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2016 Paper 1 Q11
D: 1516.0 B: 1484.7

The point \(O\) is at the top of a vertical tower of height \(h\) which stands in the middle of a large horizontal plain. A projectile \(P\) is fired from \(O\) at a fixed speed \(u\) and at an angle \(\alpha\) above the horizontal. Show that the distance \(x\) from the base of the tower when \(P\) hits the plain satisfies \[ \frac{gx^2}{u^2} = h(1+\cos 2\alpha) + x \sin 2\alpha \,. \] Show that the greatest value of \(x\) as \(\alpha\) varies occurs when \(x=h\tan2\alpha\) and find the corresponding value of \(\cos 2\alpha\) in terms of \(g\), \(h\) and \(u\). Show further that the greatest achievable distance between \(O\) and the landing point is \(\dfrac {u^2}g +h\,\).

Solution: \begin{align*} \rightarrow: && x &= u \cos \alpha t\\ \Rightarrow && t &= \frac{x}{u \cos \alpha}\\ \uparrow: && -h &= u\sin \alpha t- \frac12gt^2 \\ && - h &= x\tan \alpha - \frac12 g \frac{x^2}{u^2}\sec^2 \alpha \\ \Rightarrow && \frac{gx^2}{u^2} &= h(2\cos^2 \alpha) + x2 \tan \alpha \cos^2 \alpha \\ &&&= h(1 + \cos 2 \alpha) + x \sin 2\alpha \\ \frac{\d}{\d \alpha}: && \frac{g}{u^2} 2 x \frac{\d x}{\d \alpha} &= -2h \sin 2 \alpha + 2x \cos 2 \alpha +\frac{\d x}{\d \alpha} \sin 2 \alpha \\ \Rightarrow && \frac{\d x}{\d \alpha} \left ( \frac{2xg}{u^2} - \sin 2 \alpha \right) &= 2\cos 2 \alpha (x -h \tan 2 \alpha) \end{align*} Since the turning point will be a maximum must be \(x = h \tan 2 \alpha\). Therefore, let \(c = \cos 2 \alpha\) \begin{align*} && \frac{gh^2}{u^2} \tan^2 2 \alpha &= h(1 + \cos 2 \alpha) + h \tan 2 \alpha \sin 2 \alpha \\ \Rightarrow && \frac{gh}{u^2}(c^{-2}-1) &= 1+c+\frac{1-c^2}{c} \\ \Rightarrow && \frac{gh(1-c^2)}{u^2c^2} &= \frac{c+c^2+1-c^2}{c}\\ &&&= \frac{1+c}{c} \\ \Rightarrow && \frac{gh(1-c)}{u^2c} &= 1 \\ \Rightarrow && u^2c &= gh(1-c) \\ \Rightarrow && c(u^2+gh) &= gh \\ \Rightarrow && \cos 2 \alpha &= \frac{gh}{u^2+gh} \\ \\ \Rightarrow && d_{max}^2 &= h^2 + h^2 \tan^2 2 \alpha \\ &&&= h^2\sec^2 2 \alpha \\ &&&= h^2 \frac{(u^2+gh)^2}{g^2h^2} \\ &&&= \frac{(u^2+gh)^2}{g^2} \\ &&&= \left (\frac{u^2}{g}+h \right)^2 \\ \Rightarrow && d_{max} &= \frac{u^2}{g}+h \end{align*}

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2016 Paper 1 Q12
D: 1516.0 B: 1484.7

  1. Alice tosses a fair coin twice and Bob tosses a fair coin three times. Calculate the probability that Bob gets more heads than Alice.
  2. Alice tosses a fair coin three times and Bob tosses a fair coin four times. Calculate the probability that Bob gets more heads than Alice.
  3. Let \(p_1\) be the probability that Bob gets the same number of heads as Alice, and let~\(p_2\) be the probability that Bob gets more heads than Alice, when Alice and Bob each toss a fair coin \(n\) times. Alice tosses a fair coin \(n\) times and Bob tosses a fair coin \(n+1\) times. Express the probability that Bob gets more heads than Alice in terms of \(p_1\) and \(p_2\), and hence obtain a generalisation of the results of parts (i) and (ii).

