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STEP Maths I, II, III 1988 solutions

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kabbers
III/1:

Sketch y=x2exx+1y = \frac{x^2 e^{-x}}{x+1}

I'm not going to bother posting most of the differentiating legwork, there arent too many tricks, its just an algebra bash.

Differentiating (quotient rule is your friend), we get

dydx=exx(2x2)(x+1)2\frac{dy}{dx} = e^{-x}\frac{x(2-x^2)}{(x+1)^2}

So we have turning points at x = 0, and x = +2,2+\sqrt{2}, -\sqrt{2}


Differentiating again, we get

d2ydx2=ex(23x2)(x+1)22(x+1)(2xx3)(2xx3)(x+1)2(x+1)4\frac{d^2 y}{dx^2} = e^{-x}\frac{(2-3x^2)(x+1)^2 - 2(x+1)(2x-x^3) - (2x - x^3)(x+1)^2}{(x+1)^4}

Substituting in values of x for the turning points, we find that x = 0 is a minimum, and the other two maximums.


Noticing the denominator of y, we find that there is an asymptote at x = -1.

And because of the exponential properties of e^-x, y tends to zero as x+x \to +\infty

y tends to -\infty as xx \to -\infty since the numerator remains positive and given the exponenential properties of e^x much greater than the denominator, while the denominator becomes negative.


So the graph will look http://www.thestudentroom.co.uk/showpost.php?p=11883448&postcount=14



Prove 0<0x2ex1+xdx<1\displaystyle 0 < \int^{\infty}_0 \frac{x^2 e^{-x}}{1+x} dx < 1

First note that x2x+1=x2+xx+1xx+1=xxx+1=x(x+1x+11x+1)=x1+1x+1\frac{x^2}{x+1} = \frac{x^2+x}{x+1} - \frac{x}{x+1} = x - \frac{x}{x+1} = x - (\frac{x+1}{x+1} - \frac{1}{x+1}) = x - 1 + \frac{1}{x+1}

Hence we may split the integral into 0(x1)exdx+01x+1exdx\displaystyle\int^{\infty}_0 (x - 1)e^{-x} dx + \int^{\infty}_0 \frac{1}{x+1}e^{-x} dx

Consider 0(x1)exdx\displaystyle\int^{\infty}_0 (x - 1)e^{-x} dx

=0xexdx0exdx= \displaystyle\int^{\infty}_0 xe^{-x} dx - \int^{\infty}_0 e^{-x} dx

=([xex]0+0exdx0exdx= \displaystyle([-xe^{-x}]^{\infty}_0 + \int^{\infty}_0 e^{-x} dx - \int^{\infty}_0 e^{-x} dx

=0= 0


Now consider 01x+1exdx\displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx

I posit the inequality 1x+1ex<ex\frac{1}{x+1}e^{-x} < e^{-x} for x > 0

1x+1<1\frac{1}{x+1} < 1
1<(x+1)1 < (x+1)
0<x0 < x

So our inequality holds.

So:

01x+1exdx<0exdx\displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx < \int^{\infty}_0 e^{-x} dx

01x+1exdx<[ex]0\displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx < [-e^{-x}]^{\infty}_0

01x+1exdx<1\displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx < 1

Thus 0(x1)exdx+01x+1exdx<1\displaystyle\int^{\infty}_0 (x - 1)e^{-x} dx + \int^{\infty}_0 \frac{1}{x+1}e^{-x} dx < 1

And therefore, 0x2ex1+xdx<1\int^{\infty}_0 \frac{x^2 e^{-x}}{1+x} dx < 1


Now notice that the graph of y=x2ex1+xy = \frac{x^2 e^{-x}}{1+x} is greater than 0 for x > 0, and hence so is the infinite integral.

So, 0<0x2ex1+xdx<1\displaystyle 0 < \int^{\infty}_0 \frac{x^2 e^{-x}}{1+x} dx < 1

please point out any mistakes :smile:


I'm probably wrong but to show that the integral lies between 0 and 1 would it not be sufficient to say that

(x^2)(e^-x)/(x + 1) < (x^2)(e^-x)/x = x(e^-x) for 0 < x < infinity

then integrate x(e^-x), which equals 1. Seems a lot simpler?
Yeah, that's fine.

