Standard integrals, derivatives and trig identities.

i really fucking hate this.

sin2A =

2sinAcosA

cos2A =

cos²A - sin²A = 2cos²A - 1 = 1 - sin²A

tan2A =

2tanA/(1-tan²A)

\sec^2x=

1+\tan^2x

\operatorname{cosec}^2x=

1+\cot^2x

\int\tan^2xdx=

\tan x-x+C or use trig identity and\int\sec^2xdx

\int\cot^2xdx=

-\cot x-x+C or using cosec identity

\int\operatorname{cosec}^2xdx=

-\cot x+C

\int_{}^{}\sec xdx=

\ln\left|\sec x+\tan x\right|+C

\int_{}^{}\operatorname{cosec}xdx =

\ln\left|\operatorname{cosec}x-\cot x\right|

\int_{}^{}\sinh xdx=

\cosh x+C

\int_{}^{}\cosh xdx=

\sinh x+C

\int\tanh xdx=

\ln\left(\cosh x\right)+C

\int_{}^{}\coth xdx=

\ln\left|\sinh x\right|+C

\int_{}^{}\operatorname{sech}xdx=

\arctan\left(\sinh x\right)+C

\int\,co\operatorname{sech}xdx

\ln\left|\tanh\left(\frac{x}{2}\right)\right|+C

\int_{}^{}\operatorname{sech}^2xdx

tanhx + C

\int_{}^{}\operatorname{cosech}^2xdx =

-cothx + C

\int_{}^{}\tanh^2xdx

x - tanhx + C

\int_{}^{}\coth^2xdx=

x - cothx + C

\frac{d}{dx}(secx) =

secxtanx

\frac{d}{dx}(cosecx) =

-\operatorname{cosec}x\cot x

\frac{d}{dx}(cotx)

-\operatorname{cosec}^2x

\frac{d}{dx}(sec^2x)

2\sec^2x\tan x

\frac{d}{dx}(cosec^2 x)

-2\operatorname{cosec}^2x\cot x

\frac{d}{dx}(cot^2 x)

-2\cot x\operatorname{cosec}^2x

\frac{d}{dx}(arcsinx)

\frac{1}{\sqrt{1-x^2}}

\frac{d}{dx}(arccosx)

-\frac{1}{\sqrt{1-x^2}}

\frac{d}{dx}(arctanx)

\frac{1}{1+x^2}