STEP maths I, II, III 2007 solutions

STEP II Q6, solution by chewwy, shamelessly stolen from the STEP II thread:

http://www.thestudentroom.co.uk/attachment.php?attachmentid=42044&d=1183161671

Reply 2

I'll do Question 3 STEP I. (What a gem)

Reply 3

generalebriety

16

An excellent idea. :biggrin:

III/1:

\displaystyle \tan (\theta_1 + \theta_2 + \theta_3 + \theta_4) = \frac{\tan (\theta_1 + \theta_2) + \tan (\theta_3 + \theta_4)}{1 - \tan (\theta_1 + \theta_2)\tan (\theta_3 + \theta_4)}

\displaystyle = \frac{\frac{t_1+t_2}{1-t_1t_2} + \frac{t_3+t_4}{1-t_3t_4}}{1 - \left(\frac{t_1+t_2}{1-t_1t_2}\right) \left(\frac{t_3+t_4}{1-t_3t_4}\right)}

\displaystyle = \frac{(t_1+t_2)(1-t_3t_4) + (t_3+t_4)(1-t_1t_2)}{ (1-t_1t_2)(1-t_3t_4) - (t_1+t_2)(t_3+t_4)} .

\displaystyle at^4 + bt^3 + ct^2 + dt + e = a(t-t_1)(t-t_2)(t-t_3)(t-t_4)

\displaystyle \Rightarrow t_1+t_2+t_3+t_4 = -b/a,\quad t_1t_2 + t_1t_3 + t_1t_4 + t_2t_3 + t_2t_4 + t_3t_4 = c/a,

\displaystyle t_1t_2t_3 + t_1t_2t_4 + t_1t_3t_4 + t_2t_3t_4 = -d/a, \quad t_1t_2t_3t_4 = e/a.

\displaystyle \tan (\theta_1 + \theta_2 + \theta_3 + \theta_4) = \frac{(t_1+t_2+t_3+t_4) - (t_1t_2t_3 + t_1t_2t_4 + t_1t_3t_4 + t_2t_3t_4)}{t_1t_2t_3t_4 + 1 - (t_1t_2 + t_1t_3 + t_1t_4 + t_2t_3 + t_2t_4 + t_3t_4)}

\displaystyle = \frac{(-b/a) + (d/a)}{(e/a) + 1 - (c/a)}

\displaystyle = \frac{d-b}{a+e-c} . \quad (*)

--

\displaystyle \tan \theta_i = t_i \Rightarrow \sin \theta_i = \frac{t_i}{\sqrt{1 + t_i^2}} ,\quad \cos \theta_i = \frac{1}{\sqrt{1+t_i^2}} .

Substituting into the given equation:

\displaystyle p\cos 2\theta + \cos (\theta -\alpha ) + p = 0

\displaystyle 2p\cos^2\theta - p + \cos\theta\cos\alpha - \sin\theta\sin\alpha + p = 0

\displaystyle \frac{2p}{1+t^2} + \frac{\cos\alpha}{\sqrt{1+t^2}} - \frac{t\sin\alpha}{1 + t^2} = 0

\displaystyle 2p = -(\cos\alpha + t\sin\alpha ) \sqrt{1+t^2}

\displaystyle 4p^2 = (\cos^2\alpha + t\sin 2\alpha + t^2\sin^2\alpha )(1+t^2)

\displaystyle (\sin^2\alpha )t^4 + (\sin 2\alpha )t^3 + t^2 + (\sin 2\alpha )t + (\cos^2\alpha - 4p^2) = 0

Substituting into (*):

\displaystyle \tan (\theta_1 + \theta_2 + \theta_3 + \theta_4) = \frac{\sin 2\alpha - \sin 2\alpha}{a+e-c} = 0

\displaystyle \tan (\theta_1 + \theta_2 + \theta_3 + \theta_4) = 0 \Rightarrow \theta_1 + \theta_2 + \theta_3 + \theta_4 = n\pi

as required.

Reply 4

STEP III Q2, solution by Speleo, shamelessly stolen from the STEP III thread:

Speleo

2
i

1.3.5.7...(2n-1) = \frac{1.2.3.4...(2n-1)(2n)}{2.4.6.8...(2n)}

= \frac{(2n)!}{(2.1)(2.2)(2.3)(2.4)...(2.n)} = \frac{(2n!)}{2^nn!}

as required.