While we're nitpicking - taking the 2nd derivative to deduce the nature of the turning points is a lot more work than simply sketching x(2-x^2) to argue whether the derivative is increasing or decreasing.
1988 Paper 2 number 4
1988 Paper 2 number 8
1988 Paper 2 numbers 13 & 14
1988 Paper 2 numbers 15 & 16
Reply 67
Question 8, STEP III

Spoiler

Reply 68
Can somebody please post the solution for STEP III Q3?
Reply 69
Original post by SimonM
Done :smile:


Am I missing something or is the solution, well, missing?
Reply 70
Where can i get STEP 2 solution 2
Original post by SimonM

....

STEP III Q10
Consider the positions of the dogs as time goes on. They will form a square of decreasing side length that rotates about the centre. If we take the position of one of the dogs to be (x,y)(x,y) from the centre, then we see by symmetry that the position of one the next dogs along is (y,x)(-y,x). We can also see that the tangent at any moment goes through the position of the next dog along, so:
dydx=yxx(y)=yx11+yx\frac{dy}{dx}=\frac{y-x}{x-(-y)}=\frac{\frac{y}{x}-1}{1+\frac{y}{x}}
let:
u=yxdydx=dudxx+u[br]dudxx+u=u11+u[br]dudxx=u211+u[br]dxdu1x=1+uu21[br]1xdx=1+uu21duu=\frac{y}{x} \Rightarrow \frac{dy}{dx}=\frac{du}{dx}x+u[br]\therefore \frac{du}{dx}x+u=\frac{u-1}{1+u}[br]\Rightarrow \frac{du}{dx}x=\frac{-u^2-1}{1+u}[br]\Rightarrow \frac{dx}{du} \frac{1}{x}=\frac{1+u}{-u^2-1}[br]\Rightarrow \frac{1}{x}dx=\frac{1+u}{-u^2-1}du
let θ\theta be the angle between the x axis and the straight line to the point (x,y)(x,y). Since u=yxu=\frac{y}{x} and, by trigonometry, yx=Tanθu=Tanθ\frac{y}{x}=Tan\theta \Rightarrow u=Tan\theta. So, substituting into the RHS and integrating we get:
lnx+c1=ln(secθ)θ+c2lnx+c_1=-ln(sec\theta)-\theta+c_2.
By trigonometry, again, x=rcosθx=rcos\theta where r is the distance from the origin to the point (x,y)ln(rcosθ)+ln(secθ)=θ+cln(r)=θ+cr=eθec(x,y) \Rightarrow ln(rcos\theta)+ln(sec\theta)=-\theta +c \Rightarrow ln(r)=-\theta +c \Rightarrow r=e^{-\theta}e^c
ece^c is constant so r=eθλr=e^{-\theta}\lambda for some constant λ\lambda as required.
To generalise to n-dogs held at the vertices of an n-gon it must be noted that the next dog along will have the coordinates of the first dog bur rotated 2πn\frac{2\pi}{n} about the centre of the field i.e:
(cos(2πn)sin(2πn)sin(2πn)cos(2πn))(xy)=(xcos2πnysin2πnxsin2πn+ycos2πn)\begin{pmatrix} cos(\frac{2\pi}{n}) & -sin(\frac{2\pi}{n})\\sin(\frac{2\pi}{n}) & cos(\frac{2\pi}{n}) \end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}xcos\frac{2\pi}{n}-ysin\frac{2\pi}{n}\\xsin\frac{2 \pi}{n}+ycos\frac{2\pi}{n} \end{pmatrix}
