LaTeX Maths Equation - IamCK

# LaTeX Maths Equation

Formulae list for Engineering Mathematics

# Trigonometric

 $cos^2 \ x + sin^2 \ x = 1$ $1 + tan^2\ x = sec^2 \ x$ $cot^2 \ x + 1 = cosec^2 \ x$ $sin \ 2x = 2sin\ x \ cos \ x$ $cos \ 2x = cos^2 \ x - sin^2 \ x$ $tan \ 2x = \frac{2tan\ x}{1 - tan^2 \ x }$} $cos \ 2x = 2cos^2 \ x - 1$ $cos \ 2x = 1 - 2sin^2 \ x$ $sin \ (x \pm y) = sin\ x\ cos \ y \pm cos \ x \ sin \ y$ $cos \ (x \pm y) = cos\ x \ cos \ y \mp sin\ x\ sin\ y$ $tan(x \pm y) = \frac{tan\ x \pm tan\ y}{1 \mp tan\ x\ tan\ y}$ $2 sin \ x \ cos \ y = sin (x + y) + sin(x - y)$ $2 sin\ x \ sin\ y = cos(x + y) + cos(x - y)$ $2cos\ x\ cos\ y = cos(x+y) + cos(x-y)$

# Hyperbolic

 $sinh\ x = \frac{e^x - e^{-x}}{2}$ $cosh\ x = \frac{e^x + e^{-x}}{2}$ $cosh^2\ x - sinh^2\ x = 1$ $1 - tanh^2\ x = sech^2\ x$ $coth^2\ x - 1 = cosech^2 x$ $sinh\ 2x = 2sinh\ x cosh\ x$ $cosh \ 2x = cosh^2 \ x + sinh^2 \ x$ $cosh \ 2x = 2cosh^2 \ x - 1$ $cosh \ 2x = 1 - 2sinh^2 \ x$ $tanh\ 2x = \frac{2tanh\ x}{1 + tanh^2\ x}$ $sinh \ (x \pm y) = sinh\ x\ cosh \ y \pm cosh \ x \ sinh \ y$ $cosh \ (x \pm y) = cosh\ x \ cosh \ y \pm sinh\ x\ sinh\ y$ $tanh(x \pm y) = \frac{tanh\ x \pm tanh\ y}{1 \pm tanh\ x\ tanh\ y}$

# Logarithm

 $a^x = e^{x\ ln a}$ $log_a x = \frac{log_b x}{log_b a}$

# Inverse Hiperbolic

 $sinh^{-1}\ x = ln(x + \sqrt{x^2 + 1}), \ -\infty < x < \infty$ $cosh^{-1}\ x = ln(x + \sqrt{x^2-1}),\ x \geq 1$ $tanh^{-1}\ x = \frac{1}{2} ln( \frac{1+x}{1-x}) ,\ -1 < x < 1$

# Differentiations

 $\frac{d}{dx} \ [k] = 0,\ k = contant$ $\frac{d}{dx} \ [x^n] = nx^{n-1}$ $\frac{d}{dx} \ [ln\mid x \mid] = \frac{1}{x}$ $\frac{d}{dx} \ [cos \ x] = -sin \ x$ $\frac{d}{dx} \ [sin\ x] = cos \ x$ $\frac{d}{dx} \ [tan \ x] = sec^2 \ x$ $\frac{d}{dx} \ [cot \ x] = -cosec^2 \ x$ $\frac{d}{dx} \ [sec \ x] = sec \ x \ tan \ x$ $\frac{d}{dx} \ [cosec \ x] = -cosec \ x \ cot \ x$ $\frac{d}{dx} \ [e^x] = e^x$ $\frac{d}{dx} \ [cosh \ x] = sinh \ x$ $\frac{d}{dx} \ [sinh \ x] = cosh \ x$ $\frac{d}{dx} \ [tanh \ x] = sech^2 \ x$ $\frac{d}{dx} \ [coth \ x] = -cosech^2 \ x$ $\frac{d}{dx} \ [sech \ x] = -sech \ x \ tanh \ x$ $\frac{d}{dx} \ [cosech \ x] = -cosech \ x \ coth \ x$

# Integrations

 $\int k \ dx = kx + c$ $\int x^n \ dx = \frac{x^{n+1}}{n+1} + c , n \neq -1$ $\int \frac{1}{x} \ dx = ln\mid x \mid + c$ $\int sin \ x \ dx = -cos \ x + c$ $\int cos \ x \ dx = sin \ x + c$ $\int sec^2 \ x \ dx = tan \ x + c$ $\int cosec^2 \ x \ dx = -cot \ x + c$ $\int sec \ x \ tan \ x \ dx = sec \ x + c$ $\int cosec \ x \ cot \ x \ dx = - cosec \ x + c$ $\int e^x \ dx = e^x + c$ $\int sinh \ x \ dx = -cosh \ x + c$ $\int cosh \ x \ dx = sinh \ x + c$ $\int sech^2 \ x \ dx = tanh \ x + c$ $\int cosech^2 \ x \ dx = -coth \ x + c$ $\int sech \ x \ tanh \ x \ dx = sech \ x + c$ $\int cosech \ x \ coth \ x \ dx = - cosech \ x + c$

