Author Topic: 4U Maths Question Thread (Read 665489 times) Tweet Share

hanaacdr · « **Reply #750 on:** January 27, 2017, 07:39:08 am »

Quote from: RuiAce on January 26, 2017, 11:07:19 pm

$\text{Let }u=e^x\implies du = e^xdx$
$\begin{align*}\int \frac{e^x+e^{2x}}{1+e^{2x}}dx&= \int \frac{e^x(1+e^x)}{1+e^{2x}}dx\\ &= \int \frac{1+u}{1+u^2}du\\ &= \int \frac{1}{1+u^2}du + \int \frac{u}{1+u^2}du\\ &= \tan^{-1}u + \frac{1}{2}\ln (1+u^2)+C\\ &= \tan^{-1}(e^x) + \frac{1}{2}\ln (1+e^{2x})+C\end{align*}$

Thank you!

hanaacdr · « **Reply #751 on:** January 27, 2017, 09:19:07 am »

Hi
could please help me out with these questions

thank you so much!
really appreciate it!

RuiAce · « **Reply #752 on:** January 27, 2017, 09:42:13 am »

Quote from: hanaacdr on January 27, 2017, 09:19:07 am

Hi
could please help me out with these questions

thank you so much!
really appreciate it!

$x=\cos 2\theta \implies dx = -\sin 2\theta\, d\theta\\ \text{and we should consider some double angle expansions}$
$\begin{align*}\int \sqrt{\frac{1+x}{1-x}}dx&= -2\int \sqrt{\frac{1+\cos 2\theta}{1-\cos 2\theta}}\sin 2\theta\, d\theta\\ &= -4\int\sqrt{\frac{2\cos^2\theta}{2\sin^2\theta}}\sin \theta \cos \theta\, d\theta\\ &= -4\int \cos^2\theta \, d\theta\\ &= -2 \int \left(\frac12+\frac12\cos 2\theta\right)d\theta\tag{3U trick}\\ &= -2\theta-\sin 2\theta + C\\ &=-\cos^{-1}x-\sqrt{1-x^2}+C \end{align*}\\ \text{Subject to some little mistakes, haven't checked}$
Q9 shouldn't be as hard, unless it gets icky. Post up any progress in working.

hanaacdr · « **Reply #753 on:** January 27, 2017, 10:39:40 am »

Quote from: RuiAce on January 27, 2017, 09:42:13 am

$x=\cos 2\theta \implies dx = -\sin 2\theta\, d\theta\\ \text{and we should consider some double angle expansions}$
$\begin{align*}\int \sqrt{\frac{1+x}{1-x}}dx&= -2\int \sqrt{\frac{1+\cos 2\theta}{1-\cos 2\theta}}\sin 2\theta\, d\theta\\ &= -4\int\sqrt{\frac{2\cos^2\theta}{2\sin^2\theta}}\sin \theta \cos \theta\, d\theta\\ &= -4\int \cos^2\theta \, d\theta\\ &= -2 \int \left(\frac12+\frac12\cos 2\theta\right)d\theta\tag{3U trick}\\ &= -2\theta-\sin 2\theta + C\\ &=-\cos^{-1}x-\sqrt{1-x^2}+C \end{align*}\\ \text{Subject to some little mistakes, haven't checked}$
Q9 shouldn't be as hard, unless it gets icky. Post up any progress in working.

this really helps!
thank you!

hanaacdr · « **Reply #754 on:** January 27, 2017, 10:51:35 am »

Quote from: hanaacdr on January 27, 2017, 10:39:40 am

this really helps!
thank you!

for question 9,
i keep getting the wrong answer
i am not sure what im doing wrong

thanks

hanaacdr · « **Reply #755 on:** January 27, 2017, 11:51:10 am »

could you also help me with this question
i have no idea how to do it,
thank you!

de · « **Reply #756 on:** January 27, 2017, 11:53:56 am »

Quote from: hanaacdr on January 27, 2017, 10:51:35 am

for question 9,
i keep getting the wrong answer
i am not sure what im doing wrong

thanks

I assume you got to the point where you had $\int \frac1{4sin^2( \theta)} d \theta$ from there ignore the 4 obviously and try dividing the numerator and the denominator by cos^2.
Then you're in the position to make another substitution.

de · « **Reply #757 on:** January 27, 2017, 11:59:41 am »

Quote from: hanaacdr on January 27, 2017, 11:51:10 am

could you also help me with this question
i have no idea how to do it,
thank you!

