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What subject does this biased teacher teach. If subjective subjects, like English or History, she/he can give much higher marks to those favoured.

In using Integration by Parts, I prefer using the form: \int udv = uv - \int vdu But I think whilst this form looks simpler & cleaner, it requires you to be good with the use of differentials, like: d(x+1), d(1 + ln x), d\sqrt{1-x^2} $ or $ dx = d(\sqrt x ^2) = 2\sqrt x\cdot d\sqrt x...

For Q24 can use: d\sqrt{1-x^2} = \frac {-x}{\sqrt{1-x^2}}dx \\ \\ I_{2n+1} = \int^-_- \frac {x^{2n+1}}{\sqrt{1-x^2}} \cdot dx = - \int ^-_- x^{2n}\frac {-x}{\sqrt{1-x^2}}\cdot dx = - \int ^-_- x^{2n} d\sqrt{1-x^2}\\ \\ = -\left [ x^{2n}\cdot \sqrt{1-x^2} \right ] ^1 _0 + \int ^-_- \sqrt...

I_n = \int ^1 _0 x^n \sqrt{1-x} dx = \frac {-2}{3}\int ^1 _0 x^n d(1-x)^\frac {3}{2} \\ \\ = \frac {-2}{3}\left[x^n (1-x)^{\frac {3}{2}} \right ]^1 _0 + \frac {2}{3}\int ^1 _0 (1-x)^\frac {3}{2} n \cdot x^{n-1}dx \\ \\ = 0 + \frac {2n}{3} \int ^1 _0 (1-x) \sqrt {1-x} \cdot x^{n-1} dx \\ \\ =...

Is this thread a promotional forum for Dr Du?

Is the angle nearly 82 deg?

Correct!

Using my special method (just another way of doing substitution type integrals): \int \frac {1}{1 + \sqrt x} dx = \int \frac {1}{1 + \sqrt x} d(\sqrt x)^2 = \int \frac {1}{1 + \sqrt x} \times 2\sqrt x d \sqrt x \\ \\ = 2\int (1 - \frac {1}{1 + \sqrt x}) d\sqrt x = 2\int 1 d\sqrt x - 2\int...

Without the formal process of substitution. Of course, if question asks you to use the method of substitution, then you must use the method. \int sin^2 x cos^3 x dx = \int sin^2 x \times cos^2 x \times cos x dx = \int sin^2x (1-sin^2x)dsinx \\ \\ = \int ((sin x)^2 - (sin x)^4) dsin x = \frac...

(cis \theta)^5 = cis 5 \theta \implies cos 5\theta + i sin 5\theta = (cos\theta +i sin \theta)^5 \equiv (c + is)^5 \\ \\ = c^5 +5ic^4s +10i^2c^3s^2 + 10 i^3c^2s^3 + 5i^4cs^4 + i^5s^5 \\ \\ = (c^5 -10 c^3 s^2 +5 cs^4) + i(5c^4 s - 10c^2s^3 + s^5) Equating reals and imaginaries: cos 5\theta =...

For reduction formula for sec x: I_n = \int sec^n x dx = \frac {sec^{n-2}tanx}{n-1} + \frac {n-2}{n-1} \int sec^{n-2} xdx I_4 = \int ^- _- sec^4xdx = \frac {sec^2 x tan x}{4-1} + \frac {4-2}{4-1} \int ^- _- sec^2 x dx \\ \\ = \left [ \frac {sec^2 x tan x}{3} + \frac {2}{3} tan x...

\int ^- _- sec^4 x dx = \int ^- _- sec^2 x \cdot sec^2 x dx \int ^- _- (tan^2 x + 1) dtanx \\ \\ = \left [ \frac {tan^3 x}{3} + tan x\right ]^{\frac {\pi}{4}} _0 = \frac {1}{3} + 1 = \frac {4}{3}

sin2x - 2 cos^2 2x \\ = 2sinxcos x - 2sinx \cdot sinx \\ = 2sin x(cos x - sin x) \\ = $positive $ \times (cos x - sin x) \\ = (+ve) times (+ve) = +ve (i.e. > 0) since over the given domain, sin x > 0 and cos x > sin x.

From the font, looks like Jim Coroneos's solution?

\therefore $ $ tan x + \frac{2tan x}{1-tan^2 x} =\frac {tan x (1-tan^2 x) + 2tanx }{1 - tan^2 x} \\ \\ = \frac {tan x (1 - tan^2 x + 2)}{1-tan^2 x} = \frac {tan x(3 - tan^2 x)}{1-tan^2 x} = 0\\ \\ \therefore tanx(3-tan^2 x) = 0 $ provided $tan^2 x \neq 1 \\ \\ \therefore $ either: $ tan x...

What's given answer of the volume of the torus so-formed?

Oh ya! I mistakenly used 2 as the radius of the corresponding circle instead of the cube-root of 2. Thanks.

Area: 3\sqrt 3 ??

pethmin, why are you repeating your HSC? I thought you completed it in 1998.

I draw a unit circle, centre the origin. z = cis \theta $ is just a point on the circle and $ \frac {1}{z} = \overline z $ is just another point directly below z. $ z + \frac {1}{z} $ form a diagonal on the x-axis of the rhombus, formed by z and $\frac {1}{z} $ as adjacent sides. You can...