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If faced with sinx + icosx, just factor out "i" ---> i(cosx - isinx)

You shouldn't write your cosecs that way. Too easily mistaken with arcsin.

\frac{d}{dx}\left ( \cos e^{x} \right )=-e^{x}\sin e^{x} \\\\ \therefore \int e^{x}\sin e^{x}dx = -\cos e^{x}+C

I dunno, we've finished them a few weeks ago and wanted to see if I could still remember ^^

Go nuts - it's in your standard integrals though =)

Keke, it's Willy ^^

\int \frac{\sqrt{x^{2}+9}+\sqrt{x^{2}-9}}{\sqrt{x^{4}-81}}dx=\int \left ( \frac{1}{\sqrt{x^{2}-9}}+\frac{1}{\sqrt{x^{2}+9}} \right )dx \\\\ = \cosh^{-1}\frac{x}{3}+\sinh^{-1}\frac{x}{3} + C \\\\ = \ln \left ( x+\sqrt{x^{2}-9} \right )+\ln\left ( x+\sqrt{x^{2}+9} \right ) + C

I thought so too :P

Sup Madge?

An alternative would be to find sech using the relationship between tanh and sech and then to proceed with sinh and cosh

There's actually a "triangle" in hyperbolic space which can be used quite easily with these questions. I believe it has applications in relativity, but it also works here. c^2 = a^2 - b^2, where a > b.

Yep, that's sufficient.

No way!? I found some great girls in the strangest of classes =/

Big one for me: Whilst (x+3)/x = 1 + 3/x, x/(x+3) =/= 1 + x/3. This sort of rushed error has cost me many marks in exams.

A slight discrepancy mirakon: arg(z1z2) = arg(z1) + arg(z2) There is no such thing as the argument of an angle, only of a complex number.

These ones are awful. Absolutely cannot stand them.

\int_{0}^{\sin x}ydy=\frac{1}{2}\left [y^{2} \right ]^{\sin x}_{0}=\frac{1}{2}.\sin^{2}x \\\\ \therefore \int_{0}^{\pi}\int_{0}^{\sin x}ydydx=\frac{1}{2}\int_{0}^{\pi}\sin^{2}xdx \\\\ = \frac{1}{4}\int_{0}^{\pi}\left ( 1-\cos 2x \right )dx \\\\=\frac{1}{4}\left [x-\frac{1}{2}.\sin 2x \right...

Nvm, my careless call.

\text{Note that} \,\, -2\tan^{-1}\frac{4}{3}\neq \tan^{-1}-\frac{8}{3}

Assumption states that (1+x)^k > 1+kx. Therefore (1+x)(1+x)^k > (1+kx)(1+x); x>-1