Solution:

  1. There are several possibilities \begin{array}{c|c|c} \text{Alice} & \text{Bob} & P \\ \hline 0 & 1 & \frac1{2^2} \cdot 3 \cdot \frac{1}{2^3} = \frac{3}{2^5} \\ 0 & 2 & \frac1{2^2} \cdot 3 \cdot \frac{1}{2^3} = \frac{3}{2^5} \\ 0 & 3 & \frac1{2^2} \cdot \frac{1}{2^3} = \frac{1}{2^5} \\ 1 & 2 & 2 \cdot \frac1{2^2} \cdot 3 \cdot \frac{1}{2^3} = \frac{6}{2^5} \\ 1 & 3 & 2\cdot \frac1{2^2} \cdot \frac{1}{2^3} = \frac{2}{2^5} \\ 2 & 3 & \frac1{2^2} \cdot \frac{1}{2^3} = \frac{1}{2^5} \\ \hline && \frac{1}{2^5}(3+3+1+6+2+1) = \frac{16}{2^5} = \frac12 \end{array}
  2. There are several possibilities \begin{array}{c|c|c} A & B & \text{count} \\ \hline 0 & 1 & 4 \\ 0 & 2 & 6 \\ 0 & 3 & 4 \\ 0 & 4 & 1 \\ 1 & 2 & 3\cdot6 \\ 1 & 3 & 3\cdot4 \\ 1 & 4 & 3 \\ 2 & 3 & 3\cdot4 \\ 2 & 4 & 3 \\ 3 & 4 & 1 \\ \hline && 64 \end{array} Therefore the total probability is \(\frac12\)
  3. \(\mathbb{P}(\text{Bob more than Alice}) = p_1 \cdot \underbrace{\frac12}_{\text{he wins by breaking the tie on his last flip}} + p_2\) If \(p_3\) is the probability that Alice gets more heads than Bob, then by symmetry \(p_3 = p_2\) and \(p_1 + p_2 + p_3 = 1\). Therefore \(p_1 + 2p_2 = 1\). ie \(\frac12 p_1 + p_2 = \frac12\) therefore the answer is always \(\frac12\) for all values of \(n\).

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2016 Paper 1 Q13
D: 1500.0 B: 1500.0

An internet tester sends \(n\) e-mails simultaneously at time \(t=0\). Their arrival times at their destinations are independent random variables each having probability density function \(\lambda \e^{-\lambda t}\) (\(0\le t<\infty\), \( \lambda >0\)).

  1. The random variable \(T\) is the time of arrival of the e-mail that arrives first at its destination. Show that the probability density function of \(T\) is \[ n \lambda \e^{-n\lambda t}\,,\] and find the expected value of \(T\).
  2. Write down the probability that the second e-mail to arrive at its destination arrives later than time \(t\) and hence derive the density function for the time of arrival of the second e-mail. Show that the expected time of arrival of the second e-mail is \[ \frac{1}{\lambda} \left( \frac1{n-1} + \frac 1 n \right) \]

Solution:

  1. \(\,\) \begin{align*} && \mathbb{P}(T > t) &= \mathbb{P}(\text{all emails slower than }t) \\ &&&= \left ( \int_t^{\infty} \lambda e^{-\lambda x} \d x \right)^n \\ &&&= \left ( [- e^{-\lambda x}]_t^\infty\right)^n\\ &&&= e^{-n\lambda t} \\ \Rightarrow && f_T(t) &= n \lambda e^{-n\lambda t} \\ \end{align*} Therefore \(T \sim \text{Exp}(n \lambda)\) and \(\E[T] = \frac{1}{n \lambda}\)
  2. Let \(T_2\) be the time until the second email arrives, then. \begin{align*} && \P(T_2 > t) &= \P(\text{all emails} > t) + \P(\text{all but 1 emails} > t) \\ &&&= e^{-n\lambda t} + n \cdot e^{-(n-1)\lambda t}(1-e^{-\lambda t}) \\ &&&= (1-n)e^{-n\lambda t} + n \cdot e^{-(n-1)\lambda t} \\ \Rightarrow && f_{T_2}(t) &= - \left ( (1-n) n \lambda e^{-n \lambda t} -n(n-1)\lambda e^{-(n-1)\lambda t} \right) \\ &&&= n(n-1) \lambda \left (e^{-(n-1)\lambda t} - e^{-n\lambda t} \right) \\ \Rightarrow && \E[T_2] &= \int_0^{\infty} t \cdot n(n-1) \lambda \left (e^{-(n-1)\lambda t} - e^{-n\lambda t} \right) \d t \\ &&&= \int_0^{\infty} \left (n \cdot t (n-1) \lambda e^{-(n-1)\lambda t} -(n-1)\cdot tn \lambda e^{-n\lambda t} \right) \d t \\ &&&= \frac{n}{\lambda(n-1)} - \frac{n-1}{\lambda n} \\ &&&= \frac{1}{\lambda} \left (1+\frac{1}{n-1}- \left (1 - \frac{1}{n} \right) \right) \\ &&&= \frac{1}{\lambda} \left ( \frac{1}{n-1} + \frac{1}{n} \right) \end{align*} (We can also view this second expectation as expected time for first email + expected time (of the remaining \(n-1\) emails) for the first email, and we can see that will have that form by the memorilessness property of exponentials)