So, like the previous part,:
dydx=yxsin2πnycos2πnxxcos2πn+ysin2πn=yx1sin2πnyx1cos2πn1cos2πn+yx1sin2πn\frac{dy}{dx}=\frac{y-xsin\frac{2 \pi}{n}-ycos\frac{2 \pi}{n}}{x-xcos\frac{2 \pi}{n}+ysin\frac{2 \pi}{n}}=\frac{yx^{-1}-sin\frac{2 \pi}{n}-yx^{-1}cos\frac{2 \pi}{n}}{1-cos\frac{2 \pi}{n}+yx^{-1}sin\frac{2 \pi}{n}}
let:
u=yxdydx=dudxx+u[br]dudxx=sin2πnu2sin2πn1cos2πn+usin2πn[br]dxdu1x=cosec(2πn)1cos2πn+usin2πnu2+1u=\frac{y}{x} \Rightarrow \frac{dy}{dx}=\frac{du}{dx}x+u[br]\therefore \frac{du}{dx}x=\frac{-sin\frac{2 \pi}{n}-u^2sin\frac{2 \pi}{n}}{1-cos\frac{2 \pi}{n}+usin\frac{2 \pi}{n}}[br]\Rightarrow \frac{dx}{du}\frac{1}{x}=-cosec(\frac{2 \pi}{n})\frac{1-cos\frac{2 \pi}{n}+usin\frac{2 \pi}{n}}{u^2+1}
let:
u=Tanθdudθ=sec2θ[br]dxdu1x=cosec(2πn)1cos2πn+tanθsin2πnsec2θ[br]1xdx=(cosec(2πn)+cot2πnTanθ)dθu=Tan\theta \Rightarrow \frac{du}{d\theta}=sec^2\theta[br]\Rightarrow \frac{dx}{du}\frac{1}{x}=-cosec(\frac{2 \pi}{n})\frac{1-cos\frac{2 \pi}{n}+tan\theta sin\frac{2 \pi}{n}}{sec^2\theta}[br]\Rightarrow \frac{1}{x}dx=(-cosec(\frac{2 \pi}{n})+cot\frac{2 \pi}{n}-Tan\theta)d\theta
Integrating both sides we get:
lnx+c=(cosec(2πn)+cot2πn)θlnsecθ+clnx+c=(-cosec(\frac{2 \pi}{n})+cot\frac{2 \pi}{n})\theta-ln\sec\theta +c
And now, just like in the previous part:
ln(rcosθ)+ln(secθ)+c=(cosec(2πn)+cot2πn)θ+c[br]r=eθ(cosec(2πn)cot2πn)ec=eθ(1cos2πnsin2πn)λ=eθTan(πn)λln(rcos\theta)+ln(sec\theta)+c=(-cosec(\frac{2 \pi}{n})+cot\frac{2 \pi}{n})\theta+c[br]\Rightarrow r=e^{-\theta(cosec(\frac{2 \pi}{n})-cot\frac{2 \pi}{n})}e^c= e^{-\theta (\frac {1-cos\frac{2 \pi}{n}}{sin\frac{2 \pi}{n}})}\lambda=e^{-\theta Tan(\frac{\pi}{n})}\lambda for some constant λ\lambda. Letting n=4 we can check and see that it gives the same answer as the first part.
EDIT: Just realised that the final result simplified to the much nicer form that it is currently in.
(edited 12 years ago)
Reply 72
Where can I find the step past papers before 1997? I have been looking for them for days, but all the links do not work now. Can you help me? thanks a lot!
Original post by Dzwx777
Where can I find the step past papers before 1997? I have been looking for them for days, but all the links do not work now. Can you help me? thanks a lot!


http://www.mathshelper.co.uk/oxb.htm
Reply 74
Original post by insparato
III/3

zi=2 |z - i| = 2 is the circle on the complex plane with radius 2 and centre (0,i)

so x^2 + (y - 1)^2 = 2

and therefore in paramatrised form

x=2cosθ x = \sqrt2 cos\theta

y=1+2sinθ y = 1 + \sqrt2 sin \theta

I've omitted sketches however I have described them in sufficient detail to make an easy sketch if one chooses.

