HSC 2017 MX2 Integration Marathon (archive) (1 Viewer)

InteGrand · Mar 3, 2017

Re: HSC 2017 MX2 Integration Marathon

si2136 said:
Do you mean examples?

What I meant was like I = Int(sec^6) dx or I = Int(tan^8) dx

A trig function to the power of a high even number, which using half angle would be tedious and time-costly.

You can also try reduction formulas.

Paradoxica · Mar 3, 2017

Re: HSC 2017 MX2 Integration Marathon

si2136 said:
Do you mean examples?

What I meant was like I = Int(sec^6) dx or I = Int(tan^8) dx

A trig function to the power of a high even number, which using half angle would be tedious and time-costly.

For secants and tangents, you simply substitute t=tanx

$\int \sec^{2n+2}{x}\,\text{d}x = \int (t^2+1)^n \, \text{d}t$

For tangents, it's easier to divide into two cases:

case 1: exponent is 4n+2

case 2: exponent is 4n

that way, the long division will be very straight forward

cosecants and cotangents are dealt with identically.

sines and cosines are much tricker to deal with, which is why reduction exists.

si2136 · Mar 3, 2017

Re: HSC 2017 MX2 Integration Marathon

For Sine and Cosine, are there any simpler ways without reduction?

Mahan1 · Mar 4, 2017

Re: HSC 2017 MX2 Integration Marathon

nancy99 said:
View attachment 33823

$\int \frac{x^2+2x+1}{x^2-3x-1} dx = \int \frac{x^2-3x-1}{x^2-3x-1} dx +\frac{5}{2} \int \frac{2x-3}{x^2-3x-1} dx + \frac{23}{2} \int \frac{1}{x^2-3x-1} dx$

$= \int \frac{x^2-3x-1}{x^2-3x-1} dx +\frac{5}{2} \int \frac{2x-3}{x^2-3x-1} dx + \frac{19}{2} \int \frac{1}{(x-\frac{3+\sqrt{13}}{2})(x-\frac{3-\sqrt{13}}{2})} dx$

$= \int \frac{x^2-3x-1}{x^2-3x-1} dx + \frac{5}{2} \int \frac{2x-3}{x^2-3x-1} dx + \frac{1}{\sqrt{13}}\frac{19}{2}( ( ln (x-\frac{3+\sqrt{13}}{2}) - ln (x - \frac{3-\sqrt{13}} {2} ) )$

$= x + \frac{5}{2}ln(x^2-3x-1) + \frac{1}{\sqrt{13}}\frac{19}{2}( ( ln (x-\frac{3+\sqrt{13}}{2}) - ln (x - \frac{3-\sqrt{13}} {2} ) )$

Paradoxica · Mar 4, 2017

Re: HSC 2017 MX2 Integration Marathon

si2136 said:
For Sine and Cosine, are there any simpler ways without reduction?

It depends on your definition of simple.

You could use the polar form of a complex number combined with the binomial theorem to obtain the power of the sine/cosine in multiple angle form, free of exponents.

Then there's the option of substitution. The details of that particular method is left as an exercise to the reader. (hint: differentiation under the integral sign)

Paradoxica · Mar 6, 2017

Re: HSC 2017 MX2 Integration Marathon

Mahan1 said:
$\textrm{1.If} \ \alpha, \beta \ \textrm{ are real numbers such that}$
$\beta^{3} -6\beta^{2} + 13\beta = 19 , \ \alpha^{3}-6\alpha^{2} +13\alpha = 1$
$\textrm{Hence, find } \ (\alpha + \beta)$

$\textrm{2.If }f(x) = \frac{4^{x}}{4^{x}+2}$
$\textrm{Find} f(\frac{1}{14}) + ... + f(\frac{13}{14}) \ \textrm{without using calculator}$

This does not belong here.

Paradoxica · Mar 6, 2017

Re: HSC 2017 MX2 Integration Marathon

$\int_0^1 x \log{(\sqrt{1+x}+\sqrt{1-x})}\log{(\sqrt{1+x}-\sqrt{1-x})} \text{d}x$

$\text{Hint: }4ab \equiv (a+b)^2-(a-b)^2$

omegadot · Mar 11, 2017

Re: HSC 2017 MX2 Integration Marathon

Paradoxica said:
$\int_0^1 x \log{(\sqrt{1+x}+\sqrt{1-x})}\log{(\sqrt{1+x}-\sqrt{1-x})} \text{d}x$

$\text{Hint: }4ab \equiv (a+b)^2-(a-b)^2$

$\noindent This integral is hardly easy as one must handle the limits of integration with great care. From the hint we have$

