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Integrals From Tables of Antiderivatives

April 5th, 2009 |

Author: Merrin

Based on our table of derivatives, we can form a table of antiderivatives. If the derivative of the right hand side of the following equation matches the integrand, then it is an antiderivative.

$\displaystyle \int f(x)dx=\displaystyle \int \cfrac{dF}{dx} dx=F(x)+C$

We can start with some of the simple derivatives we know and see which integration rules we can build up.

$\cfrac{d}{dx} x^{n+1} =(n+1)x^{n} \quad \to \displaystyle \int x^{n} dx =\cfrac{x^{n+1} }{n+1} +C\qquad n\ne -1$

We know what to do for the power rule when n is equal to negative one. We use the natural logarithm.

$\cfrac{d}{dx} \ln x=\cfrac{1}{x} \to \displaystyle \int \cfrac{dx}{x} =\ln \left|x\right| +C$

An absolute value sign is necessary because the integration interval may be negative. The interval can’t cross zero.

The derivative of the exponential of x we found to be the exponential of x. It should come as no surprise that the antiderivative of this function is also itself.

$\cfrac{d}{dx} e^{x} =e^{x} \to \displaystyle \int e^{x} dx=e^{x}+C$

The integral of $a^x$ has a slight twist compared to $e^x$, remember what we did for the derivatives of each.

$\displaystyle\cfrac{d}{dx} a^{x} =a^{x} \ln a\to \int a^{x} dx=\cfrac{a^{x} }{\ln a} +C$

As a quick check, differentiate these antiderivatives.

$\cfrac{d}{dx}\left(\cfrac{x^{n+1}}{n+1}+C\right)=x^n$

$\cfrac{d}{dx}\left( \ln x+C \right)=x^{-1}$

$\cfrac{d}{dx}\left({e^x+C}\right) = e^x$

$\cfrac{d}{dx}\left( \cfrac{a^x}{\ln a} +C\right)=a^x$

We also learned a whole array of derivatives for the trigonometric and hyperbolic functions and their inverses. We can study this table to build up our knowledge of integrals. Some of the knowledge is a bit esoteric but we will find some of the inverse functions are very useful for integrating integrals that contain various forms of

$\, \sqrt{x^{2} -a^{2} }\qquad \sqrt{x^{2} +a^{2} }\qquad \sqrt{a^{2} -x^{2} }$

Here is our table of derivatives for the trigonometric and hyperbolic functions and their inverses.

$\int \cos xdx =\sin x+C} \\ {\displaystyle \int \sin xdx =-\cos x+C} \\ {\displaystyle \int \sec ^{2} xdx=\tan x+C } \\ {\displaystyle \int \csc x\cot xdx=-\csc x+C } \\ {\displaystyle \int \sec x\tan xdx=\sec x+C } \\ {\displaystyle \int \csc ^{2} xdx=-\cot x+C$

It is interesting to note that our work with derivatives did not reveal all the integrals of the trigonometric functions to the first power. Later we will learn how to find

$\displaystyle \int \tan xdx \quad \displaystyle \int \cot xdx \quad \displaystyle \int \csc xdx\quad \displaystyle \int \sec xdx$

We have a similar set of integrals that we know for hyperbolic functions from a table of antiderivatives.

$\int \cosh xdx =\sinh x+C}\\{\displaystyle \int \sinh xdx =\cosh x+C}\\ {\displaystyle \int \text{sech}^{2} xdx=\tanh x+C }\\ {\displaystyle \int \text{csch} x\text{coth} xdx=-\text{csch}x+C } \\ {\displaystyle \int \text{sech x}\tanh xdx=-\text{sech} x+C } \\ {\displaystyle \int \text{csch}^{2} xdx=-\text{coth} x+C$

We are after results that span the variety of forms that normally occur in integration problems. these forms is There are a finite number of these selected inverse functions but there are many more possible integrals. In certain cases, your integral may reduce to one of these particular integrals.

$\int \cfrac{dx}{\sqrt{1-x^{2} } } =\arcsin x+C$

$\int \cfrac{dx}{\sqrt{1+x^{2} } } =\text{arcsinh}\,\, x+C$

$\int \cfrac{dx}{1+x^{2} } =\arctan x+C$

$\int \cfrac{dx}{1-x^{2} } =\text{arctanh}\,\, x+C$

$\int \cfrac{dx}{\sqrt{x^{2} -1} } =\text{arccosh} x+C$

$\int \cfrac{dx}{x\sqrt{x^{2} -1} } =\text{arcsec} x+C$

Part of learning integration is exploring the relations between the elementary functions through integration or differentiation and knowing which integrals can and cannot be done with the known methods.

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