(i)

w=z+izi w = \frac{z + i}{z - i}

wzwi=z+i wz - wi = z + i

wzz=i+wi wz - z = i + wi

z(w1)=i(w+1) z(w - 1) = i(w + 1)

z=i(w+1)w1 z = \frac{i(w+1)}{w-1}

zi=i(w+1)w1i=i(w+1)i(w1)w1 z - i = \frac{i(w +1)}{w - 1} - i = \frac{i(w + 1) - i(w-1)}{w - 1}

=1+iw1 = \frac{1 + i}{w - 1}

zi=1+iw1 |z - i| = \left|\frac{1+i}{w-1}\right|

2=1+iw1 2 = \frac{|1 + i|}{|w - 1|}

w1=22 |w - 1| = \frac{\sqrt2}{2}

Okay so its a circle of radius 22 \frac{\sqrt2}{2} and the centre is (1,0)


(ii)

If z = x + iy and is real then y must = 0

so

w=u+iv=x+i(y+1)x+i(y1)=x+ixi w = u + iv = \frac{x + i(y + 1)}{x + i(y - 1)} = \frac{x + i}{x - i}

(u+iv)(xi)=x+i (u + iv)(x - i) = x + i

uxui+vxi+v=x+i ux - ui + vxi + v = x + i

ux+v+i(u+vx1)=x ux + v + i(-u + vx - 1) = x

-u + vx - 1 must = 0 (this is important bit on)

so

ux+v=x ux + v = x

x(1u)=v x(1 - u) = v

x=v1u x = \frac{v}{1 - u}

Now from before

u+vx1=0 -u + vx - 1 = 0

vx=u+1 vx = u + 1

x=u+1v x = \frac{u+1}{v}

so

v1u=u+1v \frac{v}{1 - u} = \frac{u+1}{v}

v2=(u+1)(1u) v^2 = (u + 1)(1 - u)

u2+v2=1 u^2 + v^2 = 1

So in the W plane, when z is real, you get a circle centre(0,0) with radius 1

(iii)

if z = x + iy and z is imaginary then x = 0

u+vi=x+i(y+1)x+i(y1) u + vi = \frac{x + i(y + 1)}{x + i(y - 1)}

u+vi=i(y+1)i(y1) u + vi = \frac{i(y + 1)}{i(y - 1)}

(u+vi)(iyi)=yi+i (u + vi)(iy - i) = yi + i

uyiuivy+v=yi+i uyi - ui - vy + v = yi + i

yiuyi=uivy+vi yi - uyi = -ui - vy + v - i

y(iui)=i(1u)vy+v y(i-ui) = i(-1 -u) - vy + v

vy+v=0 -vy + v = 0

y=i(1u)i(1u) y = \frac{i(-1-u)}{i(1 - u)}

y=1u1u y = \frac{-1 -u}{1 - u}

1=1u1u 1 = \frac{-1 - u}{1 - u}

1u=1u -1 - u = 1 -u

0=2 0 = 2

Wahey! wtf ? I'm sure i got something different when jotting it down, something that made sense! hmm have i gone wrong somewhere? I think I'm doing the right method.

erm....I have something different....
1988 STEP III number 3.
I'm afraid I disagree with the solutions published so far. Here is my solution.

zi=2x+(y1)i=2 where z=x+iy |z-i|=2 \Rightarrow |x+(y-1)i|=2 \text{ where }z=x+iy
x2+(y1)2=4 so parametric equations are x=2cosθ,y=2sinθ+1 \Rightarrow x^2+(y-1)^2=4 \text{ so parametric equations are }x=2\cos\theta, y=2\sin\theta+1
Unparseable latex formula:

(i) w=\dfrac{z+i}{z-i} \Rightarrow z=\left( \dfrac{w+1}{w-1} \right)i \text{ and }|z-i|=2 \Rightarrow \left|\dfrac{w+1}{w-1}i-i\right|=2 \Rightarrow left| \dfrac{2i}{w-1} \right|=2


