$

$\int \cot^{-1}x\text{ d}x = x\cot^{-1}x-\int x\times \frac{-1}{1+x^2}\text{ d}x$

$=x\cot^{-1}x+\int \frac{x}{1+x^2}\text{ d}x$

$=x\cot^{-1}x+\frac{1}{2}\ln (1+x^2)+c.$

InteGrand · May 24, 2015

Re: MX2 2015 Integration Marathon

$Find $\int \frac{x^2}{1+x^4}\text{ d}x$.$

leehuan · May 24, 2015

Re: MX2 2015 Integration Marathon

I have horrible memories with an integral that popped that one out. Refer to pg17.

Basically speaking, 2I appeared halfway.

I thought I wouldn't quote it so someone else may have an attempt, especially since I didn't show working out for partial fractions.

Ekman · May 25, 2015

Re: MX2 2015 Integration Marathon

Next Question:

$\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin{x} +\cos{x}}dx$

VBN2470 · May 25, 2015

Re: MX2 2015 Integration Marathon

Use the property that $$\int_{0}^{\frac{\pi}{2}} f(\theta)\text{ d}\theta=\int_{0}^{\frac{\pi}{2}} f(\frac{\pi}{2}-\theta)\text{ d}\theta$.$ then use auxiliary angle method to solve the rest.

Drsoccerball · May 26, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
$Find $\int \frac{x^2}{1+x^4}\text{ d}x$.$

$let x = \sqrt{tan\theta}$

you're left with $\frac{1}{2} \int{\sqrt{tan\theta}}$ which has been solved by lee somewhere in this forum

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