\frac{1}{\sqrt{1-4x}} = (1-4x)^{-\frac{1}{2}}

Binomial expansion valid for

|4x| < 1

i.e.

|x| < \frac{1}{4}

as given.

\frac{1}{\sqrt{1-4x}} = 1 + 2x + ... + \frac{(-\frac{1}{2})(-\frac{3}{2})...(-\frac{2r-1}{2}).(-4x)^r}{r!} + ...

Every successive term is multiplied by two extra negative terms, one fraction from the exponent and one power of -4x, so every term stays positive overall, so all minus signs can be removed.

= 1 + 2x + ... + \frac{(\frac{1}{2})(\frac{3}{2})...(\frac{2r-1}{2}).(4x)^r}{r!} + ...

Using the formula just proved and noting that there is a divisor of

2^r

in total from the product of

(\frac{3}{2})...(\frac{2r-1}{2})

:

= 1 + 2x + ... + \frac{(2r!)4^rx^r}{2^r2^r(r!)^2} + ...

= 1 + 2x + ... + \frac{(2r!)x^r}{(r!)^2} + ...

= 1 + \Sigma_1^{\infty}\frac{(2n!)x^n}{(n!)^2}

as required.
Verify that 2x is the first term by setting n = 1 and getting 2x.

ii
Differentiate both sides with respect to x:

\frac{2}{(1-4x)^{\frac{3}{2}}} = \Sigma_1^{\infty}\frac{n(2n!)x^{n-1}}{(n!)^2}

Setting

x = \frac{6}{25}

:

250 = \Sigma_1^{\infty}\frac{n(2n!)\frac{6}{25}^{n-1}}{(n!)^2}

Multiply both sides by

\frac{6}{25}

:

\Sigma_1^{\infty}\frac{n(2n!)\frac{6}{25}^n}{(n!)^2} = 60

as required.

iii
This time integrate instead:

-\frac{1}{2}\sqrt{1-4x} = x + \Sigma_1^{\infty}\frac{(2n!)x^{n+1}}{(n+1)(n!)^2} + C

Set x = 0:

-\frac{1}{2} = 0 + 0 + C \rightarrow C = -\frac{1}{2}

Set x =

\frac{2}{9}

:

-\frac{1}{6} = \frac{2}{9} + \Sigma_1^{\infty}\frac{(2n!)2^{n+1}}{3^{2n}(n+1)!(n!)} - \frac{1}{2}

\Sigma_1^{\infty}\frac{(2n!)2^{n+1}}{3^{2n+2}(n+1)!(n!)} = \frac{1}{9}

\Sigma_1^{\infty}\frac{(2n!)2^{n+1}}{3^{2n}(n+1)!(n!)} = 1

as required.

Reply 5

STEP III Q6:

Firstly,

pp^* = |p|^2 = a^2

and similarly for q.

Thus we have

pq(p^* - q^*) = pp^*q - qq^*p = a^2(q-p)

and the first result follows.

Then, note that if

PQ

and

RS

are perpendicular,

\displaystyle \frac{p-q}{s-r} = \frac{|p-q|}{|s-r|}i

or

\displaystyle \frac{p-q}{s-r} = -\frac{|p-q|}{|s-r|}i

Either way, we have

\displaystyle \frac{p^* - q^*}{s^* - r^*} = \left(\frac{p-q}{s-r}\right)^*

\displaystyle \frac{p^* - q^*}{s^* - r^*} = -\frac{p-q}{s-r}

and thus

\displaystyle \frac{s-r}{s^* - r^*} = - \frac{p-q}{p^*-q^*}

\displaystyle \frac{s-r}{s^* - r^*} + \frac{p-q}{p^*-q^*} = 0

from which it follows that

pq + rs

\displaystyle = -a^2 \left(\frac{p-q}{p*-q*} + \frac{s-r}{s^* - r^*}\right)

= 0

as required.

For the next part, we have in the case

n=3

: (let

a_i, b_i

represent

A_i, B_i

in the Argand plane, with the circle centred in the origin and with radius a)

Unparseable latex formula:

\displaystyle \left\{\begin{array} a_1a_2 = -b_1b_2 \\ a_2a_3 = - b_2b_3 \\ a_3a_1 = -b_3b_1 \end{array}\right.

Multiplying the first and the third equations, and then dividing by the second, we get

\displaystyle a_1^2 = -b_1^2

there are obviously two distinct

b_1

that satisfy this, as required.