2i=2w1 i.e. w1=1 so it is a circle, centre w=1 radius 1 \Rightarrow |2i|=2|w-1| \text{ i.e. }|w-1|=1 \text{ so it is a circle, centre }w=1 \text{ radius }1
(ii)z real Im((u+1)iv(u1)iv)= Im(((u+1)iv)((u1)iv)(u1)2+v2)=2uv(u1)2+v2=0 (ii) z \text{ real } \Rightarrow \text {Im}\left( \dfrac{(u+1)i-v}{(u-1)i-v} \right)= \text{ Im}\left( \dfrac{((u+1)i-v)((u-1)i-v)}{(u-1)^2+v^2} \right)= -\dfrac{2uv}{(u-1)^2+v^2}=0
hence, 2uv=0 i.e. It is the two axes \text{hence, }2uv=0 \text{ i.e. It is the two axes}
(iii)z imaginary Re(w+1w=1)i=0 i.e. v2(u21)(u1)2+v2=0v2u2+1=0 i.e. A hyperbola (iii) z\text{ imaginary }\Rightarrow \text{Re} \left( \dfrac{w+1}{w-=1} \right)i=0 \text{ i.e. } \dfrac {v^2-(u^2-1)}{(u-1)^2+v^2}=0 \Rightarrow v^2-u^2+1=0 \text{ i.e. A hyperbola}
(edited 12 years ago)
1988 STEP I question 1 Alternative Solution

h(x)=lnxxh(x)=1lnxx2=0 when lnx=1x=e \text{h}(x)=\dfrac{\ln x}{x} \Rightarrow \text{h}'(x)=\dfrac{1- \ln x}{x^2}=0 \text{ when } \ln x=1\Rightarrow x=\text{e}
for small h,h(1h)>0 while h(1+h)<0 so a maximum point at (e,e1 \text{for small }h, \text{h}'(1-h)>0 \text{ while h}'(1+h)<0 \text { so a maximum point at }(\text{e,e}^{-1}
x=1h(x)=0 x=1 \Rightarrow \text{h}(x)=0
h(x)0 as x \text {h}(x) \rightarrow 0 \text{ as }x \rightarrow \infty
h(x) as x0 \text {h}(x) \rightarrow -\infty \text{ as }x \rightarrow 0

nm=mnmlnn=nlnmlnnn=lnmmn^m=m^n \Rightarrow m \ln n=n \ln m \Rightarrow \dfrac{\ln n}{n}=\dfrac{\ln m}{m}
Clearly, m and n must lie on either side of the turning point \text{Clearly, }m \text{ and }n\text{ must lie on either side of the turning point}
 so the only possible values for the smaller of m and n are 1 and 2\text{ so the only possible values for the smaller of }m \text{ and }n \text{ are } 1 \text{ and }2
 there is obviously no other integer with 1m=m1 but 24=42\text{ there is obviously no other integer with }1^m=m^1 \text{ but } 2^4=4^2
 so only pair of integers is 2 and 4 \text{ so only pair of integers is }2 \text { and }4
1988 STEP I question 2 Alternative solution

g(x)= f(x)+1f(x) g(x)= f(x)f(x)(f(x))2 \text{g}(x)= \text{ f}(x)+\dfrac{1}{\text{f}(x)} \Rightarrow \text{ g}'(x)=\text{ f}'(x) -\dfrac{\text{f}'(x)}{(\text{f}(x))^2}
and g(x)= f(x)(f(x))2f(x)f(x).2f(x)f(x)(f(x))4 \text{and g}''(x)=\text{ f}''(x)- \dfrac{(\text{f}(x))^2\text{f}''(x)-\text{f}'(x).2\text{f}(x)\text{f}'(x)}{(\text{f}(x))^4}
i.e. g(x)= f(x)f(x)(f(x))2+2(f(x))2(f(x))3\text{i.e. g}''(x)=\text{ f}''(x)- \dfrac{\text{f}''(x)}{(\text{f}(x))^2}+ \dfrac{2(\text{f}'(x))^2}{(\text{f}(x))^3}
Unparseable latex formula:

\text{f}(x)=4+ \cos2x+2 \sin x \Rightarrow \text{f}'(x)=2\cosx-2 \sin2x \text{ and f}''(x)=-2 \sin x-4\cos2x


so g(x)=2cosx2sinx2cosx2sin2x(4+cos2x+2sinx)2\text{so g}'(x)=2\cos x-2\sin x-\dfrac{2\cos x-2\sin 2x}{(4+\cos 2x+2\sin x)^2}
Unparseable latex formula:

