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The Proof is Trivial!

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Reply 1120

Spoiler



I come bearing gifts :biggrin:

Problem 174*

Evaluate 0ln(x)1+x2dx\displaystyle \int_0^{\infty} \frac{\ln(x)}{1+x^2} dx

Problem 175*

Evaluate 01sin(arccos(x))dx\displaystyle\int_0^1 \sin(\arccos(x)) dx

Problem 176*

Evaluate 01x(1x)99dx\displaystyle\int_0^1 x(1-x)^{99} dx

Problem 177*

Evaluate csc(x)sin(x)dx\displaystyle\int \sqrt{\csc(x)-\sin(x)} dx

Problem 178*

Evaluate 02x51+x3dx\displaystyle\int_0^2 x^5 \sqrt{1+x^3} dx

Problem 179*

Evaluate 0π2sin(4x)sin(x)dx\displaystyle \int_0^{\frac{\pi}{2}} \frac{\sin(4x)}{\sin(x)} dx

Problem 180*

Evaluate (cos(x)ln(x)+sin(x)x)dx\displaystyle \int \left( \cos(x) \ln(x) +\frac{\sin(x)}{x} \right) dx

Problem 181*

Evaluate 1x1dx\displaystyle \int \frac{1}{\sqrt{x}-1} dx

Problem 182*

Evaluate 1(1+x)xx2dx\displaystyle \int \frac{1}{(1+\sqrt{x})\sqrt{x-x^2}} dx

Problem 183*

Evaluate 1x(x4+1)10dx\displaystyle \int \frac{1}{\sqrt{x} (\sqrt[4]{x}+1)^{10}} dx

...and finally, this one was tie breaker between the top two (how it took them a whole minute I have no idea! You can do this **** in your head :lol:)

Problem 184*

Evaluate x41x2dx\displaystyle \int \frac{x^4}{1-x^2} dx
(edited 10 years ago)
Solution 174

Let x=tanθ:x = \tan \theta :

0lnx1+x2dx=0π2ln(tanθ)dθ\displaystyle\int_{0}^{\infty} \dfrac{\ln x}{1 + x^{2}} dx = \displaystyle\int_{0}^{ \frac{\pi}{2}} \ln \left( \tan \theta \right) d \theta

=0π2ln(sinθ)dθ0π2ln(cosθ)dθ=0= \displaystyle\int_{0}^{ \frac{\pi}{2}} \ln \left( \sin \theta \right ) d \theta - \displaystyle\int_{0}^{ \frac{\pi}{2}} \ln \left( \cos \theta \right) d \theta = 0

Solution 175

01sin(arccosx)dx=011x2dx\displaystyle\int_{0}^{1} \sin \left( \arccos x \right) dx = \displaystyle\int_{0}^{1} \sqrt{1 - x^{2}} dx

Let x=sinθ:x = \sin \theta :

011x2dx=120π21+cos2θ dθ=π4\displaystyle\int_{0}^{1} \sqrt{1 - x^{2}} dx = \dfrac{1}{2} \displaystyle\int_{0}^{ \frac{\pi}{2}} 1 + \cos 2 \theta \ d \theta = \dfrac{\pi}{4}

Solution 177

cscxsinxdx=cosxsinxdx=2sinx+C\displaystyle\int \sqrt{ \csc x - \sin x} dx = \displaystyle\int \dfrac{ \cos x}{\sqrt{ \sin x}} dx = 2 \sqrt{ \sin x } + \mathcal{C}

Solution 180

(lnxcosx+sinxx)dx=lnxsinx+C\displaystyle\int \left( \ln x \cdot \cos x + \dfrac{ \sin x}{x} \right) dx = \ln x \cdot \sin x + \mathcal{C}

I'll leave the rest, 183 looks like an algebraic mess :lol:
(edited 10 years ago)
Solution 175:

I = 01sin(arccosx) dx \int_0^1 sin(arccosx) \ dx

Let arcsinu=arccosx arcsinu = arccosx

11u2du=11x2dx \dfrac{1}{\sqrt{1-u^2}} du = -\dfrac{1}{\sqrt{1-x^2}} dx

Re-arranging, we get:

dx=1x21u2du dx = -\dfrac{\sqrt{1-x^2}}{\sqrt{1-u^2}} du

1x2=1cos2arcsinu=sinarcsinu=u \sqrt{1-x^2} = \sqrt{1-cos^2arcsinu} = sinarcsinu = u

