HSC 2005 Q4 a (1 Viewer)

Carl5 · Sep 30, 2011

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I don't think I understand the step between 1 and 2.

Could someone please explain it?

SpiralFlex · Sep 30, 2011

From step 1. You can clearly write,

$\cos x = \frac{\mathrm{d} }{\mathrm{d} x}\: \sin x$

$\int_{0}^{\frac{\pi}{4}} \frac{\mathrm{d} }{\mathrm{d} x}\: \sin x * \sin^2 x\: {\mathrm{d} x}$

$\int_{0}^{\frac{\pi}{4}} \sin^2 x \: \mathrm{d} (\sin x)$

So in when we are integrating like this, we are integrating with respect to $\sin x$ . Note: If were normally integrating with respect to $x$ we would not "simply add one to the power and divide by the new power." Since we're in respect with $\sin x$ we can do so at our will.

$=\frac{(\sin x)^3}{3}$ With your limits. I am sure you can do it from here.

AAEldar · Sep 30, 2011

Carl5 said:
I don't think I understand the step between 1 and 2.

Could someone please explain it?

$\int_0^\frac{\pi}{4}cosx sin^2x dx \\\\ u=sinx\;\;\;\;\;du=cosxdx\;\;\;\;\;\;x=0,\;u=0\;\;\;\;x=\frac{\pi}{4},\;u=\frac{1}{\sqrt2} \\\\ =\int_0^\frac{1}{\sqrt2}u^2du \\\\ =\frac{1}{3}[u^3]_0^\frac{1}{\sqrt2} \\\\ =\frac{1}{3}[(\frac{1}{\sqrt2})^3-0] \\\\ =\frac{1}{3}(\frac{1}{2\sqrt2}) \\\\ =\frac{\sqrt2}{12}$

Ohhh beaten by Spiral.

TheCardician · Sep 30, 2011

In my opinion, it would seem easier to let u = sinx then du = cosxdx, and so on and so forth, but it is a tad longer. But I do see the logic in this solution now ^^ and it's quick. So I suppose it's a matter of preference on what method you use. Good explanation Spiral

taeyang · Sep 30, 2011

Never seen THAT before :O that's so cool, I'de just do ye ol'e

$$,Let u = sinx$\\\\ $When x = pi/4, u = 1/\sqrt{2}$\\ $When x = 0, u = 0\\\\ \therefore \frac{du}{dx} = cosx\\\\ \therefore du = dxcosx\\\\ \therefore \int_{0}^{\frac{1}{\sqrt{2}}} u^{2}du\\\\\\\\ Blah$

(Don't mind me, I'm just practising my latex ^.^)

Edit...
Why are you people so fast...

AAEldar · Sep 30, 2011

taeyang said:
Never seen THAT before :O that's so cool, I'de just do ye ol'e

$$,Let u = sinx$\\\\ $When x = pi/4, u = 1/\sqrt{2}$\\ $When x = 0, u = 0\\\\ \therefore \frac{du}{dx} = cosx\\\\ \therefore du = dxcosx\\\\ \therefore \int_{0}^{\frac{1}{\sqrt{2}}} u^{2}du\\\\\\\\ Blah$

(Don't mind me, I'm just practising my latex ^.^)

Edit...
Why are you people so fast...

U sub is great.

SpiralFlex · Sep 30, 2011

Note: HSCers should be studying right now.

bleakarcher · Sep 30, 2011

aaeldar made a mistake, otherwise is wrong.
integral[f(x)*f '(x)] dx=(1/2)[f(x)]^2+C

AAEldar · Sep 30, 2011

bleakarcher said:
aaeldar made a mistake, otherwise is wrong.
integral[f(x)*f '(x)] dx=(1/2)[f(x)]^2+C

LOL how embarrassing for me.

Disregard my post D:

Drongoski · Sep 30, 2011

AAEldar said:
LOL how embarrassing for me.

Disregard my post D:

Don't worry - we all make a slip now and then; that makes us all human.

Carl5 · Sep 30, 2011

SpiralFlex said:
From step 1. You can clearly write,

$\cos x = \frac{\mathrm{d} }{\mathrm{d} x}\: \sin x$

$\int_{0}^{\frac{\pi}{4}} \frac{\mathrm{d} }{\mathrm{d} x}\: \sin x * \sin^2 x\: {\mathrm{d} x}$

$\int_{0}^{\frac{\pi}{4}} \sin^2 x \: \mathrm{d} (\sin x)$

So in when we are integrating like this, we are integrating with respect to $\sin x$ . Note: If were normally integrating with respect to $x$ we would not "simply add one to the power and divide by the new power." Since we're in respect with $\sin x$ we can do so at our will.

$=\frac{(\sin x)^3}{3}$ With your limits. I am sure you can do it from here.

I still don't understand why the d(sinx) has disappeared. Also, why isn't the integration of Sin(x)^3 all fancy like the integration of Sin(x)^2 haha?

EDIT: What exactly is d(sinx) now when there isn't the 1/dx in there?

And yeah, I usually do the substitution method, but this seemed too interesting to ignore haha

SpiralFlex · Sep 30, 2011

Carl5 said:
I still don't understand why the d(sinx) has disappeared. Also, why isn't the integration of Sin(x)^3 all fancy like the integration of Sin(x)^2 haha?

EDIT: What exactly is d(sinx) now when there isn't the 1/dx in there?