For

n = 4

, it's instead

Unparseable latex formula:

\displaystyle \left\{\begin{array} a_1a_2 = -b_1b_2 \\ a_2a_3 = - b_2b_3 \\ a_3a_4 = -b_3b_4 \\ a_4a_1 = -b_4b_1 \end{array}\right.

In this case, any

b_1

can satisfy this, becayse if you choose a

b_1

, you can always choose a

b_2

so that the first equation is satisfied, and then a

b_3

so that the second one is satisfied, and then a

b_4

so that the third is satisfied, and then, and here's the catch, equation 4 follows from the other three: (multiply the first and third equations and divide by the fourth) and so the fourth equation will automatically satified.

For larger n, we have exactly the same sitution as in n = 3 when n is odd (you can use the same reasoning, just alternatingly multiply and divide all the equations together) and as in n = 4 when n is even (again the same reasoning, the last equation can always be obtained by alternatingly multiplying and dividing the first ones together.

Reply 6

Square

13

Don't suppose the papers are up on the 'net yet are they?

Reply 7

Square

Don't suppose the papers are up on the 'net yet are they?

Not really, but they are stilled linked in the first post. :tsr2:

Reply 8

Square

13

Gonna try some of the mechanics questions :smile:

Reply 9

generalebriety

16

Ooooooh. III/7. Mine. :wink:

(i)

\displaystyle t(x) = \int_0^x \frac{1}{1+u^2} \text{d} u.

u = v^{-1} \Rightarrow \text{d} u = -v^{-2} \text{d} v

\displaystyle \therefore t(x) = \int_{u=0}^{u=x} \frac{1}{1+v^{-2}} \cdot -v^{-2} \text{d} v = -\int_\infty^{1/x} \frac{v^{-2}}{1 + v^{-2}} \text{d} v

\displaystyle = \int_{1/x}^\infty \frac{1}{1+v^2} \text{d} v

\displaystyle \therefore t(1/x) + t(x) = \int_{x}^\infty \frac{1}{1+v^2} \text{d} v + \int_0^x \frac{1}{1+u^2} \text{d} u = \int_0^\infty \frac{1}{1+u^2} \text{d} u = p/2.

Choose x = 1 and result follows.

(ii)

\displaystyle y = \frac{u}{\sqrt{1+u^2}} \Rightarrow y^2(1+u^2) = u^2 \Rightarrow u = \frac{y}{\sqrt{1-y^2}}

\displaystyle \frac{\text{d} u}{\text{d} y} = (1-y^2)^{-1/2} + y^2(1-y^2)^{-3/2} = (1-y^2+y^2)(1-y^2)^{-3/2} = \frac{1}{\sqrt{(1-y^2)^3}}

Performing this substitution:

\displaystyle t(x) = \int_{u=0}^{u=x} \frac{1}{1 + \frac{y^2}{1-y^2}} \cdot \frac{1}{\sqrt{(1-y^2)^3}} \text{d} y

\displaystyle = \int_{u=0}^{u=x} \frac{1-y^2}{\sqrt{(1-y^2)^3}} \text{d} y

\displaystyle = \int_0^{x/\sqrt{1+x^2}} \frac{1}{\sqrt{1-y^2}} \text{d} y = s\left( \frac{x}{\sqrt{1+x^2}} \right) .

Again, choose x = 1 and result follows.

(iii)

\displaystyle z = \frac{u + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}} u}

\displaystyle \Rightarrow u = \frac{z - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}} z}

(detail omitted!)

\displaystyle \Rightarrow \frac{\text{d} u}{\text{d} z} = \frac{(1 + \frac{1}{\sqrt{3}} z) - (z - \frac{1}{\sqrt{3}})(\frac{1}{ \sqrt{3}})}{(1 + \frac{1}{\sqrt{3}} z)^2} = \frac{4/3}{{(1 + \frac{1}{\sqrt{3}} z)^2}}

\displaystyle t(x) = \int_{u=0}^{u=x} \frac{1}{1 + \left( \frac{z - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}} z} \right)^2} \cdot \frac{4/3}{{(1 + \frac{1}{\sqrt{3}} z)^2}} \text{d} z

\displaystyle = \int_{u=0}^{u=x} \frac{4/3}{(1 + \frac{1}{\sqrt{3}} z)^2 + (z - \frac{1}{\sqrt{3}} )^2} \text{d} z

\displaystyle = \int_{u=0}^{u=x} \frac{4/3}{\frac{4}{3} (1 + z^2)} \text{d} z

\displaystyle = \int_{1/\sqrt{3}}^{\alpha (x)} \frac{1}{1+z^2} \text{d} z

where

\displaystyle \alpha (x) = \frac{x + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}} x} .

