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 2016 MX2 Integration Marathon (archive) (4 Viewers)

Thread starter leehuan
Start date Nov 5, 2015

Status: Not open for further replies.

D

Drsoccerball

Well-Known Member

#41

Re: MX2 2016 Integration Marathon

$\int{\frac{(sin^2x +1)(sinx +1)}{(cos^2x +1)(cosx +1)}}dx$

leehuan

Well-Known Member

#42

Re: MX2 2016 Integration Marathon

Something must've went wrong to get the negative answer but I'm too tired to check.

Sent from my iPhone using Tapatalk

I

InteGrand

Well-Known Member

#43

Re: MX2 2016 Integration Marathon

leehuan said:
Something must've went wrong to get the negative answer but I'm too tired to check.

Sent from my iPhone using Tapatalk

6 pages...now that is one tedious integral!

J

Jamesw123

New Member

#44

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$Evaluate $\int_0^{\infty} \frac{\tan^{-1}{x}}{x(x^2+1)^5}dx$

$$using the substitution $ y = \tan^{-1}{x}$

$Using the substitution, the integral becomes$

$\int_{0}^{\frac{\pi}{2}} \frac{1}{ \tan y (sec^8 y)}dy$

$= \int_{0}^{\frac{\pi}{2}} cosec^{8}y coty$

$= \left[\frac{cosec^8 y}{8} \right]_{0}^{\frac{\pi}{2}}$

$= \frac{\pi ^8}{2^{11}}$

leehuan

Well-Known Member

#45

Re: MX2 2016 Integration Marathon

Jamesw123 said:
$Using the substitution, the integral becomes$

$\int_{0}^{\frac{\pi}{2}} \frac{1}{ \tan y (sec^8 y)}dy$

$= \int_{0}^{\frac{\pi}{2}} cosec^{8}y coty$

$= \left[\frac{cosec^8 y}{8} \right]_{0}^{\frac{\pi}{2}}$

$= \frac{\pi ^8}{2^{11}}$

1/sec(x) = cos(x), not cosec(x)?

leehuan

Well-Known Member

#46

Re: MX2 2016 Integration Marathon

Found my mistake, in the last page line 3, the pi/2 ln(2) should be negative. Hence at the end pi/2 ln(2) is positive.

Otherwise, yay wolfram alpha agrees!

Paradoxica

-insert title here-

#47

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$\int{\frac{(sin^2x +1)(sinx +1)}{(cos^2x +1)(cosx +1)}}dx$

Tedious, Messy, Unsimplifiable, and did not come with a bobcat. Would not buy/try again.

Sy123

This too shall pass

#48

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$\int{\frac{(sin^2x +1)(sinx +1)}{(cos^2x +1)(cosx +1)}}dx$

$\\ = \int \frac{1+\sin x + \sin^2 x + \sin^3x}{(1+\cos^2x)(1+\cos x)} \\ = \int \frac{2 - \cos^2x}{(1+\cos x)(1+\cos^2x)} \ dx + \int \sin x \frac{2 - \cos^2x}{(1+\cos x)(1+\cos^2x)} \ dx$

$\\ $Using partial fractions$ \ \frac{2-A^2}{(1+A)(1+A^2)} = \frac{\frac{1}{2}}{1+A} + \frac{-\frac{3}{2}A + \frac{3}{2}}{1+A^2}$

$\\ = \frac{1}{2}\int \frac{1}{1+\cos x} \ dx- \frac{3}{2} \int \frac{\cos x}{2 - \sin^2 x} \ dx + \frac{3}{2} \int \frac{1}{1+\cos^2 x} \ dx + \frac{3}{2} \int \frac{u}{1+u^2} \ du - \frac{3}{2} \int \frac{1}{1+u^2} \ du - \frac{1}{2} \int \frac{1}{1+u} \ du \\ $Where$ \ u = \cos x$

$\\ = \frac{1}{2}I_1 - \frac{3}{2} I_2 + \frac{3}{2} I_3 + \frac{3}{2}I_4 - \frac{3}{2} I_5 - \frac{1}{2} I_6$

$I_1 = \int \frac{1}{1+\cos x} \ dx = \tan \frac{x}{2} + \epsilon_1$

$I_2 = \int \frac{\cos x}{2 - \sin^2 x} \ dx = \frac{1}{2\sqrt{2}} \ln \left| \frac{\sqrt{2} + \sin x}{\sqrt{2} - \sin x} \right| + \epsilon_2$