\text{g}'(x)=0 \text{ when }}2\cos x-2\sin 2x=0 \text{ or }(4+\cos 2x+2\sin x)^2=1


4+cos2x+2sin2x>1 for all x so solutions are given by 2cosx2sin2x=0 4+\cos 2x+2\sin 2x>1 \text{ for all }x \text{ so solutions are given by }2 \cos x-2 \sin 2x=0
cosx2sinxcosx=0cosx=0 or sinx=12 so x=π6,π2,5π6 or 3π2\Rightarrow \cos x-2\sin x\cos x=0 \Rightarrow \cos x=0 \text{ or }\sin x =\dfrac{1}{2} \text{ so }x=\dfrac{\pi}{6},\dfrac{\pi}{2},\dfrac{5\pi}{6} \text{ or }\dfrac{3\pi}{2}
g(x) is a continuous function since 4+cos2x+2sinx cannot be zero \text{g}(x) \text{ is a continuous function since }4+\cos 2x+2\sin x \text{ cannot be zero }
 so we need only consider one of the turning points \text{ so we need only consider one of the turning points }
x=π2 f(x)=5, f(x)=0 and f(x)=2 g(x)>0x=\dfrac{\pi}{2} \Rightarrow \text{ f}(x)=5,\text{ f}'(x)=0\text{ and f}''(x)=2 \Rightarrow \text{ g}''(x)>0
so a minimum at x=π2\text{so a minimum at }x=\dfrac{\pi}{2}
Unparseable latex formula:

\text{ g}\left(\dfrac{\pi}{6}\right)= \dfrac{11}{2}+ \dfrac{2}{11}= \dfrac{125}{22}, \text{ g}\left(\dfrac{\pi}{2}\right)=5+ \dfrac{1}{5}= \dfrac{26}{5}, \text{ g} \left( \dfrac{5\pi}{6} \right)= \dfrac{11}{2}+ \dfrac{2}{11}= \dfrac{125}{22} \text{ and g} \left( \dfrac{3\pi}{2} \rihjt)=1+1=2


Unparseable latex formula:

\text{so stationary points are maxima at } \left( \dfrac{\pi}{6}, \dfrac{125}{22} \right) \text{ and }\left( \dfrac{5\pi}{6}, \dfrac{125}{22} \right) \text{ and minima at }\left( \dfrac{\pi}{2}, \dfrac{26}{5} \right) \text{ and } \left( \dfrac{3\piu}{2}, 2\right)

1988 STEP I question 7 Alternative solution

f(x)=ax2+bx+c f(x)=2ax+b \text{f}(x)=ax^2+bx+c \Rightarrow \text{ f}'(x)=2ax+b
f(1)(x+12)+ f(1)(x12)2 f(0)x=(a+b+c)(x+12)+(ab+c)(x12)2cx \text{f}(1)(x+ \frac{1}{2})+\text{ f}(-1)(x- \frac{1}{2})-2\text{ f}(0)x=(a+b+c)(x+ \frac{1}{2})+(a-b+c)(x- \frac{1}{2})-2cx
=2ax+2cx+b2cx=2ax+b= f(x)=2ax+2cx+b-2cx=2ax+b=\text{ f}'(x)
Since f(x) is linear it must attain it’s maximum value at either x=1 or x=1 \text{Since f}'(x) \text{ is linear it must attain it's maximum value at either }x=1 \text{ or }x=-1
f(1)=12f(1)32f(1)2f(0) and f(1)=32f(1)+12f(1)2f(0)\text{f}'(-1)=-\frac{1}{2}\text{f}(1)- \frac{3}{2}\text{f}(-1)-2\text{f}(0) \text{ and f}'(1)= \frac{3}{2}\text{f}(1)+ \frac{1}{2}\text{f}(-1)-2\text{f}(0)
Unparseable latex formula:

