STEP maths I, II, III 1990 solutions

1990 STEP III. no.4.pdf29.2KB

nota bene

Attempt at II/10 (not confident about using det(ab)=det(a)det(b) without proving it)

To show that for two square matrices A and B and AB=0 implies that |A|=0 or |B|=0 or |A|=|B|=0, remember that |AB|=|A||B| for square matrices, and the determinant of the zero matrix is 0. Hence, taking det(AB)=det(A)det(B)=0 then follows det(A)=0 or det(B)=0 or det(A)=det(B)=0. The last condition is not sufficient for the original equation to hold, a counterexample is

\begin{pmatrix}4&2\\2&1\end{pmatrix} \begin{pmatrix}1&0\\0&0 \end{pmatrix}= \begin{pmatrix} 4&0\\2&0 \end{pmatrix}

. If

|B| \not=0

A must be singular.

It's hard to know, but I think you can assume the det AB = det A det B result.

If I'm not mistaken, I think that if det B is non-zero then A must be the zero matrix. Because we know the image of every row of A upon right-multiplication by B is zero, and as B isn't singular this implies every row of A is zero.

M^3+2M^2-5M-6I=(M+pI)(M+qI)(M+rI)=(M^2+MIq+pIM+pqI^2)(M+rI)=(M^2+Mq+pM+pqI)(M+rI)=M^3+M^2r+M^2q+Mqr+Mpr+M^2p+Mpq+pqrI)=M^3+M^2(r+p+q)+M(pq+qr+pr)+pqrI

Equating coefficients we have

pqr=-6 \rightarrow pr=\frac{6}{q} \newline pq+pr+qr=-5 \newline r+p+q=2 \rightarrow p+r=2-q

Sunstituting these two into the one left over we have

q(p+r)+pr=-5 \newline q(2-q)-\frac{6}{q}=-5

multiply by q

-q^3+2q^2+5q-6=0

This isn't really necessary: You could just write down M^3+2M^2-5M-6I = (M-aI)(M-bI)(M-cI) where a,b,c are the roots of x^3+2x^2-5x-6 = 0.

We can spot the trivial solution q=1 and thus factorise

(q-1)(-q^2+q+6)=0

using the quadratic formula the other solutions are -2 and 3, thus the factorisation is:

(M+I)(M-2I)(M+3I)

Now, to solve

M^3+2M^2-5M-6I=(M+I)(M-2I)(M+3I)=0

where

\begin{pmatrix}a&c\\0&b\end{pmatrix}=M

multiplying this together we have

\begin{pmatrix}a+1&c \\0&b+1 \end{pmatrix} \begin{pmatrix}a-2&c\\0&b-2 \end{pmatrix} \begin{pmatrix}a+3&c\\0&b+3\end{pmatrix}

= \begin{pmatrix}(a+1)(a-2)(a+3)&(a+1)(a-2)c+c(b+3)(a+b-1) \\ 0&(b+1)(b-2)(b+3)\end{pmatrix}

For this to equal the zero matrix a=-1, a=2, a=-3 and b=-1, b=2, b=-3 are potential candidates, no matter what the value of c is (and if c=0 any combination of the above will give a solution).
a=-1 together with b=2 respectively a=-1 together with b=-3 will make any value of c possible. (a=-1 with b=-1 gives no solution)
a=2 together with b=-3 also makes it possible for c to take any value (a=2 with b=-1 or a=2 with b=2 gives no soluition)
a=-3 gives no solution for either of the three relevant values of b - therefore all possible matrices M of the type

\begin{pmatrix}a&c\\0&b\end{pmatrix}

that satisfy the given equation have been found.

Hope this is right, and I'm aware that I probably need to show explicitly why the cases don't work, but latexing the algebra doesn't appeal very much to me!

I think what you've done is OK - might lose you a mark, but nothing more. It's probably a little easier if you use the "det B is non-zero implies A = 0" result.

@brian: I'm not ignoring your posts, but I'm not as confident on the stats, and I'd basically have to print your solutions out and do the questions myself to check them. I don't have the time right now I'm afraid.

Reply 102

brianeverit

9

Dystopia

Attempt at STEP III, Q8 - error suspected.

Spoiler

The only error I can see is a missing lnx in the numerator of the second term in your solution

Reply 103

brianeverit

9

Here is solution to Paper III question 4.

Reply 104

brianeverit

9

Paper fmb number 10
Attached is my part solution,. Can anyone tell me how to complete it?

