Finished your HSC? Check out our University Discussion, Non-School forums or help out next year's HSC students!
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page

HSC 2017 MX2 Integration Marathon (archive) (2 Viewers)

Thread starter leehuan
Start date Oct 18, 2016

Status: Not open for further replies.

leehuan

Well-Known Member

#261

Re: HSC 2017 MX2 Integration Marathon

$\begin{align*}I_n&=\int_0^1\left(\frac{x^n+x^{n-1}-x^{n-1}}{\sqrt{x+1}}\right)dx\\ &= \int_0^1 \frac{x^{n-1}(x+1)}{\sqrt{x+1}}-I_{n-1}\\ &\stackrel{IBP}{=}\frac{x^n}n \sqrt{x+1} \bigg|_0^1 - \frac{1}{2n}\int_0^1\frac{x^n}{\sqrt{x+1}}-I_{n-1}\\ \implies \left(1+\frac{1}{2n}\right)I_n&=\frac{\sqrt2}{n} - I_{n-1}\\ (2n+1)I_n&=2\sqrt2 - 2nI_{n-1} \end{align*}$

Last edited: Jun 27, 2017

S

stupid_girl

Active Member

#262

Re: HSC 2017 MX2 Integration Marathon

Try this one.
$\cot 3x\sec^2(ln \sqrt{sin 3x})$

pikachu975

Premium Member

#263

Re: HSC 2017 MX2 Integration Marathon

stupid_girl said:
Try this one.
$\cot 3x\sec^2(ln \sqrt{sin 3x})$

d(lnsqrt(sin3x)) = 3/2 * cot3x * dx
2/3 * d(lnsqrt(sin3x)) = cot3x dx

Therefore integrand becomes 2/3 * sec^2 (lnsqrt(sin3x)) * d(lnsqrt(sin3x))
Integrate:
= 2/3 * tan(lnsqrt(sin3x)) + C

D

Drongoski

Well-Known Member

#264

Re: HSC 2017 MX2 Integration Marathon

pikachu975 said:
d(lnsqrt(sin3x)) = 3/2 * cot3x * dx
2/3 * d(lnsqrt(sin3x)) = cot3x dx

Therefore integrand becomes 2/3 * sec^2 (lnsqrt(sin3x)) * d(lnsqrt(sin3x))
Integrate:
= 2/3 * tan(lnsqrt(sin3x)) + C

Not bad.

J

jathu123

Active Member

#265

Re: HSC 2017 MX2 Integration Marathon

long time no integrals
here's a couple of good ones to try out!

$$\noindent $ \int \frac{x^2-2}{\left ( x^4+5x^2+4 \right )\tan^{-1}\left ( x+\frac{2}{x} \right )}$ d$x \\ \\ \\ \int e^x\left ( \frac{1-x}{1+x^2} \right ) ^2$ d$x \\ \\ \\ \int \frac{\sin x - \cos x}{\sqrt{\sin(2x)}}$ d$x \\ \\ \\\int \left ( \frac{1}{\ln x}+\ln \left ( \ln x \right ) \right )$ d$x$

Last edited: Aug 12, 2017

pikachu975

Premium Member

#266

Re: HSC 2017 MX2 Integration Marathon

jathu123 said:
long time no integrals
here's a couple of good ones to try out!

$$\noindent $ \int \frac{x^2-2}{\left ( x^4+5x^2+4 \right )\tan^{-1}\left ( x+\frac{2}{x} \right )}$ d$x \\ \\ \\ \int e^x\left ( \frac{1-x}{1+x^2} \right ) ^2$ d$x \\ \\ \\ \int \frac{\sin x - \cos x}{\sqrt{\sin(2x)}}$ d$x \\ \\ \\\int \left ( \frac{1}{\ln x}+\ln \left ( \ln x \right ) \right )$ d$x$