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2016 Paper 2 Q1
D: 1600.0 B: 1516.0

The curve \(C_1\) has parametric equations \(x=t^2\), \(y= t^3\), where \(-\infty < t < \infty\,\). Let \(O\) denote the point \((0,0)\). The points \(P\) and \(Q\) on \(C_1\) are such that \(\angle POQ\) is a right angle. Show that the tangents to \(C_1\) at \(P\) and \(Q\) intersect on the curve \(C_2\) with equation \(4y^2=3x-1\). Determine whether \(C_1\) and \(C_2\) meet, and sketch the two curves on the same axes.

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2016 Paper 2 Q2
D: 1600.0 B: 1516.0

Use the factor theorem to show that \(a+b-c\) is a factor of \[ (a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2) +8(a^3+b^3+c^3) \,. \tag{\(*\)} \] Hence factorise (\(*\)) completely.

  1. Use the result above to solve the equation \[ (x+1)^3 -3 (x+1)(2x^2 +5) +2(4x^3+13)=0\,. \]
  2. By setting \(d+e=c\), or otherwise, show that \((a+b-d-e)\) is a factor of \[ (a+b+d+e)^3 -6(a+b+d+e)(a^2+b^2+d^2+e^2) +8(a^3+b^3+d^3+e^3) \, \] and factorise this expression completely. Hence solve the equation \[ (x+6)^3 - 6(x+6)(x^2+14) +8(x^3+36)=0\,. \]

Solution: Suppose \(c = a+b\) then \begin{align*} (a+b+c)^3 &-6(a+b+c)(a^2+b^2+c^2) +8(a^3+b^3+c^3) \\ &= (2(a+b))^3-6(2(a+b))(a^2+b^2+(a+b)^2) + 8(a^3+b^3+(a+b)^3) \\ &=16(a+b)^3 - 24(a+b)(a^2+b^2+ab)+8(a^3+b^3) \\ &= 8(a+b)(2(a+b)^2-3(a^2+b^2+ab)+(a^2-ab+b^2)) \\ &= 0 \end{align*} Therefore \(a+b-c\) is a factor. By symmetry \(a-b+c\) and \(-a+b+c\) are also factors. Since our polynomial is degree \(3\) it must be \(K(a+b-c)(b+c-a)(c+a-b)\) for some \(K\). Since the coefficient of \(a^3\) is \(3\), \(K = 3\). so we have: \(3(a+b-c)(b+c-a)(c+a-b)\)