1990.III.10(part).pdf28.2KB

Reply 105

ivydengys

Corrections to 1990 II No.1

nota bene

Factorise

x^4-2x^3+x^2

to

(x^2-1)^2

Just a very small error.

x^4-2x^3+x^2

=

(x^2-x)^2

Reply 106

OP

Question 4

Unparseable latex formula:

\begin{array} {l}[br]\displaystyle \cos \alpha + \cos ( \alpha + 2 \beta ) + \ldots + \cos ( \alpha + 2 n \beta ) \\[br]\displaystyle = \rm Re \, \left ( e^{i\alpha} + e^{i(\alpha+2\beta)} + \ldots + e^{i(\alpha+2n\beta)} \right ) \\[br]\displaystyle = \rm Re \, \left ( e^{i\alpha} ( 1+e^{i2\beta}+\ldots+e^{i2n\beta}) \right ) \\[br]\displaystyle = \rm Re \, \left ( e^{i\alpha} \left (\frac{e^{i2(n+1)\beta} - 1}{e^{i2\beta}-1} \right ) \right ) \\[br]\displaystyle = \rm Re \, \left ( e^{i\alpha} \frac{e^{i(n+1)\beta}}{e^{i\beta}} \frac{e^{i(n+1)\beta}-e^{-i(n+1)\beta}}{e^{i\beta}-e^{-i\beta}} \right ) \\[br]\displaystyle = \rm Re \, \left ( e^{i (\alpha + n\beta)} \frac{\sin(n+1)\beta}{\sin \beta} \right ) \\[br]\displaystyle = \frac{\sin (n+1)\beta \cos (\alpha + n \beta)}{\sin \beta} \end{array}

Unparseable latex formula:

\begin{array}{l}[br]\displaystyle \cos \alpha + \binom{n}{1} \cos ( \alpha + 2\beta ) + \ldots + \binom{n}{r} \cos ( \alpha + 2r \beta ) + \ldots + \cos ( \alpha + 2n \beta ) \\[br]\displaystyle = \rm Re \, \left ( e^{i\alpha} \left (1+e^{i2\beta} \right )^n \right ) \\[br]\displaystyle = \rm Re \, \left ( e^{i \alpha} \left ( 2 \cos \beta e^{i \beta} \right )^n \right ) \\[br]\displaystyle = \rm Re \, \left ( e^{i( \alpha + n \beta )} 2^n cos ^n \beta \right ) \\[br]\displaystyle = 2^n \cos^n \beta \cos ( \alpha + n \beta) [br]\end{array}[br]

For our final one, we proceed by induction.

Theorem

\displaystyle 1+\cos \theta \sec \theta + \ldots \cos r \theta \sec^r \theta + \ldots + \cos n \theta \sec ^n \theta = \frac{\sin (n+1)\theta \sec^n \theta}{\sin \theta}

Proof (by Induction)

Base Case

When

n=0

LHS = 1

\displaystyle RHS = \frac{\sin \theta \times 1}{\sin \theta} = 1 = LHS

as required

Inductive Hypothesis

Assume that, for all

k\leq n

,

\displaystyle 1+\cos \theta \sec \theta + \ldots \cos r \theta \sec^r \theta + \ldots + \cos k \theta \sec ^k \theta = \frac{\sin (k+1)\theta \sec^k \theta}{\sin \theta}

Inductive Step

If

\displaystyle 1+\cos \theta \sec \theta + \ldots + \cos r \theta \sec^r \theta + \ldots + \cos n \theta \sec ^n \theta = \frac{\sin (n+1)\theta \sec^n \theta}{\sin \theta}

Then by adding

\cos (n+1)\theta \sec^{n+1} \theta

to each side we get

Unparseable latex formula:

\begin{array}{l} \displaystyle 1+\cos \theta \sec \theta + \ldots + \cos n \theta \sec ^n \theta +\cos (n+1)\theta \sec^{n+1} \theta \\ [br]\displaystyle = \frac{\sin (n+1)\theta \sec^n \theta}{\sin \theta} + \cos (n+1)\theta \sec^{n+1} \theta \\[br]= \displaystyle \frac{\sin(n+1)\theta \sec^n \theta + \sin \theta \cos (n+1) \theta \sec^{n+1}}{\sin \theta} \\[br]= \displaystyle \frac{\sec^{n+1} \theta}{\sin \theta} \left ( \sin(n+1)\theta \cos \theta + \sin \theta \cos (n+1) \theta \right ) \\[br]= \displaystyle \frac{\sec^{n+1} \theta}{\sin \theta} \sin(n+2) \end{array}

