
# 8.0: Prelude to Techniques of Integration

• Page ID
3478
•

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

Over the next few sections we examine some techniques that are frequently successful when seeking antiderivatives of functions. Sometimes this is a simple problem, since it will be apparent that the function you wish to integrate is a derivative in some straightforward way. For example, faced with

$\int x^{10}\,dx$

we realize immediately that the derivative of $$x^{11}$$ will supply an $$x^{10}$$: $$(x^{11})'=11x^{10}$$. We don't want the "11'', but constants are easy to alter, because differentiation "ignores'' them in certain circumstances, so

${d\over dx}{1\over 11}{x^{11}}={1\over 11}11{x^{10}}=x^{10}.$

From our knowledge of derivatives, we can immediately write down a number of antiderivatives. Here is a list of those most often used:

$\displaylines{ \int x^n\,dx={x^{n+1}\over n+1}+C, \quad\hbox{if n\not=-1}\cr \int x^{-1}\,dx = \ln |x|+C\cr \int e^x\,dx = e^x+C\cr \int \sin x\,dx = -\cos x+C\cr \int \cos x\,dx = \sin x+C\cr \int \sec^2 x\,dx = \tan x+C\cr \int \sec x\tan x\,dx = \sec x+C\cr \int {1\over1+x^2}\,dx = \arctan x+C\cr \int {1\over \sqrt{1-x^2}}\,dx = \arcsin x+C\cr }$