Then

\displaystyle \alpha \left( \frac{1}{\sqrt{3}} \right) = \sqrt{3}

and so

\displaystyle t\left( \frac{1}{\sqrt{3}} \right) = \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{1}{1+z^2} \text{d} z.

\displaystyle 3t\left( \frac{1}{\sqrt{3}} \right) = \int_0^{1/\sqrt{3}} \frac{1}{1+u^2} \text{d} u + \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{1}{1+z^2} \text{d} z + \int_{\sqrt{3}}^\infty \frac{1}{1+v^2} \text{d} v

\displaystyle = \int_0^\infty \frac{1}{1+u^2} \text{d} u = \frac{1}{2} p.

Reply 10

STEP I Question 3

cos^4\theta - sin^4\theta = (cos^2\theta + sin^2\theta)(cos^2\theta - sin^2\theta) = cos2\theta

cos^4\theta + sin^4\theta = \frac{1}{4}(cos^22\theta + 2cos2\theta + 1) + \frac{1}{4}(cos^22\theta - 2cos2\theta +1)

= \frac{1}{2}cos^22\theta + \frac{1}{2} = \frac{1}{2}(1-sin^22\theta) + \frac{1}{2}

= 1 - \frac{1}{2}sin^22\theta

Right this stinks of I + J adding and substracting these integrals using the identities. But scrap that theres sin^6x so im going for reduction formula.

Unparseable latex formula:

I_n = \displaystyle \int_0^{\frac{\pi}{2}} sin^n\theta \hspace5 d\theta

Unparseable latex formula:

I_n = \displaystyle \int sin^{n-1}\theta sin \theta \hspace5 d\theta

By parts

\displaystyle u = sin^{n-1}\theta

\displaystyle \frac{du}{dx} = (n-1)sin^{n-2}\theta cos\theta

\displaystyle \frac{dv}{dx} = sin\theta

\displaystyle v = -cos\theta

Unparseable latex formula:

\displaystyle I_n = [-sin^{n-1}\theta cos\theta]_0^{\frac{\pi}{2}} - \int_0^{\frac{pi}{2}} -(n-1)sin^{n-2}\theta cos^2\theta \hspace5 d\theta

Unparseable latex formula:

\displaystyle = (n-1)\int_0^{\frac{\pi}{2}}sin^{n-2}\theta (1-sin^2\theta) \hspace5 d\theta

Unparseable latex formula:

\displaystyle = (n-1)\int_0^{\frac{\pi}{2}}sin^{n-2}\theta - sin^n\theta \hspace5 d\theta

\displaystyle I_n = (n-1)(I_{n-2} - I_n)

\displaystyle I_n = nI_{n-2} - nI_n - I_{n-2} + I_n

\displaystyle nI_n = (n-1)I_{n-2}

\displaystyle I_n = \frac{n-1}{n} I_{n-2}

Let

Unparseable latex formula:

\displaystyle I_n = \int_0^{\frac{\pi}{2}} cos^n\theta \hspace5 d\theta = \int_0^{\frac{\pi}{2}} cos^{n-1}\theta cos\theta \hspace5 d\theta

\displaystyle u = cos^{n-1}\theta

\displaystyle \frac{du}{dx} = -(n-1)cos^{n-2}sin\theta

\displaystyle \frac{dv}{dx} = cos \theta

\displaystyle v = sin\theta

Unparseable latex formula:

\displaystyle I_n = [cos^{n-1}\theta sin\theta]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} -(n-1)cos^{n-2}sin^2\theta \hspace5 d\theta

Unparseable latex formula:

\displaystyle I_n = (n-1)\int_0^{\frac{\pi}{2}} cos^{n-2}\theta(1-cos^2\theta) \hspace5 d\theta

Unparseable latex formula:

\displaystyle I_n = (n-1)\int_0^{\frac{\pi}{2}} cos^{n-2}\theta - cos^n\theta \hspace5 d\theta

\displaystyle I_n = (n-1)(I_{n-2} - I_n)