Therefore, if our inductive hypothesis is true for n, it is true for n+1. Since it is true for 0, by the principle of mathematical induction, it is true for all nonnegative integers

Reply 107

OP

Question 15 STEP III

(Bear in mind I have never used any of this stats stuff before, and I'm not sure if it's all valid)

i)

\displaystyle \Pr(M=m,N=n) = \left ( \frac{1}{4} \right )^m \left ( \frac{3}{4} \right )^n \frac{1}{4} = \left ( \frac{1}{4} \right )^{m+1} \left ( \frac{3}{4} \right )^n

\displaystyle \boxed{\Pr(M=m)} = \sum_{n} \Pr (M=m, N=n) = \sum_{n=1}^{\infty} \left ( \frac{1}{4} \right )^{m+1} \left ( \frac{3}{4} \right )^n [br]\\ = \left ( \frac{1}{4} \right )^{m+1} \sum_{n=1}^{\infty} \left ( \frac{3}{4} \right )^n = 3\left ( \frac{1}{4} \right )^{m+1}

Unparseable latex formula:

\displaystyle \boxed{\Pr(N=n)} = \sum_{m} \Pr(M=m,N=n) = \sum_{m=0}^{\infty} \left ( \frac{1}{4} \right )^{m+1} \left ( \frac{3}{4} \right )^n \\[br]= \left ( \frac{3}{4} \right )^n \sum_{m=0}^{\infty} \left ( \frac{1}{4} \right )^{m+1} = \left ( \frac{3}{4} \right )^n \frac{1}{3}

ii)
Lemma 1

\displaystyle \boxed{\sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2}}

(Proof omitted, but it basically follows from differentiating the geometric progression)

\displaystyle E(M) = \sum_{m} m \Pr (M=m) = \sum_{m=0}^{\infty} m 3\left ( \frac{1}{4} \right )^{m+1} \\ = \frac{3}{4} \sum_{m=0}^{\infty} m \left ( \frac{1}{4} \right )^m = \frac{3}{4} \frac{\frac{1}{4}}{(\frac{1}{4}-1)^2} = \frac{1}{3}

\displaystyle E(N) = \sum_{n} n \Pr (N=n) = \sum_{n=1}^{\infty} n \frac{1}{3} \left ( \frac{3}{4} \right )^n = \frac{1}{3} \frac{\frac{3}{4}}{(1-\frac{3}{4})^2} = 4

Independence

We say two events are Independent if

\Pr ( A \cap B ) = \Pr ( A ) \Pr(B)

\displaystyle \Pr (M=m, N=n) = \left ( \frac{1}{4} \right )^{m+1} \left ( \frac{3}{4} \right )^n

\displaystyle \Pr (M=m) \times \Pr(N=n) = 3\left ( \frac{1}{4} \right )^{m+1} \times \left ( \frac{3}{4} \right )^n \frac{1}{3} = \left ( \frac{1}{4} \right )^{m+1} \left ( \frac{3}{4} \right )^n = \Pr (M=m, N=n)

Therefore they are independent. This is to be expected, since each toss is independent.

Last bit

\displaystyle \Pr ( N > k ) = 1 - \Pr ( N \leq k ) = 1 - \sum_{n=1}^{k} \Pr ( N = n ) = 1 - \sum_{n=1}^{k} \left ( \frac{3}{4} \right )^n \frac{1}{3} = 1 - \frac{1}{4} \frac{1-\left ( \frac{3}{4} \right )^k}{1-\frac{3}{4}} = 1- (1-\left (\frac{3}{4} \right)^k) = \boxed{\left (\frac{3}{4} \right)^k}

\displaystyle \Pr ( N > M ) = \sum_m \Pr (M=m) \Pr(N>m) = \sum_{m=0}^{\infty} \left (\frac{3}{4} \right)^m 3\left ( \frac{1}{4} \right )^{m+1} = \frac{3}{4} \sum_{m=0}^{\infty} \left ( \frac{3}{16} \right)^m = \frac{12}{13}

\displaystyle \Pr (N = M) = \sum_{m} \Pr(N=m) \Pr (M=m) = \sum_{m=1}^{\infty} \left ( \frac{1}{4} \right )^{m+1} \left ( \frac{3}{4} \right )^m = \frac{1}{4} \sum_{m=1}^{\infty} \left ( \frac{3}{16} \right )^m = \frac{3}{52}