For this $\int_0^{\pi/3} \frac1{1- \sin{x}}dx=\int_0^{\pi/3} \frac{1+\sin{x}}{1- \sin^2{x}}dx=\int_0^{\pi/3} \frac{1}{\cos^2{x}}+\frac{\sin{x}}{\cos^2{x}}dx$

The second part is straightforward. The first try a similar trick to what I recommended before (divide numerator and denominator by sin^2).

RuiAce · « **Reply #758 on:** January 27, 2017, 12:00:00 pm »

Quote from: de on January 27, 2017, 11:53:56 am

I assume you got to the point where you had $\int \frac1{4sin^2( \theta)} d \theta$ from there ignore the 4 obviously and try dividing the numerator and the denominator by cos^2.
Then you're in the position to make another substitution.

$\text{Assuming that is correct}\\ \int \frac{1}{4\sin^2\theta}d\theta=\frac14\int \text{cosec}^2\,\theta\,d\theta = -\frac{1}{4}\cot \theta+C\\ \text{No substitution required.}$
The integral of cosec-squared is known, and is negative cot. Just like how the integral of sec-squared is tan.

de · « **Reply #759 on:** January 27, 2017, 12:02:23 pm »

Quote from: RuiAce on January 27, 2017, 12:00:00 pm

$\text{Assuming that is correct}\\ \int \frac{1}{4\sin^2\theta}d\theta=\frac14\int \text{cosec}^2\,\theta\,d\theta = -\frac{1}{4}\cot \theta+C\\ \text{No substitution required.}$
The integral of cosec-squared is known, and is negative cot. Just like how the integral of sec-squared is tan.

Didn't know what was assumed or not in HSC, apologies!

RuiAce · « **Reply #760 on:** January 27, 2017, 12:07:16 pm »

Quote from: hanaacdr on January 27, 2017, 11:51:10 am

could you also help me with this question
i have no idea how to do it,
thank you!

In general, you have to make the substitution t=tan(x/2)

Because of how that's written, i.e. a + b sin(x) where a and b are either 1 or -1, you can manipulate the Pythagorean trigonometric identity that de used.

Quote from: de on January 27, 2017, 12:02:23 pm

Didn't know what was assumed or not in HSC, apologies!

All g

hanaacdr · « **Reply #761 on:** January 27, 2017, 01:37:57 pm »

Quote from: RuiAce on January 27, 2017, 12:07:16 pm

In general, you have to make the substitution t=tan(x/2)

Because of how that's written, i.e. a + b sin(x) where a and b are either 1 or -1, you can manipulate the Pythagorean trigonometric identity that de used.All g

thank you so much!

hanaacdr · « **Reply #762 on:** January 27, 2017, 01:38:47 pm »

Quote from: de on January 27, 2017, 12:02:23 pm

Didn't know what was assumed or not in HSC, apologies!

thank youuu
much appreciated!

cmadeleine · « **Reply #763 on:** January 27, 2017, 11:42:29 pm »

Hi! Could you please help me with this question: find the Cartesian equation of the locus of arg(z-4)-arg(z+4) = pi/4?

Thank you!

RuiAce · « **Reply #764 on:** January 27, 2017, 11:59:55 pm »

Quote from: cmadeleine on January 27, 2017, 11:42:29 pm

Hi! Could you please help me with this question: find the Cartesian equation of the locus of arg(z-4)-arg(z+4) = pi/4?

Thank you!

For now, I am going to redirect you to the discussion here.

Use the resources I have linked to.

Search the forums now!

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