Author Topic: 4U Maths Question Thread (Read 665180 times) Tweet Share

katherine123 · « **Reply #195 on:** May 08, 2016, 02:00:59 pm »

how to solve these
1.integral of x*inverse sin^-1(x^2)
2.integral of ln[x+(x^2+1)^1/2] after proving d/dx[ln[x+(x^2+a^2)] = 1/[(x^2+a^2)^1/2]

jakesilove · « **Reply #196 on:** May 08, 2016, 02:19:16 pm »

Quote from: katherine123 on May 08, 2016, 02:00:59 pm

how to solve these
1.integral of x*inverse sin^-1(x^2)
2.integral of ln[x+(x^2+1)^1/2] after proving d/dx[ln[x+(x^2+a^2)] = 1/[(x^2+a^2)^1/2]

Hey Katherine!

Below is my solution to the first question (Will get to the second in a sec). This is just a straightforward application of integration by parts; it's an important method to learn, as its very standard in the extension 2 course.

Jake

jakesilove · « **Reply #197 on:** May 08, 2016, 02:39:29 pm »

Quote from: katherine123 on May 08, 2016, 02:00:59 pm

how to solve these
1.integral of x*inverse sin^-1(x^2)
2.integral of ln[x+(x^2+1)^1/2] after proving d/dx[ln[x+(x^2+a^2)] = 1/[(x^2+a^2)^1/2]

Here is my solution to the second question!

Jake

amandali · « **Reply #198 on:** May 09, 2016, 11:01:18 pm »

1st ques: im asked to find the integral of (x^2+4)^1/2 and i used the substitution x=2tanB which gave me integral of 4*sec^3(B)

2nd ques: integral of e^x*sin(4x)cos(2x) with no limits

i changed the above integral to e^x*[sin(2x)+sin(6x)] using trig property

answer for ques 2 = e^x/370*[37sin(2x)+5sin6x-74cos(2x)-30cos(6x)] +C

RuiAce · « **Reply #199 on:** May 10, 2016, 09:18:36 am »

Quote from: amandali on May 09, 2016, 11:01:18 pm

1st ques: im asked to find the integral of (x^2+4)^1/2 and i used the substitution x=2tanB which gave me integral of 4*sec^3(B)

2nd ques: integral of e^x*sin(4x)cos(2x) with no limits

i changed the above integral to e^x*[sin(2x)+sin(6x)] using trig property

answer for ques 2 = e^x/370*[37sin(2x)+5sin6x-74cos(2x)-30cos(6x)] +C

$\text{That integral is notoriously difficult for MX2 students as hyperbolic substitutions are not allowed.}\\ \text{To integrate the classic secant-cubed:}$
$\begin{align*}I&=\int{\sec{\theta}\sec^2{\theta}\,d\theta}\\&\stackrel{IBP}{=}\tan{\theta}\sec{\theta}-\int{\tan{\theta}\cdot\sec{\theta}\tan{\theta}\,d\theta}\\&\stackrel{A}{=}\tan{\theta}\sec{\theta}-\int{\sec{\theta}(\sec^2{\theta}-1)d\theta}\\&=\tan{\theta}\sec{\theta}-\int{\sec^3{\theta}\,d\theta}+\int{\sec{\theta}\,d\theta}\\&=\tan{\theta}\sec{\theta}-I+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\\ \Leftrightarrow 2I&=\tan{\theta}\sec{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}+C\end{align*}\\ \text{as statement is true to within a constant}$
$\\\text{A - Pythagorean identity}$
$\text{Then all we must realise is that }\\ \tan{\theta}=\frac{x}{2}\\ \sec{\theta}=\sqrt{1+\tan^2{\theta}}=\sqrt{1+\frac{x^2}{4}}\\ \text{Note: Drawing a triangle is also acceptable.}$