\displaystyle \Pr ( M<N ) = 1 - \Pr ( N>M) - \Pr (M=N) = 1- \frac{12}{13} - \frac{3}{52} = \boxed{\frac{1}{52}}

Reply 108

OP

STEP II Question 7

f'(t)+f(t)-kf(t-1)=0

If there is a solution of the form

f(t) = Ae^{pt}

then

Ape^{pt} + Ae^{pt} -kAe^{p(t-1)} = 0

That is

Unparseable latex formula:

pe^{pt} +e^{pt} = ke^{p(t-1}}

Or

\boxed{p+1 = ke^{-p}}

as required.

If

k > 0

there will be one solution. (I'm not going to stick my graphs up, (no scanner)) (However, visualise the graphs

y=x+1

and

y = ke^{-x}

If our two graphs are tangential, there will be one solution (this is also the critical point between two solutions and no solutions for

k<0

So

\frac{d}{dx} (x+1) = 1

and

\frac{d}{dx} (ke^{-x}) = -ke^{-x}

That gives us

1 = -ke^{-x} \Rightarrow x = \ln (-k)

Substituting back in gives us

\ln (-k)+1 = ke^{-\ln (-k)} \Rightarrow \ln(-k)+1 = -1 \Rightarrow \boxed{k = -e^{-2}}

So we have

0 solutions if

k < -e^{-2}

1 solution if

k = -e^{-2}

or

k >0

2 solutions if

0>k>-e^{-2}

The final question asks us to find where the solutions are negative.

Consider the critical values when x = 0 is a solution. This gives us

k = 1

When

k<0

,

ke^{-x} < 0

so for this to equal

x+1

, we must have

x<-1

, which gives a negative p

When

k>0

, if

ke^{-x}<x+1

, consider at

x=0

, then the solution will be negative, that is

k<1

So this gives negative p for

\boxed{k<1}

Reply 109

OP

STEP I, Question 7

\dfrac{dy}{dx} +Py=Q

Let

y = uv

So

\displaystyle uv' + vu' + Puv = Q \Rightarrow v' + v \left (\frac{u'}{u} + P \right) = \frac{Q}{u}

Therefore, we can express

v'

in terms of

Q, u, x

if

P+\dfrac{u'}{u} = 0

That is

\boxed{u'+Pu = 0}

.

Solving

\displaystyle \frac{dy}{dx} - \frac{2y}{x+1} = (x+1)^{\frac{5}{2}}

Let

y = uv

, to find a suitable u, solve

u'+Pu =0

, that is

\frac{du}{dx} -\frac{2u}{x+1} = 0

Some separating variables gives

u = A(x+1)^2

We must now solve

v' = \frac{(x+1)^{\frac{5}{2}}}{A(x+1)^2}

, that is

v' = \frac{1}{A}(x+1)^{\frac{1}{2}}

, so

v = \frac{2}{3A}(x+1)^{\frac{3}{2}}+D

Therefore

y = uv = \frac{2}{3}(x+1)^{\frac{7}{2}}+F(x+1)^2

y(0) = \frac{2}{3}+G = 1 \Rightarrow G = \frac{1}{3}

Therefore

\boxed{y = \frac{1}{3} \Big (2(x+1)^{\frac{7}{2}}+(x+1)^{2}\Big)}

Valid for all

x>-1

(I think)

Reply 110

OP

STEP I, Question 15

The probability of obtaining k heads after exactly n tosses of a coin (where heads appears with probability p) is the probability that k-1 heads appear after n-1 tosses of a coin, and then the nth toss comes up heads. (These two cases are independent)

The probability that we have k-1 heads after n-1 tosses is

\displaystyle \binom{n-1}{k-1}p^{k-1}(1-p)^{n-1-(k-1)}

, (standard binomial distribution)

So the probability is

\boxed{\binom{n-1}{k-1} p^k (1-p)^{n-k}}

The probability that we get another head after an even number of tosses is

(1-p)p+(1-p)^3p+(1-p)^5p+\cdots = \frac{(1-p)p}{(1-(1-p)^2)}

(I'm personally not convinced by this, shouldn't the conditional probability have more of an affect on p? (I guess they stipulate p though))

We can have, oee, eoe, eeo. The probility the next head comes up on a number of the opposite parity is