\displaystyle I_n = nI_{n-2} - nI_n - I_{n-2} + I_n

\displaystyle nI_n = (n-1)I_{n-2}

\displaystyle I_n = \frac{(n-1)}{n} I_{n-2}

Unparseable latex formula:

\displaystyle \int_0^{\frac{\pi}{2}} sin^4\theta \hspace5 d\theta

\displaystyle I_4 = \frac{3}{4}I_2

\displaystyle I_2 = \frac{1}{2}I_0

\displaystyle I_0 = \frac{\pi}{2}

\displaystyle I_4 = \frac{3}{4}.\frac{1}{2}.\frac{\pi}{2}

\displaystyle I_4 = \frac{3\pi}{16}

Unparseable latex formula:

\displaystyle \int_0^{\frac{\pi}{2}} cos^4\theta \hspace5 d\theta

\displaystyle I_4 = \frac{3}{4}I_2

\displaystyle I_2 = \frac{1}{2}I_0

\displaystyle I_0 = \frac{\pi}{2}

\displaystyle I_4 = \frac{3}{4}.\frac{1}{2}.\frac{\pi}{2}

\displaystyle I_4 = \frac{3\pi}{16}

Unparseable latex formula:

\displaystyle \int_0^{\frac{\pi}{2}} sin^6\theta \hspace5 d\theta

\displaystyle I_6 = \frac{5}{6}I_4

\displaystyle I_4 = \frac{3}{4}I_2

I_2 = \frac{1}{2}I_0

I_0 = \frac{\pi}{2}

I_6 = \frac{5}{6}.\frac{3}{4}.\frac{1}{2}.\frac{\pi}{2}

I_6 = \frac{5\pi}{32}

Unparseable latex formula:

\displaystyle \int_0^{\frac{\pi}{2}} cos^6\theta \hspace5 d\theta

\displaystyle I_6 = \frac{5}{6}I_4

\displaystyle I_4 = \frac{3}{4}I_2

I_2 = \frac{1}{2}I_0

I_0 = \frac{\pi}{2}

I_6 = \frac{5}{6}.\frac{3}{4}.\frac{1}{2}.\frac{\pi}{2}

I_6 = \frac{5\pi}{32}

Reply 11

generalebriety

16

I'm also claiming III/8. :wink:

Might be my last contribution of the night then. Urgh, this will be messy to latex.

(i) Putting u = x:

0 + \text{a} (x) + x\text{b} (x) = 0

Putting u = exp(-x):

1 - \text{a} (x) + \text{b} (x) = 0

Adding:

1 + (x+1)\text{b} (x) = 0 \Rightarrow \text{b} (x) = \frac{-1}{x+1}

\therefore \text{a} (x) = \frac{x}{x+1}

General solution:

u = Ax + Be^{-x} .

\displaystyle \frac{\text{d} y}{\text{d} x} + 3y^2 + \frac{x}{1+x} y = \frac{1}{3(1+x)} .

\displaystyle y = \frac{1}{3u} \frac{\text{d} u}{\text{d} x} \Rightarrow y^2 = \frac{1}{9u^2} \left(\frac{\text{d} u}{\text{d} x}\right)^2, \quad \frac{\text{d} y}{\text{d} x} = \frac{-1}{3u^2}\frac{\text{d} u}{\text{d} x} + \frac{1}{3u} \frac{\text{d}^2 u}{\text{d} x^2}

Applying substitution,

\displaystyle \frac{-1}{3u^2}\left(\frac{\text{d} u}{\text{d} x}\right)^2 + \frac{1}{3u} \frac{\text{d}^2 u}{\text{d} x^2} + \frac{1}{3u^2} \left(\frac{\text{d} u}{\text{d} x}\right)^2 + \frac{x}{1+x} \frac{1}{3u} \frac{\text{d} u}{\text{d} x} = \frac{1}{3(1+x)}

\displaystyle \frac{\text{d}^2 u}{\text{d} x^2} + \frac{x}{1+x} \frac{\text{d} u}{\text{d} x} - \frac{1}{1+x} u = 0

which has general solution

u = Ax + Be^{-x} ,

i.e.

\displaystyle \frac{1}{3u} = \frac{1}{3(Ax + Be^{-x})} ,

\displaystyle \frac{\text{d} u}{\text{d} x} = A - Be^{-x} ,

\displaystyle y = \frac{A - Be^{-x}}{3(Ax + Be^{-x})} .

y(0) = 0:

\displaystyle 0 = A - B \Rightarrow A = B

\displaystyle \therefore y = \frac{A - Ae^{-x}}{3(Ax + Ae^{-x})} = \frac{1 - e^{-x}}{3(x + e^{-x})}.