And yeah, I usually do the substitution method, but this seemed too interesting to ignore haha

It's like treating it as a variable. I just cancelled the $dx$ . I can integrate with respect to $\sin x$ .

It's like this,

$dx$ the $x$ has been replaced by $\sin x$ .

Carl5 · Sep 30, 2011

is d(sinx) just (sinx)?

Also, I think what I don't understand is how you're integrating both sin^2x and dsinx... I'm just really confused, sorry.

AAEldar · Sep 30, 2011

Carl5 said:
I still don't understand why the d(sinx) has disappeared. Also, why isn't the integration of Sin(x)^3 all fancy like the integration of Sin(x)^2 haha?

EDIT: What exactly is d(sinx) now when there isn't the 1/dx in there?

And yeah, I usually do the substitution method, but this seemed too interesting to ignore haha

$\frac{d}{dx}sinx$ means the derivative of sinx with respect to x, which is cosx. So $\frac{d}{dx}sinx=cosx$ - they're the same thing.

If the integral was just $\int sin^3x dx$ then it would be 'fancy', but because it's $\int cosx sin^2x dx$ , where the derivative of the function is there, we can use a substitution or just the general rule that $\int f'(x) f(x)dx=\frac{1}{2}[f(x)]^2$ .

Carl5 · Oct 1, 2011

AAEldar said:
$\frac{d}{dx}sinx$ means the derivative of sinx with respect to x, which is cosx. So $\frac{d}{dx}sinx=cosx$ - they're the same thing.

If the integral was just $\int sin^3x dx$ then it would be 'fancy', but because it's $\int cosx sin^2x dx$ , where the derivative of the function is there, we can use a substitution or just the general rule that $\int f'(x) f(x)dx=\frac{1}{2}[f(x)]^2$ .

I understand that d/dx(sinx) is cosx but I don't understand what it is when it becomes d(sinx) after multiplying by dx,

Carl5 · Oct 1, 2011

basically, I don't understand

$\int_{0}^{\frac{\pi}{4}} \sin^2 x \: \mathrm{d} (\sin x)$

How this step works. What happens to the d(sinx). I see that we've integrated sin^2(x) (or at least, that's what I think you've done), but then what happened to d(sinx). Does it get cancelled out by something or what?

AAEldar · Oct 1, 2011

Carl5 said:
basically, I don't understand

$\int_{0}^{\frac{\pi}{4}} \sin^2 x \: \mathrm{d} (\sin x)$

How this step works. What happens to the d(sinx). I see that we've integrated sin^2(x) (or at least, that's what I think you've done), but then what happened to d(sinx). Does it get cancelled out by something or what?

If you look at it closely it's not just $dsinxsin^2xdx$ , but rather it's $\frac{d}{dx}sinx\;sin^2x\;dx$ . There is no new multiplication happening but rather where the $cosx$ was in the original question, they have substituted in $\frac{d}{dx}sinx$ . By doing this it is showing that the integral is in the form of $\int f'(x) f(x) dx=\frac{1}{2}[f(x)]^2$ .

So in this case, $f(x)=sinx$ and the derivative, $\frac{d}{dx}$ , is $cosx$ .

Carl5 · Oct 1, 2011

Ohhh. I get you now! Thank you so much for your help! I'll have to try to find some more similar questions in past exams to practice this considering I've never actually seen it in my text book...

I think what confused me was when you had 1/2[f(x)]^2 rather than something like 1/(n+1)[f(x)]^(n+1). But regardless, thank you for helping me despite how annoying it probably is when I keep asking the same thing when the answer is right in front of me!

AAEldar · Oct 1, 2011

Carl5 said:
Ohhh. I get you now! Thank you so much for your help! I'll have to try to find some more similar questions in past exams to practice this considering I've never actually seen it in my text book...

I think what confused me was when you had 1/2[f(x)]^2 rather than something like 1/(n+1)[f(x)]^(n+1). But regardless, thank you for helping me despite how annoying it probably is when I keep asking the same thing when the answer is right in front of me!

I'm just glad that you understand it now!

$\int_0^\frac{\pi}{4} 3sinx\;cos^2x\;dx$

$\int tan^2x\;sec^2x\;dx$

Try those.

taeyang · Oct 1, 2011

Now guys ^.^, please don't think I'm a faggot trying to "prove my math skills" in a sea of genius's haha, cause I realise I'm only doing ext1

, I am just trying my luck at lots of questions and building my Mathematics vocabulary (very slowly ^.^) (and attempting to get better at Latex.. pfft)

So then, Eldar! can you please tell me if these are right?

$$(1)........$\\ \int_{0}^{\frac{\pi}{4}}3sinxcos^{2}xdx\\ -3\int_{0}^{\frac{\pi}{4}}\frac{dcosx}{dx}cos^{2}xdx\\ -3\int_{0}^{\frac{\pi}{4}}cos^{2}x\:dcosx\\ -3\:[\frac{cos^{3}\frac{\pi}{4}}{3}-\frac{cos^{3}{0}}{3}]\\\\ \therefore = -3(\frac{1}{6\sqrt{2}}- \frac{1}{3})\\\\\\\\ $(2)........$\\ \int tan^{2}xsec^{2}dx\\ \int \frac{dtanx}{dx}tan^{2}xdx\\ \int tan^{2}x{dtanx}\\ \therefore \: = \: \frac{tan^3{x}}{3}+C$

HSC 2005 Q4 a (1 Viewer)

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