\dfrac{p}{1-(1-p)^2}

and the same parity is

\dfrac{p(1-p)}{1-(1-p)^2}

these occur with probability

\dfrac{p}{1-(1-p)^2} \times \dfrac{p}{1-(1-p)^2} \times \dfrac{p(1-p)}{1-(1-p)^2} + \dfrac{p(1-p)}{1-(1-p)^2} \times \dfrac{p}{1-(1-p)^2} \times \dfrac{p}{1-(1-p)^2} + \dfrac{p(1-p)}{1-(1-p)^2} \times \dfrac{p(1-p)}{1-(1-p)^2} \times \dfrac{p}{1-(1-p)^2} =

\displaystyle \left (\dfrac{p}{1-(1-p)^2} \right)^3 \left ( 2 (1-p) +(1-p)^2 \right ) =

\frac{(1-p)(p-3)}{(p-2)^3}

Reply 111

OP

STEP II, Question 9

We have the curves

r(1+\cos \theta) = 1

and

r = 4(1+\cos \theta)

We seek the intersection points, therefore

r^2 = 4\times 1 = 4 \Rightarrow r = \pm 2

. So

\cos \theta = \pm \frac{1}{2} - 1 \Rightarrow (r, \theta) = (2,\pm \frac{2 \pi}{3})

Our first length will be

\displaystyle \int_{-\frac{2 \pi}{3}}^{\frac{2 \pi}{3}} \sqrt{(4(1+\cos \theta))^2+(4(-\sin \theta))^2} \, d \theta = 4 \int_{-\frac{2 \pi}{3}}^{\frac{2 \pi}{3}} \sqrt{2+2\cos \theta} \, d \theta = 4\sqrt{2}\int_{-\frac{2 \pi}{3}}^{\frac{2 \pi}{3}} \sqrt{2 \cos^2 \frac{\theta}{2}} \, d \theta =

\displaystyle 8 \int_{-\frac{2 \pi}{3}}^{\frac{2 \pi}{3}} |cos \frac{\theta}{2}| \, d \theta = \boxed{16 \sqrt{3}}

The second,

\displaystyle \int_{-\frac{2 \pi}{3}}^{\frac{2 \pi}{3}} \sqrt{\left( \frac{1}{(1+\cos \theta)} \right)^2+\left ( \frac{\sin \theta}{(1+\cos \theta)^2} \right )^2} \, d \theta =

\displaystyle \int_{-\frac{2 \pi}{3}}^{\frac{2 \pi}{3}} \sqrt{\left( \frac{1}{(1+\cos \theta)} \right)^2\left ( 1+ \frac{\sin^2 \theta}{(1+\cos \theta)^2} \right )} \, d \theta =

\displaystyle \int_{-\frac{2 \pi}{3}}^{\frac{2 \pi}{3}} \left | \frac{1}{(1+\cos \theta)^2} \right | \sqrt{2+2 \cos \theta} \, d \theta =

\displaystyle 2 \int_{-\frac{2 \pi}{3}}^{\frac{2 \pi}{3}} \left | \frac{\cos \frac{\theta}{2}}{(1+\cos \theta)^2} \right | \, d \theta =

\displaystyle \frac{1}{2} \int_{-\frac{2 \pi}{3}}^{\frac{2 \pi}{3}} \left | \frac{1}{\cos^3 \frac{\theta}{2}} \right | \, d \theta =

\displaystyle \frac{1}{2} \int_{-\frac{2 \pi}{3}}^{\frac{2 \pi}{3}} \left | \mathrm{sec}^2 \frac{\theta}{2}\mathrm{sec} \frac{\theta}{2} \right | \, d \theta = \boxed{ 2\sqrt{3}+\ln (2+\sqrt{3})}

Man my integrating skills are rusty

Reply 112

OP

STEP III, Question 7

I'm not 100% confident in this solution

xy^2+x^3+a^2y-a^3 = 0

Differentiating once wrt x

y^2+2xyy'+3x^2+a^2y'=0

and again

2yy'+2yy'+2xy'^2+2xyy''+6x+a^2y''=0

At

P(0,a)

y(0)=a

a^2+0+0+a^2y' = 0 \Rightarrow y'(0) = -1

-2a-2a+0+0+0+a^2y'' = 0\Rightarrow y''(0) = \frac{4}{a}

So the Taylor series near P is

\boxed{y(h) \approx a-h+\frac{2}{a}h^2}

At

Q(a,0)

