\begin {align*} \sin nx & =\frac {e^{inx}-e^{-inx}}{2i}\\ \cos nx & =\frac {e^{inx}+e^{-inx}}{2} \end {align*}
\begin {align*} \int \frac {\sin x}{\cos x}dx & =\ln \left (\sin x\right ) \\ \csc x & =\frac {1}{\sin x}\\ \text {average value of }f\relax (x) \text { over }[b,a] & =\frac {\int _{a}^{b}f\relax (x) dx}{b-a} \end {align*}
\begin {align*} \cos ^{2}kx & =\frac {1+\cos \left (2k\right ) }{2}\\ \sin ^{2}kx & =\frac {1-\cos \left (2k\right ) }{2} \end {align*}
\begin {align*} \sin A\ \sin B & =\frac {1}{2}\left [ \ \cos \left (A-B\right ) -\cos \left ( A+B\right ) \right ] \\ \cos A\ \cos B & =\frac {1}{2}\left [ \ \cos \left (A-B\right ) +\cos \left ( A+B\right ) \right ] \\ \sin A\ \cos B & =\frac {1}{2}\left [ \ \sin \left (A-B\right ) +\sin \left ( A+B\right ) \right ] \end {align*}
I need a geometric way to visualize these equations, but for now for the exam remember them as follows: they all start with \(A-B\) , and when the functions being multiplied are different on the LHS, we get sin on the RHS, else we get cos (think of cos as nicer, since even function :).
\begin {align*} \int \tanh \relax (x) \ & =\allowbreak \ln \left (\cosh \ x\right ) \\ \int \tan x & =-\ln \left (\cos \ x\right ) \end {align*}
\(\ \int _{a}^{b}\cos ^{2}kx\ dx=\frac {b-a}{2}\). if \(k\left (b-a\right ) \) is an integer multiple of \(\pi \). (the same for \(\sin ^{2}kx\)), for example \(\int _{-1}^{1}\cos ^{2}\pi x\ dx=1\,\ \), \(\int _{-1}^{1}\cos ^{2}5\pi x\ dx=1,\) \(\int _{-5}^{1}\cos ^{2}7\pi x\ dx=3\,\), \(\int _{-1}^{1}\sin ^{2}\pi x\ dx=1\), etc... this can be very useful so remember it!
\(\int _{a}^{b}\cos kx\ dx=0\) if over a complete period. same for \(\sin x\), for example \(\int _{-\pi }^{\pi }\cos kx\ dx=0\)
\begin {align*} \sinh x & =-i\ \sin \left (ix\right ) \\ \cosh x & =\cos \left (ix\right ) \\ \tanh x & =-i\ \tan \left (ix\right ) \end {align*}
\begin {align*} e^{\ln z} & =z\\ z^{b} & =e^{b\ln z} \end {align*}
\[ \frac {1}{1-x}=1+x+x^{2}+x^{3}+\cdots \]
\[ \frac {1}{1+x}=1-x+x^{2}-x^{3}+\cdots \]
\begin {align*} \arctan x & =x-\frac {x^{3}}{3}+\frac {x^{5}}{5}-\cdots \\ \sin x & =x-\frac {x^{3}}{3!}+\frac {x^{5}}{5!}-\cdots \\ \cos x & =1-\frac {x^{2}}{2!}+\frac {x^{4}}{4!}-\cdots \\ \sinh x & =x+\frac {x^{3}}{3!}+\frac {x^{5}}{5!}+\cdots \\ \cosh x & =1+\frac {x^{2}}{2!}+\frac {x^{4}}{4!}+\cdots \\ e^{x} & =1+x+\frac {x^{2}}{2!}+\frac {x^{3}}{3!}-\cdots \\ \ln \left (1+x\right ) & =x-\frac {x^{2}}{2}+\frac {x^{3}}{3}-\cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ -1<x\leq 1\\ \ \left (1+x\right ) ^{p} & =1+px+\frac {p\left (p-1\right ) }{2!}x^{2}+\ \cdots +\ \ \ \ \ \ \ \ \ \left \vert x\right \vert <1 \end {align*}
Leibinz rule for differentiation of integrals
\[ \frac {d}{dx}\int _{u\relax (x) }^{v\relax (x) }f\left ( x,t\right ) \ dt=f\left (x,v\relax (x) \right ) \ \frac {d}{dx}v\left ( x\right ) -f\left (x,u\relax (x) \right ) \ \frac {d}{dx}u\left ( x\right ) +\int _{u}^{v}\frac {\partial }{\partial x}f\left (x,t\right ) \ dt \]