\text{f}(-1),\tewxt{ f}(0) \text{ and f}(1) \text{ are all between }\pm{1} \Rightarrow |\text{f}'(x)| \leq4


Consider 2x21 then 1 f(x0)1 i.e. f(x)1 for all x with x1 \text{Consider }2x^2-1 \text{ then }-1 \leq \text{ f}(x0) \leq1 \text{ i.e. }\text{f}(x)| \leq1 \text{ for all }x \text { with }|x| \leq1
and f(x)=4x lies between ±4 so this function satisfies the requirements \text{and f}'(x)=4x \text{ lies between }\pm4 \text{ so this function satisfies the requirements}
Original post by nota bene
I/9

i) 1elogxx2dx\displaystyle\int_1^e \frac{\log x}{x^2} dx Recall that logxdx=x(log(x)1)+c\int \log x dx= x(\log (x)-1)+c, now we can use this to integrate by parts.
I=logxx2dx=1x2x(log(x)1)2x(log(x)1)x3dxI=\displaystyle\int \frac{\log x}{x^2}dx=\frac{1}{x^2}x(\log(x)-1)-\displaystyle\int-2\frac{x(\log(x)-1)}{x^3}dx
I=1x(log(x)1)+21x=logx+1x-I=\frac{1}{x}(\log(x)-1)+2\frac{1}{x}=\frac{\log x +1}{x}
Putting in the limits gives I=2e1=-I=\frac{2}{e}-1= So I=12eI=1-\frac{2}{e}

ii) cos(x)sin(x)1+sin(x)dx\displaystyle\int \frac{\cos(x)}{\sin(x)\sqrt{1+\sin(x)}}dx Let u2=1+sin(x)2ududx=cos(x)dx=2ucos(x)duu^2=1+\sin(x)\Rightarrow 2u\frac{du}{dx}=\cos(x)\Leftrightarrow dx=\frac{2u}{\cos(x)}du
So the integral becomes
Unparseable latex formula:

\displastyle\int \frac{2}{\sin x}du=\displaystyle\int\frac{2}{u^2-1}du=\displaystyle\int\frac{1}{u-1}du-\displaystyle\int\frac{1}{u+1}du=\log|u-1|-\log|u+1|+c=\log|\frac{u-1}{u+1}|+c


Going back to x we have logu1u+1+c=log1+sinx11+sinx+1+c\log|\frac{u-1}{u+1}|+c=\log|\frac{\sqrt{1+\sin x}-1}{\sqrt{1+\sin x}+1}|+c


Alternative solution for part (b)

cosxsinx1+sinxdx \int \dfrac { \cos x}{ \sin x \sqrt{1+ \sin x}}dx
 so putting sinx=tan2θ so cosxdxdθ=2tanxsec2x\text{ so putting }\sin x= \tan^2 \theta \text{ so } \cos x \dfrac {dx}{d \theta}=2 \tan x\sec^2 x
integral becomes 2tanθsec2θdθtan2θsecθ=2secθtanθdθ=2cscθdθ=lncscθ+cotθ+c \text{integral becomes } \int \dfrac{2\tan \theta \sec^2 \theta d\theta}{\tan^2 \theta \sec\theta }=\int \dfrac{2\sec\theta}{\tan\theta}d\theta=\int 2\csc\theta d\theta=-\ln{|\csc\theta+\cot\theta|}+c
tanθ=xcotθ=1x,sinθ=x1+x so cscθ=1+xx \tan\theta=\sqrt{x} \Rightarrow \cot\theta=\dfrac{1}{\sqrt{x}},\sin\theta= \dfrac{\sqrt{x}}{\sqrt{1+x}} \text{ so }\csc\theta=\dfrac{\sqrt{1+x}}{ \sqrt{x}}
Hence cosxsinx1+sinxdx=lnA1+1+xx=lnAx1+1+x\text{Hence }\int \dfrac{ \cos x}{ \sin x \sqrt{1+\sin x}}dx=-\ln A \left| \dfrac{1+ \sqrt{1+x}}{\sqrt{x}} \right|=\ln A\left| \dfrac{\sqrt x}{1+\sqrt{1+x}} \right|

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