  1. We want \(x + a + b = x+1\), \(x^2 + a^2 + b^2 = x^3+\frac52, x^3 + a^3 + b^3 = x^3+ \frac{13}{4}\). \(a+b = 1, a^2 + b^2 = 5/2\) so \(a = \frac32, b = -\frac12\) \begin{align*} 0 &= (x+1)^3 - 3(x+1)(2x^2+5)+2(4x^3+13) \\ &= 3(x +\frac{3}{2}+\frac{1}{2})(x - \frac{3}{2} - \frac{1}{2})(-x + \frac{3}{2} - \frac{1}{2}) \\ &= 3(x+2)(x-2)(1-x) \end{align*} and so the roots are \(x = 1, 2, -2\)
  2. Letting \(c = d+e\) we have \begin{align*} (a+b+d+e)^3 &-6(a+b+d+e)(a^2+b^2+d^2+e^2) +8(a^3+b^3+d^3+e^3) \\ &= (a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2-2de) +8(a^3+b^3+c^3 - 3cde) \\ &= (a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2)+8(a^3+b^3+c^3)+12(a+b+c)de - 24cde \\ &= \underbrace{(a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2)+8(a^3+b^3+c^3)}_{\text{has a factor of }a+b-c} + 12(a+b-c)de \end{align*} Therefore there is a factor of \(a+b-c\) or \(a+b-d-e\). By symmetry we must have the factors: \((a+b-d-e)(a-b-d+e)(a-b+d-e)\) and so the final expression must be: \(K(a+b-d-e)(a-b-d+e)(a-b+d-e)\) The coefficient of \(a^3\) is \(3\), therefore \(K = 3\) We want \(x+a+b+c = x + 6\), \(x^2+a^2+b^2+c^2 = 14\) and \(x^3 + a^3+b^3+c^3 = 36\), ie \(a = 1,b=2,c=3\) would work, so \begin{align*} 0 &= (x+6)^3 - 6(x+6)(x^2+14) +8(x^3+36) \\ &= 3(x+1-2-3)(x-1+2-3)(x-1-2+3) \\ &= 3x(x-4)(x-2) \end{align*} ie the roots are \(x = 0, 2, 4\)

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2016 Paper 2 Q3
D: 1600.0 B: 1517.4

For each non-negative integer \(n\), the polynomial \(\f_n\) is defined by \[ \f_n(x) = 1 + x + \frac{x^2}{2!} + \frac {x^3}{3!} + \cdots + \frac{x^n}{n!} \]

  1. Show that \(\f'_{n}(x) = \f_{n-1}(x)\,\) (for \(n\ge1\)).
  2. Show that, if \(a\) is a real root of the equation \[\f_n(x)=0\,,\tag{\(*\)}\] then \(a<0\).
  3. Let \(a\) and \(b\) be distinct real roots of \((*)\), for \(n\ge2\). Show that \(\f_n'(a)\, \f_n'(b)>0\,\) and use a sketch to deduce that \(\f_n(c)=0\) for some number \(c\) between \(a\) and \(b\). Deduce that \((*)\) has at most one real root. How many real roots does \((*)\) have if \(n\) is odd? How many real roots does \((*)\) have if \(n\) is even?

Solution:

  1. \(\,\) \begin{align*} && f'_n(x) &= 0 + 1 + \frac{2x}{2!} + \frac{3x^2}{3!} + \cdots + \frac{nx^{n-1}}{n!} \\ &&&= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{n-1}}{(n-1)!} \\ &&&= f_{n-1}(x) \end{align*}
  2. Claim: \(f_n(x) > 0\) for all \(x > 0\) Proof: (By induction) Base case: (\(n = 1\)) \(f_1(x) = 1 + x > 1\) therefore \(f_1(x) > 0\) Suppose it's true for \(n = k\), then consider \(f_{k+1}\), if we differentiate it, we find it is increasing on \((0, \infty)\) by our inductive hypothesis. But then \(f_{k+1}(0) = 1 > 0\). Therefore \(f_{k+1}(x) > 0\) as well. Therefore by the principle of mathematical induction we are done. Since \(f_n(x) > 0\) for non-negative \(x\), if \(a\) is a root it must be negative.
  3. Suppose \(f_n(a) = f_n(b) = 0\) then \(f'_n(a) = -\frac{a^n}{n!}\) and \(f'_n(b) = -\frac{b^n}{n!}\), but then \(f_n'(a) f_n'(b) = \frac{(-a)^n(-b)^n}{(n!)^2} > 0\) since \(a < 0, b < 0\). \(_n'(a) f_n'(b)\) is positive, the two gradients must have the same sign (and not be zero). Therefore if they are both increasing, at some point the curve must cross the axis in between. Therefore there is some root \(c\) between \(a\) and \(b\). But then there is also a root between \(c\) and \(a\) and \(c\) and \(b\), and very quickly we find more than \(n\) roots which is not possivel. Therefore there must be at most \(1\) root. If \(n\) is odd there must be exactly one root, since \(f_n\) changes sign as \(x \to -\infty\) vs \(x = 0\). If \(n\) is even then there can't be any roots, since if it crossed the \(x\)-axis there would be two roots (not possible) and it cannot touch the axis, since \(f'_n(a) \neq 0\) unless \(a = 0\), and we know \(a < 0\)

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2016 Paper 2 Q4
D: 1600.0 B: 1500.0

