HSC 2005 Q4 a (1 Viewer)

SpiralFlex · Oct 1, 2011

taeyang said:
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$

Correct. You might want to put a "/" in your Latex code before your trigonometric functions.

someth1ng · Oct 1, 2011

I never knew that method was allowed...I would normally have used double angles (LOL).

SpiralFlex · Oct 1, 2011

someth1ng said:
I never knew that method was allowed...I would normally have used double angles (LOL).

Substitution for the win. It's way quicker than rewriting the integral.

someth1ng · Oct 1, 2011

This question was from 3U paper right?

SpiralFlex · Oct 1, 2011

someth1ng said:
This question was from 3U paper right?

Not too sure, must be I would think.

someth1ng · Oct 1, 2011

I'm on a computer without Adobe so I can't check =/

SpiralFlex · Oct 1, 2011

someth1ng said:
I'm on a computer without Adobe so I can't check =/

I can confirm it is now.

someth1ng · Oct 1, 2011

I can't seem to get it to work with double angles so sub it is

Carl5 · Oct 1, 2011

taeyang said:
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$

Sweet! I got the same! Thanks for the help everyone! I rather like this method.

SpiralFlex · Oct 1, 2011

someth1ng said:
I can't seem to get it to work with double angles so sub it is

If you want to over complicate it.

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

$=\int_{0}^{\frac{\pi}{4}}(\cos x (\frac{1-\cos 2x}{2}))\: {\mathrm{d} x}$

$=\int_{0}^{\frac{\pi}{4}}(\frac{1}{2}\cos x -\frac{1}{2}\cos x \cos 2x)\: {\mathrm{d} x}$

$=\frac{1}{2} \int_{0}^{\frac{\pi}{4}}(\cos x -\cos x \cos 2x)\: {\mathrm{d} x}$

$=\frac{1}{4} \int_{0}^{\frac{\pi}{4}}(2\cos x -2\cos x \cos 2x)\: {\mathrm{d} x}$

By products to sums,

$=\frac{1}{4} \int_{0}^{\frac{\pi}{4}}(2\cos x -(\cos x + \cos 3x)) {\mathrm{d} x}$

$=\frac{1}{4} \int_{0}^{\frac{\pi}{4}}(\cos x - \cos 3x) {\mathrm{d} x}$

$=\frac{1}{4}[\sin x - \frac{1}{3} \sin 3x]$ With the limits.

$=\frac{\sqrt{12}}{2}$

Drongoski · Oct 1, 2011

someth1ng · Oct 1, 2011

I'll take substitution.

SpiralFlex · Oct 1, 2011

someth1ng said:
I'll take substitution.

Wise choice.

someth1ng · Oct 1, 2011

Apparently you're supposed to do this in one line using common sense.

integral[sin^3(x)]=3sin^2(x)cos(x) and work from there...

SpiralFlex · Oct 1, 2011

someth1ng said:
Apparently you're supposed to do this in one line using common sense.

integral[sin^3(x)]=3sin^2(x)cos(x) and work from there...

What are you talking about? Who are you talking to?

someth1ng · Oct 1, 2011

wrong way, switch the eqn around LOL somethingg like that

SpiralFlex · Oct 1, 2011

someth1ng said:
wrong way, switch the eqn around LOL somethingg like that

I am still confused about who you are talking to and what you are talking about.

someth1ng · Oct 1, 2011

integral[3*sin^2(x)*cos(x)]=sin^3(x)+C
OR
d/dx [sinx]^3 = 3 (sinx)^2 (cosx)

So obviously if "integral[3*sin^2(x)*cos(x)]=sin^3(x)+C" then integrating sin^2(x)*cos(x)=[1/3](sinx)^3

That's better.

taeyang · Oct 1, 2011

someth1ng said:
integral[3*sin^2(x)*cos(x)]=sin^3(x)+C
OR
d/dx [sinx]^3 = 3 (sinx)^2 (cosx)

So obviously if "integral[3*sin^2(x)*cos(x)]=sin^3(x)+C" then integrating sin^2(x)*cos(x)=[1/3](sinx)^3

That's better.

You still didn't specify why you are actually posting this as spiral said.

someth1ng · Oct 1, 2011

"Apparently you're supposed to do this in one line using common sense."

The Excel Fast Track mini-book says that it can be done straight from memory.

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