$\begin{align*}4 \ln (\sqrt{1 + x} + \sqrt{1 - x}) \ln (\sqrt{1 + x} - \sqrt{1 - x}) &= \left [\ln(\sqrt{1 + x} + \sqrt{1 - x}) + \ln (\sqrt{1 + x} - \sqrt{1 - x}) \right ]^2\\& \quad - \left [\ln (\sqrt{1 + x} + \sqrt{1 - x}) - \ln (\sqrt{1 + x} - \sqrt{1 - x}) \right ]^2\\&= \ln^2 (2x) - \ln^2 \left (\frac{1 + \sqrt{1 - x^2}}{x} \right ).\end{align*}$

$\noindent So our integral can be rewritten as$

$\begin{align*}I &=\int_0^1 x \ln{(\sqrt{1+x}+\sqrt{1-x})}\log{(\sqrt{1+x}-\sqrt{1-x})} \, dx\\&= \frac{1}{4} \int^1_0 x \ln^2 (2x) \, dx - \frac{1}{4} \int^1_0 x\ln^2 \left (\frac{1 + \sqrt{1 - x^2}}{x} \right ) \, dx\\&= \frac{1}{4} I_1 - \frac{1}{4} I_2.\end{align*}$

$\noindent For the first integral, IBP twice, after exercising great care with the limits of integration, leads to$

$I_2 = \frac{1}{2} \ln^2 2 - \frac{1}{2} \ln 2 + \frac{1}{4},$

$\noindent while for the second integral IBP twice, after again exercising great care with the limits of integration, leads to$

$I_2 = \ln 2.$

$\noindent Thus$

$\begin{align*}\int_0^1 x \ln{(\sqrt{1+x}+\sqrt{1-x})}\log{(\sqrt{1+x}-\sqrt{1-x})} \, dx &= \frac{1}{4} \left [\frac{1}{2} \ln^2 2 - \frac{1}{2} \ln 2 + \frac{1}{4} \right ] - \frac{1}{4} \ln 2 = \frac{1}{8} \ln^2 2 - \frac{3}{8} \ln 2 + \frac{1}{16}.\end{align*}$

omegadot · Mar 11, 2017

Re: HSC 2017 MX2 Integration Marathon

$$\noindent A slightly easier one. Evaluate $\int^3_1 \frac{\ln x}{3 + x^2} \, dx.$

jathu123 · Mar 11, 2017

Re: HSC 2017 MX2 Integration Marathon

omegadot said:
$$\noindent A slightly easier one. Evaluate $\int^3_1 \frac{\ln x}{3 + x^2} \, dx.$

$$let $x=\sqrt3 \tan \theta \\ $d$x=\sqrt3 \sec ^2 \theta$ d$\theta \\ \\ I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\ln \left |\sqrt3 \tan \theta \right |}{3 +3\tan ^2 \theta}\cdot \sqrt3 \sec ^2 \theta$ d$\theta \\ \\ = \frac{\sqrt3}{3}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left ( \ln \sqrt3 + \ln \left | \tan \theta \right | \right )$ d$\theta \\ =\frac{\sqrt3}{3}\left [ x\ln \sqrt3 \right ]_{\frac{\pi}{6}}^\frac{\pi}{3}+\frac{\sqrt3}{3} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\ln\left | \tan \theta \right |$ d$\theta \\ $ Note that $ \ln( \tan \frac{\pi}{4})=0 $ and it is symmetric about $x=\frac{\pi}{4} \\ \\ $ie. $\ln\left | \tan \left ( \frac{\pi}{4}-\alpha \right ) \right |+\ln\left | \tan \left ( \frac{\pi}{4}+\alpha \right ) \right |=0 $ where $ |\alpha| < \frac{\pi}{4} \\ \\ \therefore $ as $ \frac{\pi}{3}= \frac{\pi}{4}+\frac{\pi}{12}$ and $ \frac{\pi}{6}= \frac{\pi}{4}-\frac{\pi}{12}, \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\ln\left | \tan \theta \right |$ d$\theta=0 \\$
$\therefore I=\frac{\sqrt3 \ln \sqrt3}{3}\left (\frac{\pi}{3}-\frac{\pi}{6} \right ) = \frac{\sqrt3 \pi\ln3}{36}$

jathu123 · Mar 11, 2017

Re: HSC 2017 MX2 Integration Marathon

Find:

$\int \frac{\sin x\cos x}{\sin^4 x+\cos^4 x}$ d$x$

fluffchuck · Mar 11, 2017

Re: HSC 2017 MX2 Integration Marathon

jathu123 said:
Find:

