NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Integration of various functions

Lesson

Integration is the process of finding a function variously called a *primitive* or an *antiderivative* or an *indefinite integral*.

We let $F$`F` be a function that is differentiable on an open interval and let $f=F'$`f`=`F`′ be its derivative. Then, we say $F$`F` is an antiderivative of $f$`f` and the words *primitive *and *indefinite integral* are used with the same meaning.

The usual notation for an antiderivative is

$F=\int f\ \mathrm{d}x$`F`=∫`f` `d``x`

We speak of *an *antiderivative rather than *the *antiderivative because any function $F+c$`F`+`c` where $c$`c` is a constant is also an antiderivative for $f$`f`.

Thus, for example, the function $2x$2`x` is the derivative of every function $x^2+c$`x`2+`c`. So, to specify the general antiderivative for $2x$2`x`, we write $x^2+c$`x`2+`c`. The constant $c$`c` is often called the constant of integration.

There are many techniques for finding the indefinite integral of a given continuous function when it exists. Unfortunately, there are infinitely many functions whose indefinite integral cannot be expressed in terms of the standard functions.

In practice, we may often look up a table of integrals when the required antiderivative is not obvious. In situations where there is no standard way of expressing an antiderivative, a numerical method will usually be employed to solve a problem, approximately, for which the integral was needed.

Determine $\int\frac{6\sin2x}{\cos2x}\ \mathrm{d}x$∫6`s``i``n`2`x``c``o``s`2`x` `d``x`

When a function is a fraction whose denominator is (almost) the derivative of the numerator, we call on our experience in differentiation to realise that the logarithm function is involved. In this case, if we were to differentiate $\log_e\cos2x$`l``o``g``e``c``o``s`2`x`, we would obtain $-\frac{-2\sin2x}{\cos2x}$−−2`s``i``n`2`x``c``o``s`2`x`

It follows that $\int\frac{6\sin2x}{\cos2x}\ \mathrm{d}x=-3\log_e\cos2x+c$∫6`s``i``n`2`x``c``o``s`2`x` `d``x`=−3`l``o``g``e``c``o``s`2`x`+`c` and this can be confirmed by differentiating the expression on the right.

Find the general primitive for $f(x)=\sin x\cos x$`f`(`x`)=`s``i``n``x``c``o``s``x`.

We could do this by first rewriting the function as $f(x)=\frac{1}{2}\sin2x$`f`(`x`)=12`s``i``n`2`x`. But, another way is to observe that $\cos x$`c``o``s``x` is the derivative of $\sin x$`s``i``n``x`. Then, after putting $u=\sin x$`u`=`s``i``n``x` and therefore $\frac{\mathrm{d}u}{\mathrm{d}x}=\cos x$`d``u``d``x`=`c``o``s``x`, we can rewrite the integral as

$F(x)=\int u\frac{\mathrm{d}u}{\mathrm{d}x}\ \mathrm{d}x$`F`(`x`)=∫`u``d``u``d``x` `d``x`

We omit the details, but it can be shown that this is equivalent to $\int u\ \mathrm{d}u$∫`u` `d``u`.

Thus, $F(x)=\frac{u^2}{2}+c=\frac{1}{2}\sin^2x+c$`F`(`x`)=`u`22+`c`=12`s``i``n`2`x`+`c` which, again, can be confirmed by differentiation.

Find the primitive of $\frac{\cos x}{\sin x}$`c``o``s``x``s``i``n``x`.

You may use $c$`c` as a constant.

Find the primitive of $\frac{\sin x}{1-\cos x}$`s``i``n``x`1−`c``o``s``x`.

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

Apply integration methods in solving problems