Integral Calculus- Recap

 $\int_{ }^{ }x^{n\ }dx\ =\ \frac{x^{n+1}}{n\ +\ 1}\ +\ C$  

 $\int_{ }^{ }\frac{1}{x}\ dx\ =\ \ln\left|x\right|\ +\ C$  

 $\int_{ }^{ }a^{x\ }dx\ =\ \frac{a^x}{\ln a}\ +\ C$  

 $\int_{ }^{ }e^{x\ }dx\ =\ e^x\ +\ C$  

 $\frac{d\left(\sin x\right)}{dx}\ =\ \cos x$  

 $\frac{d\left(\cos x\right)}{dx}\ =\ -\sin x$  

 $\frac{d\left(\tan x\right)}{dx}\ =\ \sec^2x$  

 $\frac{d\left(\sec x\right)}{dx}\ =\ \sec x\ \tan x$  

 $\frac{d\left(\operatorname{cosec}x\right)}{dx}\ =\ -\operatorname{cosec}x\ \cot x$  

 $\frac{d\left(\cot x\right)}{dx}\ =\ -\operatorname{cosec}^2x$  

 $\ \int_{ }^{ }\cos x\ =\ \sin x\ +\ C$  

 $\ \int_{ }^{ }-\sin x\ =\ \cos x\ +\ C$  

 $\ \int_{ }^{ }\sec^2x\ =\ \tan x\ +\ C$  

 $\ \int_{ }^{ }\sec x\ \tan x\ =\ \sec x\ +\ C$  

 $\ \int_{ }^{ }-\operatorname{cosec}x\ \cot x\ =\ \operatorname{cosec}x\ +\ C$  

 $\ \int_{ }^{ }-\operatorname{cosec}^2x\ =\ \cot x\ +\ C$  

 $\int_{ }^{ }\sec x\ dx\ =\ \ln\left|\tan x\ +\ \sec x\right|\ +\ C$  

 $\int_{ }^{ }\operatorname{cosec}x\ dx\ =-\ln\left|\cot x\ +\ \operatorname{cosec}x\right|\ +\ C\ =\ \ln\left|\tan\left(\frac{x}{2}\right)\right|\ +\ C$  

 $\int_{ }^{ }\tan x\ dx\ =\ \ln\left|\sec x\right|\ +\ C$  

 $\int_{ }^{ }\cot x\ dx\ =\ \ln\left|\sin x\right|\ +\ C$  

 $\int\ \frac{dx}{\sqrt{a^2\ -\ x^2}}\ =\ \sin^{-1}\left(\frac{x}{a}\right)\ +\ C$  

 $\int\ \frac{dx}{x\sqrt{x^2\ -\ a^2}}\ =\ \frac{1}{a}\sec^{-1}\left(\frac{x}{a}\right)+\ C$  

 $\int\ \frac{dx}{x^2\ +\ a^2}\ =\frac{1}{a}\ \tan^{-1\ }\left(\frac{x}{a}\right)\ +\ C$  

 $\int\ \frac{dx}{x^{2\ }-\ a^2}\ =\ \frac{1}{2a}\ \ln\left|\frac{x\ -a}{x\ +\ a}\right|\ +\ C$  

 $\int\ \frac{dx}{-x^{2\ }+\ a^2}\ =\ \frac{1}{2a}\ \ln\left|\frac{x\ +a}{-x\ +\ a}\right|\ +\ C$  

Notice how the numerator's 'a' sign depends and denominator's 'x' sign depends

 $\int\ \frac{dx}{\sqrt{x^{2\ }\pm\ a^2}}\ =\ \ \ln\left|x\ +\sqrt{x^{2\ }\pm\ a^2}\ \right|\ +\ C$ 

 $\ \int_{ }^{ }\sqrt{x^{2\ }\pm\ a^2}\ dx\ =\ \frac{x}{2}\sqrt{x^{2\ }\pm\ a^2}\ \ \pm\ \frac{a^2}{2}\ln\left|x\ +\ \sqrt{x^{2\ }\pm\ a^2}\right|\ +\ C$  

 $\int_{ }^{ }\sqrt{-x^{2\ }+\ a^2}\ dx\ =\ \frac{x}{2}\sqrt{-x^{2\ }+\ a^2}\ \ +\ \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)+\ C$  

Notice how the function behaves differently when the coefficient of x is negative.

The signs of each term are heavily dependant on the Integrand.

 $\int\left[f\left(x\right)\ \pm\ g\left(x\right)\right]dx\ =\ \int f\left(x\right)dx\ \pm\ \int g\left(x\right)dx$  

 $\int kf\left(x\right)dx\ =\ k\int f\left(x\right)dx$  

 $\frac{\text{d}\int f\left(x\right)\ dx}{\text{d}x}\ =\ f\left(x\right)$  

 $\int f\left(ax\ +\ b\right)dx\ =\ \frac{1}{a}f\left(ax\ +\ b\right)\ +\ C$  

 $\left(a\ \pm\ b\right)^3\ =\ a^3\ \pm\ b^{3\ }\pm\ 3ab\left(a\ \pm b\right)$  

 $a^{3\ }\ -\ \ b^3\ \ =\ \left(a\ -\ b\right)\left(a^{2\ }+\ b^{2\ }+\ ab\right)$  

 $a^{3\ }\ +\ \ b^3\ \ =\ \left(a\ +\ b\right)\left(a^{2\ }+\ b^{2\ }-\ ab\right)$  

Carefully observe the signs in the differences of cubes identity.

 $\int\tan x\ \tan2x\ \tan\ 3x\ dx\$  

 $\int\ \frac{\left(x^{3\ }+\ 8\right)\ \left(x\ -\ 1\right)}{x^{2\ }-2x\ +\ 4}dx$