$I_3 = \int \frac{1}{1+\cos^2 x} \ dx = \frac{1}{2} \int \frac{\csc^2x}{\frac{1}{2} + \cot^2x} \ dx = \frac{1}{2} \tan^{-1} (\sqrt{2} \cot x) + \epsilon_3$

$I_4 = \int \frac{u}{1+u^2} \ du = \frac{1}{2} \ln |1 + \cos^2 x| + \epsilon_4$

$I_5 = \int \frac{1}{1+u^2} \ du = \tan^{-1} \cos x + \epsilon_5$

$I_6 = \int \frac{1}{1+u} \ du = \ln |1 + \cos x| + \epsilon_6$

Last edited: Nov 11, 2015

D

Drsoccerball

Well-Known Member

#49

Re: MX2 2016 Integration Marathon

Sy123 said:
$\\ = \int \frac{1+\sin x + \sin^2 x + \sin^3x}{(1+\cos^2x)(1+\cos x)} \\ = \int \frac{2 - \cos^2x}{(1+\cos x)(1+\cos^2x)} \ dx + \int \sin x \frac{2 - \cos^2x}{(1+\cos x)(1+\cos^2x)} \ dx$

$\\ $Using partial fractions$ \ \frac{2-A^2}{(1+A)(1+A^2)} = \frac{\frac{1}{2}}{1+A} + \frac{-\frac{3}{2}A + \frac{3}{2}A}{1+A^2}$

$\\ = \frac{1}{2}\int \frac{1}{1+\cos x} \ dx- \frac{3}{2} \int \frac{\cos x}{2 - \sin^2 x} \ dx + \frac{3}{2} \int \frac{1}{1+\cos^2 x} \ dx + \frac{3}{2} \int \frac{u}{1+u^2} \ du - \frac{3}{2} \int \frac{1}{1+u^2} \ du - \frac{1}{2} \int \frac{1}{1+u} \ du \\ $Where$ \ u = \cos x$

$\\ = \frac{1}{2}I_1 - \frac{3}{2} I_2 + \frac{3}{2} I_3 + \frac{3}{2}I_4 - \frac{3}{2} I_5 - \frac{1}{2} I_6$

$I_1 = \int \frac{1}{1+\cos x} \ dx = \tan \frac{x}{2} + \epsilon_1$

$I_2 = \int \frac{\cos x}{2 - \sin^2 x} \ dx = \frac{1}{2\sqrt{2}} \ln \left| \frac{\sqrt{2} + \sin x}{\sqrt{2} - \sin x} \right| + \epsilon_2$

$I_3 = \int \frac{1}{1+\cos^2 x} \ dx = \frac{1}{2} \int \frac{\csc^2x}{\frac{1}{2} + \cot^2x} \ dx = \frac{1}{2} \tan^{-1} (\sqrt{2} \cot x) + \epsilon_3$

$I_4 = \int \frac{u}{1+u^2} \ du = \frac{1}{2} \ln |1 + \cos^2 x| + \epsilon_4$

$I_5 = \int \frac{1}{1+u^2} \ du = \tan^{-1} \cos x + \epsilon_5$

$I_6 = \int \frac{1}{1+u} \ du = \ln |1 + \cos x| + \epsilon_6$

I seperated it into :

$I= \int{ \frac{3(1+sinx)}{(cos^2x +1)(cosx+1)}- \frac{sinx}{cosx +1} - \frac{1}{cosx +1}}dx$

$Couldn't integrate the first term :P$

I

InteGrand

Well-Known Member

#50

Re: MX2 2016 Integration Marathon

Sy123 said:
$\\ $Using partial fractions$ \ \frac{2-A^2}{(1+A)(1+A^2)} = \frac{\frac{1}{2}}{1+A} + \frac{-\frac{3}{2}A + \frac{3}{2}A}{1+A^2}$

Is this just a typo (and doesn't affect the rest of the working)? (The second fraction on the R.H.S. is just 0.) Ceebs carefully reading through the whole thing haha.

D

Drsoccerball

Well-Known Member

#51

Re: MX2 2016 Integration Marathon

InteGrand said:
Is this just a typo (and doesn't affect the rest of the working)? (The second fraction on the R.H.S. is just 0.) Ceebs carefully reading through the whole thing haha.

:O my nigga

Sy123

This too shall pass

#52

Re: MX2 2016 Integration Marathon

InteGrand said:
Is this just a typo (and doesn't affect the rest of the working)? (The second fraction on the R.H.S. is just 0.) Ceebs carefully reading through the whole thing haha.