y(a)=0

0+0+3a^2+a^2y'=0\Rightarrow y'(a) = -3

0+0+2a(-3)^2+0+6a+a^2y''=0 \Rightarrow y''(a) = \frac{24}{a}

So the Taylor series near Q is

\boxed{y(h) \approx a-3h+\frac{12}{a}h^2}

At

R(a,-a)

y(a)=-a

(-a)^2+2a(-a)y'+3a^2+a^2y'=0 \Rightarrow y'(a) = 4

2(-a)4+2(-a)4+2a4^2+2a(-a)y''+6a+a^2y''=0 \Rightarrow y''(a) = \frac{22}{a}

So the Taylor series near R is

\boxed{y(h) \approx a+4h+\frac{11}{a}h^2}

Considering it as a quadratic in y, the discriminant must be non-negative

That is

a^4-4(x)(x^3-a^3) \geq 0 \Rightarrow \boxed{4x^4 -4a^3x-a^4 \leq 0}

Reply 113

SimonM

STEP III, Question 7

I'm not 100% confident in this solutionWhat bit are you unsure about?

Reply 114

OP

DFranklin

What bit are you unsure about?

How "rigorously" I've treated those Taylor polynomials. (I've never been taught them properly, I've always just gone from guessing off wikipedia)

Reply 115

SimonM

How "rigorously" I've treated those Taylor polynomials. (I've never been taught them properly, I've always just gone from guessing off wikipedia)

I think what you've done is fine (assuming no algebra errors - I confess I haven't checked).

If you wanted to be 'formal' about it, you could put +o(x^2) at the end of each expansion, but to be honest, most people wouldn't bother even in a formal setting.

Incidentally, is there some trick to evaluating

\displaystyle \int_{-2\pi/3}^{2\pi/3} |sec^3 \frac{x}{2}|\, dx

or have you skipped rather a lot of working out? (I'm far too lazy to do the integral myself, but Mathematica comes up with something faintly horrendous that neither looks like something you could just write down, nor like something that easily simplifies).

Reply 116

OP

STEP III, Question 16

Part i)

We have an isosceles triangle, side lengths

X, \frac{1-X}{2}, \frac{1-X}{2}

The only angle which can be obtuse is the unique angle it is given as

2\sin^{-1} \frac{X/2}{\frac{1-X}{2}} \ge 90^{\circ}

So

\frac{X}{1-X} \ge \frac{1}{\sqrt{2}}

So

X \ge \frac{1}{\sqrt{2}+1}

For our triangle to exist,

\frac{1-X}{2}+\frac{1-X}{2} \ge X

(triangle inequality)

So we need to find

\frac{\frac{1}{2}-\frac{1}{\sqrt{2}+1}}{\frac{1}{2}} = \boxed{3-2\sqrt{2}}

Part ii)

We must show the triangle inequality satisfies for each side

Therefore x+y>2-x-y => x+y>1 (which is true since x+y=1 is a lower bound)

wlog x+2-x-y>y => y<1 (which is true given the boundaries)

\cos B = \frac{a^2+c^2-b^2}{2ac}

(Cosine rule)

Since B is obtuse we obtain

\frac{(2-x-y)^2+x^2-y^2}{2x(2-x-y)} < 0

So

(2-x-y)^2+x^2-y^2 <0

or

2x^2-4x+4 -4y+2xy < 0

Which is

\boxed{y> \frac{x^2-2x+2}{2-x}}

Reply 117

OP

DFranklin

Incidentally, is there some trick to evaluating

\displaystyle \int_{-2\pi/3}^{2\pi/3} |sec^3 \frac{x}{2}|\, dx

or have you skipped rather a lot of working out? (I'm far too lazy to do the integral myself, but Mathematica comes up with something faintly horrendous that neither looks like something you could just write down, nor like something that easily simplifies).

Not really any trick to it, just some more working that I couldn't be bothered to post up

(Incidentally mathematica evaluates nicely for me?)

Reply 118

SimonM

Not really any trick to it, just some more working that I couldn't be bothered to post up

(Incidentally mathematica evaluates nicely for me?)

The online integrator gives:

\displaystyle \tanh^{-1}(\tan(x/2)) + \frac{4}{1-\sin x} - \frac{1}{4(\cos(x/2)+\sin(x/2)^2}

Edit: thinking about it, the definite limits probably save you a fair bit of hassle when doing the standard IBP approach, compared with the indefinite answer.

Reply 119

OP