$\noindent$ $ 1) By inspection $ \frac{\left(x^2-2\right)}{\left(x^4+5x^2+4\right)} $ is the derivative of $ tan^{-1}\left(x+\frac{2}{x}\right) $ hence the integral becomes: $ \\ \\ \quad \frac{d\left(tan^{-1}\left(x+\frac{2}{x}\right)\right)}{tan^{-1}\left(x+\frac{2}{x}\right)} $ which equals $ ln\left|tan^{-1}\left(x+\frac{2}{x}\right)\right|+C$

$$\noindent$ $ 4) let$ \quad u = lnx , \quad e^u du = dx \\ $ Integral becomes $ \int \:\left(\frac{1}{u}+lnu\right)e^udu $ and applying reverse product rule you get \quad xln(lnx) + C.$

Last edited: Aug 12, 2017

O

omegadot

Active Member

#267

Re: HSC 2017 MX2 Integration Marathon

jathu123 said:
long time no integrals
here's a couple of good ones to try out!

$$\noindent $ \int \frac{x^2-2}{\left ( x^4+5x^2+4 \right )\tan^{-1}\left ( x+\frac{2}{x} \right )}$ d$x \\ \\ \\ \int e^x\left ( \frac{1-x}{1+x^2} \right ) ^2$ d$x \\ \\ \\ \int \frac{\sin x - \cos x}{\sqrt{\sin(2x)}}$ d$x \\ \\ \\\int \left ( \frac{1}{\ln x}+\ln \left ( \ln x \right ) \right )$ d$x$

$\noindent The second one can also be found using the reverse product rule. Begin by noting that$

$\begin{align*}\left (\frac{1 - x}{1 + x^2} \right )^2 &= \frac{(1 - x)^2}{(1 + x^2)^2} = \frac{1 + x^2}{(1 + x^2)^2} - \frac{2x}{(1 + x^2)^2} = \frac{1}{1 + x^2} - \frac{2x}{(1 + x^2)^2} = f(x) + f'(x),\end{align*}$

$$\noindent where $f(x) = \frac{1}{1 + x^2}.$

$\noindent So the integral becomes$

$\begin{align*}\int e^x \left (\frac{1 - x}{1 + x^2} \right )^2 \, dx &= \int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C = \frac{e^x}{1 + x^2} + C.\end{align*}$

O

omegadot

Active Member

#268

Re: HSC 2017 MX2 Integration Marathon

jathu123 said:
long time no integrals
here's a couple of good ones to try out!

$$\noindent $ \int \frac{x^2-2}{\left ( x^4+5x^2+4 \right )\tan^{-1}\left ( x+\frac{2}{x} \right )}$ d$x \\ \\ \\ \int e^x\left ( \frac{1-x}{1+x^2} \right ) ^2$ d$x \\ \\ \\ \int \frac{\sin x - \cos x}{\sqrt{\sin(2x)}}$ d$x \\ \\ \\\int \left ( \frac{1}{\ln x}+\ln \left ( \ln x \right ) \right )$ d$x$

$\noindent There is quite a nice way to do the third one. We begin by noting that$

$\begin{align*}\sin 2x &= 2 \sin x \cos x = 1 + 2 \sin x \cos x - 1 = (\cos^2 x + \sin^2 x) + 2\sin x \cos x - 1 = (\cos x + \sin x)^2 - 1.\end{align*}$

$\noindent So, for the integral we have$

$\begin{align*}\int \frac{\sin x - \cos x}{\sqrt{\sin 2x}} \, dx &= \int \frac{\sin x - \cos x}{\sqrt{(\cos x + \sin x)^2 - 1}} \, dx.\end{align*}$

$\noindent Letting $u = \cos x + \sin x, du = -(\sin x - \cos x) \, dx $ so$

$\begin{align*}\int \frac{\sin x - \cos x}{\sqrt{\sin 2x}} \, dx &= -\int \frac{du}{\sqrt{u^2 - 1}}\\ &=-\ln |u + \sqrt{u^2 - 1}| + C\\ &= -\ln |\cos x + \sin x + \sqrt{\sin 2x}| + C.\end{align*}$

pikachu975

Premium Member

#269

Re: HSC 2017 MX2 Integration Marathon

Integrate 1/[(sinx)sqrt(sin2x)]

dan964

what

#270

Re: HSC 2017 MX2 Integration Marathon

2017 thread is now closed.
The new 2018 marathon has been created.

dan

Status: Not open for further replies.

Share:

Facebook Twitter Reddit Pinterest Tumblr WhatsApp Email Link

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

Top