Let \[ y=\dfrac{x^2+x\sin\theta+1}{x^2+x\cos\theta+1} \,.\]

  1. Given that \(x\) is real, show that \[ (y\cos\theta -\sin\theta)^2 \ge 4 (y-1)^2 \,. \] Deduce that \[ y^2+1 \ge 4(y-1)^2 \,, \] and hence that \[ \dfrac {4-\sqrt7}3 \le y \le \dfrac {4+\sqrt7}3 \,. \]
  2. In the case $y= \dfrac {4+\sqrt7}3 \,$, show that \[\sqrt{y^2+1}=2(y-1)\] and find the corresponding values of \(x\) and \(\tan\theta\).

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2016 Paper 2 Q5
D: 1600.0 B: 1484.0

In this question, the definition of \(\displaystyle\binom pq\) is taken to be \[ \binom pq = \begin{cases} \dfrac{p!}{q!(p-q)!} & \text{ if } p\ge q\ge0 \,,\\[4mm] 0 & \text{ otherwise } . \end{cases} \]

  1. Write down the coefficient of \(x^n\) in the binomial expansion for \((1-x)^{-N}\), where \(N\) is a positive integer, and write down the expansion using the \(\Sigma\) summation notation. By considering $ (1-x)^{-1} (1-x)^{-N} \, ,$ where \(N\) is a positive integer, show that \[ \sum_{j=0}^n \binom { N+j -1}{j} = \binom{N+n}{n}\,. \]
  2. Show that, for any positive integers \(m\), \(n\) and \(r\) with \(r\le m+n\), \[ \binom{m+n} r = \sum _{j=0}^r \binom m j \binom n {r-j} \,. \]
  3. Show that, for any positive integers \(m\) and \(N\), \[ \sum_{j=0}^n(-1)^{j} \binom {N+m} {n-j} \binom {m+j-1}{j } = \displaystyle \binom N n . \]

Solution:

  1. \(\frac{(-N)(-N-1)\cdots(-N-n+1)}{n!} = \binom{N+n-1}{n}\), so \[ (1-x)^{-N} = \sum_{n=0}^{\infty} \binom{N+n-1}{n} x^n\] \begin{align*} && (1-x)^{-N-1} &= (1-x)^{-1}(1-x)^{-N} \\ &&&= (1 + x + x^2 + \cdots)\left ( \sum_{n=0}^{\infty} \binom{N+n-1}{n} x^n\right)\\ [x^{n}]: && \binom{N+1+n-1}{n} &= \sum_{j=0}^n \underbrace{1}_{x^{n-j} \text{ from 1st bracket}}\cdot\underbrace{\binom{N+j-1}{j}}_{x^j\text{ from second bracket}} \\ \Rightarrow && \binom{N+n}{n} &= \sum_{j=0}^n \binom{N+j-1}{j} \end{align*}
  2. Consider \((1+x)^{m+n} = (1+x)^m(1+x)^n\) and consider the coefficient of \(x^r\) from each side. On the left hand side this is clearly \(\binom{m+n}{r}\) on the right hand side we can take \(x^j\) from \((1+x)^m\) and \(x^{n-j}\) from \((1+x)^n\) and \(j\) can take any value from \(0\) to \(r\), ie \[ \binom{m+n} r = \sum _{j=0}^r \binom m j \binom n {r-j} \]
  3. Consider \((1-x)^{-(N+m+1)} = (\)

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2016 Paper 2 Q6
D: 1600.0 B: 1484.0

This question concerns solutions of the differential equation \[ (1-x^2) \left(\frac{\d y}{\d x}\right)^2 + k^2 y^2 = k^2\, \tag{\(*\)} \] where \(k\) is a positive integer. For each value of \(k\), let \(y_k(x)\) be the solution of \((*)\) that satisfies \(y_k(1)=1\); you may assume that there is only one such solution for each value of \(k\).

  1. Write down the differential equation satisfied by \(y_1(x)\) and verify that \(y_1(x) = x\,\).
  2. Write down the differential equation satisfied by \(y_2(x)\) and verify that \(y_2(x) = 2x^2-1\,\).
  3. Let \(z(x) = 2\big(y_n(x)\big)^2 -1\). Show that \[ (1-x^2) \left(\frac{\d z}{\d x}\right)^2 +4n^2 z^2 = 4n^2\, \] and hence obtain an expression for \(y_{2n}(x)\) in terms of \(y_n(x)\).
  4. Let \(v(x) = y_n\big(y_m(x)\big)\,\). Show that \(v(x) = y_{mn}(x)\,\).