I=10u21u2 du \Rightarrow I = - \int_1^0 \dfrac{u^2}{\sqrt{1-u^2}} \ du

I=0111u221u2 du I = \int_0^1 \dfrac{ 1-\sqrt{1-u^2}^2}{\sqrt{1-u^2}} \ du

=0111u21u2 du = \int_0^1 \dfrac{1}{\sqrt{1-u^2}}- \sqrt{1-u^2} \ du

The first part of this is a standard arcsin integral and yields π2 \frac{\pi }{2} . Let's look at the second part:

J=011u2 du J = \int_0^1 \sqrt{1-u^2} \ du

Let u=sinx u = sinx

J=0π2cos2x dx \Rightarrow J = \int_0^{\frac{\pi }{2}} cos^2x \ dx

=120π2cos2x+1 dx = \frac{1}{2} \int_0^{\frac{\pi }{2}} cos2x+1 \ dx

=12[12sin2x+x]0π2 = \frac{1}{2} [ \frac{1}{2}sin2x+x ]_0^{\frac{\pi }{2}}

=π4 = \frac{\pi }{4}

Therefore our final answer is π4\frac{\pi }{4}
Original post by Felix Felicis
Problem 171, 172


Felix you need to relable your questions, there already is a 171 :biggrin:



Problem 185*

Find all f:  RRf:\;\mathbb{R}\to\mathbb{R} such that:

Unparseable latex formula:

\begin{aligned}(\text{i})\;f(x)>0\qquad (\text{ii})\; f(x)\geq x+1\qquad(\text{iii})\; f(x)f(-x)=1\qquad (\text{iv})\;f(2x)=f^2(x)



Problem 186*

A nice problem. Prove/disprove that there exists f:  NNf:\;\mathbb{N}\to\mathbb{N} such that:

ff(n)(n)=n+1f^{f(n)}(n)=n+1


fkf^k means ff applied kk times in this case.



Well, to those who were unhappy about the lack of (*)'s, you now have loads. :biggrin: Go go go go!
Original post by Jkn


Problem 178*

Evaluate 02x51x3dx\displaystyle\int_0^2 x^5 \sqrt{1-x^3} dx



Are you sure the upper limit is supposed to be 2? Otherwise the integral is not defined for real numbers past 1...
Reply 1125
Original post by Felix Felicis
Solution 174

Let x=tanθ:x = \tan \theta :

0lnx1+x2dx=0π2ln(tanθ)dθ\displaystyle\int_{0}^{\infty} \dfrac{\ln x}{1 + x^{2}} dx = \displaystyle\int_{0}^{ \frac{\pi}{2}} \ln \left( \tan \theta \right) d \theta

=0π2ln(sinθ)dθ0π2ln(cosθ)dθ=0= \displaystyle\int_{0}^{ \frac{\pi}{2}} \ln \left( \sin \theta \right ) d \theta - \displaystyle\int_{0}^{ \frac{\pi}{2}} \ln \left( \cos \theta \right) d \theta = 0

In light of it being a "speed" competition, allow me to direct you toward my solution :rolleyes:

Solution 174
(2)

Let xx1x \to x^{-1},

0ln(x)1+x2dx=0ln(x)1+x2dx=0 \displaystyle \Rightarrow \int_0^{\infty} \frac{\ln(x)}{1+x^2} dx = \int_0^{\infty} -\frac{\ln(x)}{1+x^2} dx =0 \ \square
Reply 1126
Original post by DJMayes
Are you sure the upper limit is supposed to be 2? Otherwise the integral is not defined for real numbers past 1...

Sorry typo, 'tis now fixed (that's what you get for setting two questions at once) :lol:
Original post by Jkn
In light of it being a "speed" competition, allow me to direct you toward my solution :rolleyes:

Solution 174
(2)

Let xx1x \to x^{-1},

0ln(x)1+x2dx=0ln(x)1+x2dx=0 \displaystyle \Rightarrow \int_0^{\infty} \frac{\ln(x)}{1+x^2} dx = \int_0^{\infty} -\frac{\ln(x)}{1+x^2} dx =0 \ \square

It takes like 10 seconds less :dontknow: Neat, nonetheless :lol:
Reply 1128
Original post by Lord of the Flies
Felix you need to relable your questions, there already is a 171 :biggrin:



Problem 185*

Find all f:  RRf:\;\mathbb{R}\to\mathbb{R} such that:

Unparseable latex formula:

\begin{aligned}(\text{i})\;f(x)>0\qquad (\text{ii})\; f(x)\geq x+1\qquad(\text{iii})\; f(x)f(-x)=1\qquad (\text{iv})\;f(2x)=f^2(x)



Problem 186*

A nice problem. Prove/disprove that there exists f:  NNf:\;\mathbb{N}\to\mathbb{N} such that:

ff(n)(n)=n+1f^{f(n)}(n)=n+1


fkf^k means ff applied kk times in this case.