$\int \frac{\sin x\cos x}{\sin^4 x+\cos^4 x}$ d$x$

$\int \frac{\sin x\cos x}{\sin^4 x+\cos^4 x}$ d$x$

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

$= \int \frac{\frac{1}{2}\sin 2x}{1 - 2\sin^2 x \cos^2 x}$ d$x$

$= \int \frac{\frac{1}{2}\sin 2x}{1 - 2(\frac{1}{2} sin 2x)^2}$ d$x$

$= \int \frac{\frac{1}{2}\sin 2x}{1 - \frac{1}{2}sin^2 2x}$ d$x$

$= \int \frac{\frac{1}{2}\sin 2x}{1 - \frac{1}{2}(1 - cos^2 2x)}$ d$x$

$= \int \frac{\frac{1}{2}\sin 2x}{ \frac{1}{2} + \frac{1}{2}cos^2 2x }$ d$x$

$= \int \frac{\sin 2x}{1 + \cos^2 2x}$ d$x$

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

InteGrand · Mar 11, 2017

Re: HSC 2017 MX2 Integration Marathon

$\noindent Find $\int \frac{\cos x + \sin x}{3\cos x + 4\sin x}\, \mathrm{d}x$.$

jathu123 · Mar 11, 2017

Re: HSC 2017 MX2 Integration Marathon

InteGrand said:
$\noindent Find $\int \frac{\cos x + \sin x}{3\cos x + 4\sin x}\, \mathrm{d}x$.$

There might be a way to do this by playing around with the numerator, anyways here's how I did it

$$ Divide each term by $\sin x$
$I=\int \frac{1+\tan x}{3+4 \tan x}$ d$x \\ $ let $u= \tan x \\ $ d$u=\sec^2 x $ d$x \\ \Rightarrow $ d$x=\frac{1}{1+u^2}$ d$u \\ \\ \therefore I=\int \frac{1+u}{(3+4u)(1+u^2)}$ d$u \\ =\frac{1}{25}\int \left ( \frac{7-u}{1+u^2}+\frac{4}{3+4u} \right )$ d$u \\ =\frac{1}{25}\left ( 7\tan^{-1}u-\frac{1}{2}\ln \left | 1+u^2 \right |+\ln\left | 3+4u \right | \right ) \\ = \frac{1}{25}\left (7x-\frac{1}{2} \ln\left | \sec^2 x \right |+\ln \left | 3+4 \tan x \right | \right ) \\ $ Upon further tidying up, the integral simplifies to $\\ \\ \frac{1}{25}\left ( 7x+ \ln \left | 3\cos x +4\sin x \right | \right )+C$

InteGrand · Mar 11, 2017

Re: HSC 2017 MX2 Integration Marathon

$\noindent Find $\int \sqrt{1 + e^{x}}\, \mathrm{d}x$.$

BenHowe · Mar 11, 2017

Re: HSC 2017 MX2 Integration Marathon

fluffchuck said:
$\int \frac{\sin x\cos x}{\sin^4 x+\cos^4 x}$ d$x$

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

$= \int \frac{\frac{1}{2}\sin 2x}{1 - 2\sin^2 x \cos^2 x}$ d$x$

$= \int \frac{\frac{1}{2}\sin 2x}{1 - 2(\frac{1}{2} sin 2x)^2}$ d$x$

$= \int \frac{\frac{1}{2}\sin 2x}{1 - \frac{1}{2}sin^2 2x}$ d$x$

$= \int \frac{\frac{1}{2}\sin 2x}{1 - \frac{1}{2}(1 - cos^2 2x)}$ d$x$

$= \int \frac{\frac{1}{2}\sin 2x}{ \frac{1}{2} + \frac{1}{2}cos^2 2x }$ d$x$

$= \int \frac{\sin 2x}{1 + \cos^2 2x}$ d$x$

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

I dont understand the last line. Why isn't it ln?

InteGrand · Mar 11, 2017

Re: HSC 2017 MX2 Integration Marathon

BenHowe said:
I dont understand the last line. Why isn't it ln?

Since the denominator has a cos-squared, the numerator isn't going to be like the derivative of the denominator (it would've been though if the denominator just had cos without the square).

Instead, we have a function that is (proportional to) u'/(1 + u²), where u = cos(2x), and u' is proportional to sin(2x), so this is of the form of an inverse-tan answer.

BenHowe · Mar 11, 2017

Re: HSC 2017 MX2 Integration Marathon

Is this just like something that is known? I've never heard of anything like this lol...

InteGrand · Mar 11, 2017

Re: HSC 2017 MX2 Integration Marathon

BenHowe said:
Is this just like something that is known? I've never heard of anything like this lol...

It's just reverse chain rule (use a substitution u = cos(2x) if in doubt).

jathu123 · Mar 11, 2017

Re: HSC 2017 MX2 Integration Marathon

InteGrand said:
$\noindent Find $\int \sqrt{1 + e^{x}}\, \mathrm{d}x$.$

$$ let $u^2=1+e^x \\2u$ d$u=e^x $ d$x \\ \\ \therefore I=\int \frac{2u^2}{u^2-1}$ d$u\\ = 2\int $ d$u+\int \frac{2}{u^2-1}$ d$u \\ \\ =2u + \ln \left | u-1 \right |-\ln\left | u+1 \right | \\ \\ = 2\sqrt{1+e^x}+\ln\left | \frac{\sqrt{1+e^x} -1}{\sqrt{1+e^x} +1} \right |+C$

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