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2016 Paper 2 Q7
D: 1600.0 B: 1516.0

Show that \[ \int_0^a \f(x) \d x= \int _0^a \f(a-x) \d x\,, \tag{\(*\)} \] where f is any function for which the integrals exist.

  1. Use (\(*\)) to evaluate \[ \int_0^{\frac12\pi} \frac{\sin x}{\cos x + \sin x} \, \d x \,. \]
  2. Evaluate \[ \int_0^{\frac14\pi} \frac{\sin x}{\cos x + \sin x} \, \d x \,. \]
  3. Evaluate \[ \int_0^{\frac14\pi} \ln (1+\tan x) \, \d x \,. \]
  4. Evaluate \[ \int_0^{\frac14 \pi} \frac x {\cos x \, (\cos x + \sin x)}\, \d x \,. \]

Solution: \begin{align*} u = a-x, \d u = - \d x: && \int_0^a f(x) \d x &= \int_{u=a}^{u=0} f(a-u) (-1) \d u \\ &&&= \int_0^a f(a-u) \d u \\ &&&= \int_0^a f(a-x) \d x \end{align*}

  1. \begin{align*} && I &= \int_0^{\frac12 \pi} \frac{\sin x}{\cos x + \sin x } \d x\\ &&&= \int_0^{\frac12 \pi} \frac{\sin (\frac12 \pi - x)}{\cos (\frac12 \pi-x) + \sin (\frac12 \pi-x) } \d x\\ &&&= \int_0^{\frac12 \pi} \frac{\cos x}{\sin x + \cos x } \d x\\ \Rightarrow && 2I &= \int_0^{\frac12 \pi} 1 \d x \\ \Rightarrow && I &= \frac{\pi}{4} \end{align*}
  2. \begin{align*} && I &= \int_0^{\frac14 \pi} \frac{\sin x}{\cos x + \sin x } \d x\\ &&&= \int_0^{\frac14 \pi} \frac{\sin (\frac14 \pi - x)}{\cos (\frac14 \pi-x) + \sin (\frac14 \pi-x) } \d x\\ &&&= \int_0^{\frac14 \pi} \frac{\frac1{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x}{\frac1{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x + \frac1{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x} \d x \\ &&&= \int_0^{\frac14 \pi} \frac{\cos x - \sin x}{2 \cos x} \d x \\ &&&= \left [\frac12 x + \ln(\cos x) \right]_0^{\pi/4} \\ &&&= \frac{\pi}{8} -\frac12\ln2 - 1 \end{align*}
  3. \begin{align*} && I &= \int_0^{\frac14\pi} \ln (1+\tan x) \, \d x \\ &&&= \int_0^{\frac14 \pi} \ln \left (1 + \tan \left(\frac{\pi}{4} - x\right) \right) \, \d x\\ &&&= \int_0^{\frac14 \pi} \ln \left (1 +\frac{1 - \tan x}{1+ \tan x} \right) \, \d x\\ &&&= \int_0^{\frac14 \pi} \ln \left (\frac{2}{1+ \tan x} \right) \, \d x\\ &&&= \frac{\pi}{4} \ln 2 - I \\ \Rightarrow && I &= \frac{\pi}{8} \ln 2 \end{align*}
  4. \begin{align*} && I &= \int_0^{\frac14 \pi} \frac x {\cos x \, (\cos x + \sin x)}\, \d x \\ &&&= \int_0^{\frac14 \pi} \frac {\frac14 \pi - x} {(\frac1{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x) \, (\frac{2}{\sqrt{2}}\cos x)}\, \d x \\ &&&= \int_0^{\frac14 \pi} \frac {\frac14 \pi - x} {\cos x \, (\cos x + \sin x)}\, \d x \\ \\ \Rightarrow && I &= \frac{\pi}{8} \int_0^{\pi/4} \frac{\sec^2 x}{1 + \tan x} \d x\\ &&&= \frac{\pi}{8} \left [\ln (1 + \tan x) \right]_0^{\pi/4} \\ &&&= \frac{\pi}{8} \ln 2 \end{align*}

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