Well, to those who were unhappy about the lack of (*)'s, you now have loads. :biggrin: Go go go go!

No way these look exactly like the functional equations questions you get in Core 4!

Spoiler

Reply 1129
Original post by Felix Felicis
It takes like 10 seconds less :dontknow: Neat, nonetheless :lol:

Meh :dontknow:

Please post some interesting series and/or integration questions :smile:
Original post by Jkn
Meh :dontknow:

Please post some interesting series and/or integration questions :smile:

I'll have a look around and see what I have :lol:

Edit: I thought the first integral you posted was 0lnx(x2+1)2 dx\displaystyle\int_{0}^{\infty} \dfrac{ \ln x}{(x^{2} + 1)^{2}} \ dx and I thought no way in hell can anyone do that in 20 minutes :lol:
(edited 10 years ago)
Original post by Jkn
Lol jk functional equations are not * questions


Yes they are. In fact a lot of functional equations only require GCSE knowledge.
Reply 1132
Original post by Lord of the Flies
Yes they are. In fact a lot of functional equations only require GCSE knowledge.

But the concept of a function isn't introduced until a-level :rollseyes:
Original post by Jkn
But the concept of a function isn't introduced until a-level :rollseyes:


New rule. In your head always add another * to LOTF's questions :tongue:
Original post by Jkn
But the concept of a function isn't introduced until a-level :rollseyes:


I'm pretty sure I've seen function-related GCSE questions on the maths forum but whatever. Anyhow, can we stop cluttering the thread with this chatter, both questions are * the end.
Original post by Jkn


Problem 178*

Evaluate 02x51+x3dx\displaystyle\int_0^2 x^5 \sqrt{1+x^3} dx



Let x3=sinh2u x^3 = sinh^2u

3x2 dx=2sinhucoshu du 3x^2 \ dx = 2sinhucoshu \ du

I=23sinh3ucosh2u dx \Rightarrow I = \frac{2}{3} \int sinh^3ucosh^2u \ dx

=23sinhu(cosh2u1)cosh2u dx = \frac{2}{3} \int sinhu(cosh^2u-1)cosh^2u \ dx

=23sinhucosh4usinhucosh2u dx = \frac{2}{3} \int sinhucosh^4u - sinhucosh^2u \ dx

==23[15cosh5u13cosh3u] = = \frac{2}{3} [ \frac{1}{5}cosh^5u -\frac{1}{3}cosh^3u ]

x = 0 so u = 0, coshu = 1, x = 2 so coshu = 3

Finally plugging those in I get 1192/45, although this isn't really a nice answer...
Original post by Lord of the Flies
I'm pretty sure I've seen function-related GCSE questions on the maths forum but whatever.


Yes, but the way GCSEs are taught means they wouldn't have a clue what a function was even if they do stuff related to them....
Original post by DJMayes
Let x3=sinh2u x^3 = sinh^2u

3x2 dx=2sinhucoshu du 3x^2 \ dx = 2sinhucoshu \ du

I=23sinh3ucosh2u dx \Rightarrow I = \frac{2}{3} \int sinh^3ucosh^2u \ dx

=23sinhu(cosh2u1)cosh2u dx = \frac{2}{3} \int sinhu(cosh^2u-1)cosh^2u \ dx

=23sinhucosh4usinhucosh2u dx = \frac{2}{3} \int sinhucosh^4u - sinhucosh^2u \ dx

==23[15cosh5u13cosh3u] = = \frac{2}{3} [ \frac{1}{5}cosh^5u -\frac{1}{3}cosh^3u ]

x = 0 so u = 0, coshu = 1, x = 2 so coshu = 3

Finally plugging those in I get 1192/45, although this isn't really a nice answer...


Or u =1+x^3 ?
Reply 1138
Original post by Felix Felicis

Edit: I thought the first integral you posted was 0lnx(x2+1)2 dx\displaystyle\int_{0}^{\infty} \dfrac{ \ln x}{(x^{2} + 1)^{2}} \ dx and I thought no way in hell can anyone do that in 20 minutes :lol:

Lightweight, you can get it form applying integration by parts to the other one by integrating ln(x)\ln(x) (and differentiating the denominator) and then evaluating a fairly trivial other part with a tanx substitution :wink:
Original post by bananarama2
Or u =1+x^3 ?


Hyperbolics are more fun. :tongue:

(What answer do you get with that sub though?)

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