(ii)

\displaystyle \frac{\text{d} y}{\text{d} x} + y^2 + \frac{x}{1-x} y = \frac{1}{1-x} .

Applying the substitution

\displaystyle y = \frac{1}{u} \frac{\text{d} u}{\text{d} x}

gives (detail omitted):

\displaystyle \frac{\text{d}^2u}{\text{d} x^2} + \frac{x}{1-x} \frac{\text{d} u}{\text{d} x} - \frac{1}{1-x} u = 0

to which "obvious" solutions are u = x, u = exp x, and so the general solution is

u = Ax + Be^x.

This gives:

\displaystyle \frac{1}{u} = \frac{1}{Ax + Be^x} ,\quad \frac{\text{d} u}{\text{d} x} = A + Be^x

\displaystyle \therefore y = \frac{A + Be^x}{Ax + Be^x} .

y(0) = 2:

\displaystyle 2 = \frac{A + B}{B} \Rightarrow A = B

\displaystyle \therefore y = \frac{A + Ae^x}{Ax + Ae^x} = \frac{1 + e^x}{x + e^x} .

Reply 12

calcium878

18

Alternative I/3. As inspirato for the beginning.

insparato

STEP I Question 3

cos^4\theta - sin^4\theta = (cos^2\theta + sin^2\theta)(cos^2\theta - sin^2\theta) = cos2\theta

(*)

cos^4\theta + sin^4\theta = \frac{1}{4}(cos^22\theta + 2cos2\theta + 1) + \frac{1}{4}(cos^22\theta - 2cos2\theta +1)

= \frac{1}{2}cos^22\theta + \frac{1}{2} = \frac{1}{2}(1-sin^22\theta) + \frac{1}{2}

= 1 - \frac{1}{2}sin^22\theta

(**)

Now, (*) + (**) gives

2cos^4\theta = cos2\theta + 1 - \frac{1}{2}sin^22\theta

so that

cos^4\theta = \frac{1}{2}(cos2\theta + 1 - \frac{1}{2}sin^2 2\theta)

And, (*)-(**) gives

sin^4\theta = \frac{1}{2}(1 - \frac{1}{2}sin^2 2\theta - cos2\theta)

So,

\displaystyle \int_0^{\frac{\pi}{2}} cos^4 \theta \text{d}\theta = \frac{1}{2} \int_0^{\frac{\pi}{2}} cos2\theta + 1 - \frac{1}{2}sin^2 2\theta \text{d}\theta

-\frac{1}{2}sin^2 2\theta = \frac{1}{4}cos4 \theta - \frac{1}{4}

So that

\displaystyle \frac{1}{2} \int_0^{\frac{\pi}{2}} cos2\theta + \frac{1}{4}cos4\theta + \frac{3}{4} = \frac{1}{2} \left[\frac{1}{2}sin2\theta + \frac{3}{4}\theta + \frac{1}{16}sin4\theta\right]_0^{\frac{\pi}{2}}

= \frac{1}{2}([\frac{3\pi}{8}]-[0]) = \frac{3}{16}\pi

The one with

sin^4 \theta

is very nearly the same, just a sign change here and there (skipping the intermediate rearrangements):

\displaystyle \int_0^{\frac{\pi}{2}} sin^4 \theta \text{d}\theta = \frac{1}{2} \int_0^{\frac{\pi}{2}} -cos2\theta + \frac{1}{4}cos4\theta + \frac{3}{4} = \frac{1}{2} \left[-\frac{1}{2}sin2\theta + \frac{3}{4}\theta + \frac{1}{16}sin4\theta\right]_0^{\frac{\pi}{2}}

= \frac{1}{2}([\frac{3\pi}{8}]-[0]) = \frac{3}{16}\pi

Then a traditional mash through the trig for the 6-th power ones, which I won't type up :p:

Reply 13

Yep i did it that way originally on paper. But i thought im going to have to do reduction formula anyways so might as well follow through. Btw I was going to post that but the Latex was a bitch anyways so i couldnt be bothered lol.

Reply 14

calcium878

18

I'm still 4 chapters away from the reduction formula in FP2 ;dry; :frown:

Reply 15