# Differentiations of Inverse Functions

 $\frac{d}{dx} [sin^{-1} u] = \frac{1}{\sqrt{1-u^2}} \frac{du}{dx},\ \mid u\mid\ <\ 1$ $\frac{d}{dx} [cos^{-1} u] = \frac{-1}{\sqrt{1-u^2}} \frac{du}{dx},\ \mid u\mid\ <\ 1$ $\frac{d}{dx} [tan^{-1} u] = \frac{1}{1+u^2} \frac{du}{dx}$ $\frac{d}{dx} [cot^{-1} u] = \frac{-1}{1+u^2} \frac{du}{dx}$ $\frac{d}{dx} [sec^{-1} u] = \frac{1}{\mid u \mid\sqrt{u^2 - 1}} \frac{du}{dx},\ \mid u \mid\ >\ 1$ $\frac{d}{dx} [cosec^{-1} u] = \frac{-1}{\mid u \mid\sqrt{u^2 - 1}} \frac{du}{dx},\ \mid u \mid\ >\ 1$ $\frac{d}{dx} [sinh^{-1} u] = \frac{1}{\sqrt{u^2 + 1}} \frac{du}{dx}$ $\frac{d}{dx} [cosh^{-1} u] = \frac{1}{\sqrt{u^2 - 1}} \frac{du}{dx},\ \mid u\mid\ >\ 1$ $\frac{d}{dx} [tanh^{-1} u] = \frac{1}{1-u^2} \frac{du}{dx},\ \mid u\mid\ <\ 1$ $\frac{d}{dx} [coth^{-1} u] = \frac{1}{1-u^2} \frac{du}{dx},\ \mid u\mid\ >\ 1$ $\frac{d}{dx} [sech^{-1} u] = \frac{-1}{u\sqrt{1 - u^2}} \frac{du}{dx},\ 0\ <\ u\ <\ 1$ $\frac{d}{dx} [cosech^{-1} u] = \frac{-1}{\mid u \mid\sqrt{1 + u^2}} \frac{du}{dx},\ u\ \neq\ 0 1$

# Integrations Resulting in Inverse Functions

 $\int \frac{dx}{\sqrt{a^2 - x^2}} = sin^{-1}\ \frac{x}{a} + C$ $\int \frac{dx}{a^2 + x^2} = \frac{1}{a}tan^{-1}\frac{x}{a} + C$ $\int \frac{dx}{\mid u \mid \sqrt{x^2 - a^2}} = \frac{1}{a}sec^{-1} \frac{x}{a} + C$ $\int \frac{dx}{\sqrt{x^2 + a^2}} = sinh^{-1} \frac{x}{a} + C,\ a\ >\ 0$ $\int \frac{dx}{\sqrt{x^2 - a^2}} = cosh^{-1} \frac{x}{a} + C,\ x \ >\ 0$ $\int \frac{dx}{\sqrt{x^2 + a^2}} = sinh^{-1} \frac{x}{a} + C,\ a > 0$ $\int \frac{dx}{a^2 - x^2 } = \frac{1}{a}tanh^{-1} \frac{x}{a} + C,\ \mid x \mid\ >\ a$ $\int \frac{dx}{a^2 - x^2 } = \frac{1}{a}coth^{-1} \frac{x}{a} + C,\ \mid x \mid\ >\ a$ $\int \frac{dx}{x \sqrt{a^2 - x^2}} = -\frac{1}{a}sech^{-1}\frac{x}{a} + C,\ 0\ <\ x\ <\ a$ $\int \frac{dx}{x \sqrt{a^2 +x^2}} = -\frac{1}{a}cosech^{-1}\mid \frac{x}{a} \mid\ 0\ <\ x\ <\ a$

# Multiple Integral

$\int \int \int f(x,y,z) \ dV = \int\int[\int f（r\ cos \ \theta, r\ sin \theta, z)dz]dA$
$\int\int\int f(x,y,z)dV = \int\int\int f(\rho sin \ \phi \ cos \ \theta, \rho sin \ \phi\ sin\ \theta, \rho cos\ \phi)\rho ^2 \ sin \ \phi\ d\rho d\phi d\theta$