Yep the second term on the numerator is just 3/2, changed, (didn't affect rest of the working)

leehuan

Well-Known Member

#53

Re: MX2 2016 Integration Marathon

Can I just give the 16'ers something at a reasonable difficulty...

$\\ \int { \sin ^{ 6 }{ x } \cos ^{ 6 }{ x } dx }$

Paradoxica

-insert title here-

#54

Re: MX2 2016 Integration Marathon

leehuan said:
Can I just give the 16'ers something at a reasonable difficulty...

$\\ \int { \sin ^{ 6 }{ x } \cos ^{ 6 }{ x } dx }$

Hint: Double and Triple Angle identities.

leehuan

Well-Known Member

#55

Re: MX2 2016 Integration Marathon

Paradoxica said:
Hint: Double and Triple Angle identities.

Triple angle identity is the clever way to finish the integral. But in the scope of the 4U course unless they ask for it to be derived first using compound angles, the 4U approach to sin^3 is using a Pythagorean identity instead

D

Drsoccerball

Well-Known Member

#56

Re: MX2 2016 Integration Marathon

Or you can create a reduction formula for $\int sin^n(\theta)d\theta$

leehuan

Well-Known Member

#57

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Or you can create a reduction formula for $\int sin^n(\theta)d\theta$

Boring. I didn't even say to do one

Nah but to the 16ers that is appropriate.

G

glittergal96

Active Member

#58

Re: MX2 2016 Integration Marathon

leehuan said:
Can I just give the 16'ers something at a reasonable difficulty...

$\\ \int { \sin ^{ 6 }{ x } \cos ^{ 6 }{ x } dx }$

$I=\frac{1}{2^6}\int \sin^6(2x)\, dx\\ \\ = -\frac{1}{2^{12}} \int(\textrm{cis}(2x)-\textrm{cis}(-2x))^6 \, dx\\ \\=-\frac{1}{2^{12}}\int (\textrm{cis}(12x)+\textrm{cis}(-12x))-6(\textrm{cis}(8x)+\textrm{cis}(-8x))+15( \textrm{cis}(4x) +\textrm{cis}(-4x))-20 \, dx\\ \\ = -\frac{1}{2^{11}}\int \cos(12x)-6\cos(8x)+15\cos(4x)-10\, dx\\ \\ = \frac{1}{2^{11}}\left(10x-\frac{15}{4}\sin(4x)+\frac{3}{4}\sin(8x)-\frac{1}{12}\sin(12x)\right)+C.$

Last edited: Nov 11, 2015

leehuan

Well-Known Member

#59

Re: MX2 2016 Integration Marathon

glittergal96 said:
$I=\frac{1}{2^6}\int \sin^6(2x)\, dx\\ \\ = -\frac{1}{2^{12}} \int(\textrm{cis}(2x)-\textrm{cis}(-2x))^6 \, dx\\ \\=-\frac{1}{2^{12}}\int (\textrm{cis}(12x)+\textrm{cis}(-12x))-6(\textrm{cis}(8x)+\textrm{cis}(-8x))+15( \textrm{cis}(4x) +\textrm{cis}(-4x))-20 \, dx\\ \\ = -\frac{1}{2^{11}}\int \cos(12x)-6\cos(8x)+15\cos(4x)-10\, dx\\ \\ = \frac{1}{2^{11}}(10x-\frac{15}{4}\sin(4x)+\frac{3}{4}\sin(8x)-\frac{1}{12}\sin(12x))+C.$

On the other hand, fun
--------------------------------
$\int_0^1 \frac{x^4(1-x)^4}{1+x^2}\, dx $ and hence show that 22/7 $> \pi$

Last edited: Nov 11, 2015

G

glittergal96

Active Member

#60

Re: MX2 2016 Integration Marathon

leehuan said:
On the other hand, fun
--------------------------------
$\int _{ 0 }^{ 1 }{ \frac { { x }^{ 4 }\left( 1-{ x }^{ 4 } \right) }{ 1+{ x }^{ 2 } } dx } $ and hence show that 22/7 $> \pi$

Your integral should be

$\int_0^1 \frac{x^4(1-x)^4}{1+x^2}\, dx.$

Will leave this for a current student to do though.

Status: Not open for further replies.

Share:

Facebook Twitter Reddit Pinterest Tumblr WhatsApp Email Link

Users Who Are Viewing This Thread (Users: 0, Guests: 4)

Top