example:
\begin {align*} \frac {d}{dx}\int _{x}^{2x}\frac {e^{xt}}{t}\ dt & =\frac {e^{x\left ( 2x\right ) }}{2x}\ \frac {d}{dx}\left (2x\right ) -\frac {e^{x\relax (x) }}{x}\ \frac {d}{dx}\relax (x) +\int _{x}^{2x}\frac {\partial }{\partial x}\left (\frac {e^{xt}}{t}\right ) \ \ dt\\ & =\frac {e^{2x^{2}}}{x}\ -\frac {e^{x^{2}}}{x}\ +\int _{x}^{2x}\frac {te^{xt}}{t}\ \ dt\\ & =\frac {e^{2x^{2}}}{x}\ -\frac {e^{x^{2}}}{x}\ +\left [ \frac {e^{xt}}{x}\right ] _{x}^{2x}\ \ \ \end {align*}
To help remember the above 2 formulas, notice that when \(+x\) we get a \(-\) shown (i.e. terms flip flop), but when we have \(-x\) the series is all positive terms. These are very important to remember for problems when finding Laurent expansion of a function.
Expansion of \(\cos \) and \(\sin \) around a point different than \(0\)
expand \(\cos \relax (z) \) around \(a\), we get
\[ \left (\cos \relax (a) -\frac {\cos \relax (a) \left (z-a\right ) ^{2}}{2!}+\frac {\cos \relax (a) \left (z-a\right ) ^{4}}{4!}-\cdots \right ) +\left (-\sin \relax (a) \left (z-a\right ) +\frac {\sin \relax (a) \left (z-a\right ) ^{3}}{3!}\ \cdots \right ) \]
For example to expand \(\cos \relax (x) \) about \(\pi \) we get
\begin {align*} & \left (\cos \left (\pi \right ) -\frac {\cos \left (\pi \right ) \left ( z-\pi \right ) ^{2}}{2!}+\frac {\cos \left (\pi \right ) \left (z-\pi \right ) ^{4}}{4!}-\cdots \right ) +\overbrace {\left (-\sin \left (\pi \right ) \left ( z-\pi \right ) +\frac {\sin \relax (a) \left (z-\pi \right ) ^{3}}{3!}\ \cdots \right ) }^{=0}\\ & =-1+\frac {1}{2}\left (\pi -z\right ) ^{2}-\frac {1}{24}\left (\pi -z\right ) ^{4}+\ \cdots \end {align*}
so above is easy to remember. The \(\cos \relax (z) \) part is the same as around zero, but it has \(\cos \relax (a) \) multiplied to it, and the \(\sin \) part is the same as the \(\sin \relax (z) \) about zero but has \(\sin \relax (a) \) multiplied to it, and the signs are reversed.
For expansion of \(\sin \relax (z) \) use
\[ \left (\sin \relax (a) -\frac {\sin \relax (a) \left (z-a\right ) ^{2}}{2!}+\frac {\sin \relax (a) \left (z-a\right ) ^{4}}{4!}-\cdots \right ) +\left (\cos \relax (a) \left (z-a\right ) -\frac {\cos \relax (a) \left (z-a\right ) ^{3}}{3!}\ \cdots \right ) \]
This is the same as the expansion of \(\cos \relax (z) \) but the roles are reversed and notice the cos part start now with positive not negative term. SO all what I need to remember is that expansion of \(\cos \left ( z\right ) \) starts with \(\cos \relax (a) \) terms while expansion of \(\sin \relax (z) \) start with the \(\sin \relax (a) \) term. This is faster than having to do Taylor series expansion to find these series in the exam. \(\ \)\begin {align*} \Gamma \left (\frac {1}{2}\right ) & =\sqrt {\pi }\\ \Gamma \left (P+1\right ) & =P\Gamma \relax (P) \end {align*}
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