$\text{For question 2, you probably had the right idea but you made a mistake. You dropped the one half.}\\ \sin{4x}\cos{2x}=\frac{1}{2}(\sin{6x}+\sin{2x})$

katherine123 · « **Reply #200 on:** May 12, 2016, 12:02:22 pm »

question 1 : integral of (x^2+2x+2)^1/2

i used integration by parts which gives me
[x*(x^2+2x+2)^1/2] - integral of (x^2+x)/(x^2+2x+2)^1/2

im not sure how to integrate this part (x^2+x)/(x^2+2x+2)^1/2

question2 : integral of [x*(x+1)]^1/2

RuiAce · « **Reply #201 on:** May 12, 2016, 06:48:43 pm »

I've already done the integral of secant cubed previously

RuiAce · « **Reply #202 on:** May 12, 2016, 06:51:31 pm »

These are really just unnecessarily tedious integrals if you ask me

jamonwindeyer · « **Reply #203 on:** May 12, 2016, 07:13:35 pm »

Quote from: RuiAce on May 12, 2016, 06:51:31 pm

These are really just unnecessarily tedious integrals if you ask me(Image removed from quote.)

RuiAce · « **Reply #204 on:** May 12, 2016, 08:18:58 pm »

Quote from: jamonwindeyer on May 12, 2016, 07:13:35 pm

(Image removed from quote.)

Are you ok Jamon

HSC integrals - only partial fractions are tedious

(Except for I think 2014 when the last question of the paper was a huge surprise)

jakesilove · « **Reply #205 on:** May 12, 2016, 08:52:25 pm »

Quote from: RuiAce on May 12, 2016, 08:18:58 pm

Are you ok Jamon

HSC integrals - only partial fractions are tedious

(Except for I think 2014 when the last question of the paper was a huge surprise)

As a Class of 2014 graduate, I concur

RuiAce · « **Reply #206 on:** May 12, 2016, 09:27:52 pm »

Quote from: jakesilove on May 12, 2016, 08:52:25 pm

As a Class of 2014 graduate, I concur

To be fair that one was a huge surprise. Here's the question done in two methods for anybody interested whilst I'm here

$\text{This was the last question:}\\ \int{\frac{\ln{x}}{{\left(1+\ln{x}\right)}^{2}}dx}$

$\text{The too-much-of-a-genius method: Reverse the QUOTIENT rule}\\ \begin{align*}\int{\frac{\ln{x}}{{\left(1+\ln{x}\right)}^{2}}dx}&=\int{\frac{1+\ln{x}-1}{{\left(1+\ln{x}\right)}^{2}}dx}\\&=\int{\frac{\left(1+\ln{x}\right)-x\cdot\frac{1}{x}}{{\left(1+\ln{x}\right)}^{2}}dx}\\&=\frac{x}{1+\ln{x}}+C\end{align*}$

$\text{The standard method:}\\ \text{Let }x=e^u \Rightarrow dx=e^u\, du\\ \begin{align*}\int{\frac{\ln{x}}{{\left(1+\ln{x}\right)}^{2}}dx}&=\int{\frac{ue^u}{{(1+u)}^2}du}\\&=-\frac{ue^u}{1+u}+\int{\frac{(1+u)e^u}{1+u}du} \quad \text{ sneakiest by parts ever} \\&=-\frac{x\ln{x}}{1+\ln{x}}+e^u+C \end{align*}$
And the rest is easy to finish off.

The IBP was ridiculous dodgy as you had to pick the one to be differentiated: d/du ue^u = (u+1)e^u

jamonwindeyer · « **Reply #207 on:** May 13, 2016, 11:43:12 am »

Quote from: jakesilove on May 12, 2016, 08:52:25 pm

As a Class of 2014 graduate, I concur

Though I didn't take MX2, I actually heard about that question. It's interesting because in my opinion I actually think the MX1 paper for 2014 was a little easier than the standard, or maybe I just got lucky with the subjects

RuiAce · « **Reply #208 on:** May 13, 2016, 01:24:45 pm »

Quote from: jamonwindeyer on May 13, 2016, 11:43:12 am

Though I didn't take MX2, I actually heard about that question. It's interesting because in my opinion I actually think the MX1 paper for 2014 was a little easier than the standard, or maybe I just got lucky with the subjects

That's cause it was easier

amandali · « **Reply #209 on:** May 15, 2016, 12:17:37 am »

im stuck after making sin^n(x) to sin^n-2(x)sin^2(x)

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