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New Zealand
Level 8 - NCEA Level 3

Integration of various functions


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 dx

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$2x is the derivative of every function $x^2+c$x2+c. So, to specify the general antiderivative for $2x$2x, we write $x^2+c$x2+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.


Example 1

Determine $\int\frac{6\sin2x}{\cos2x}\ \mathrm{d}x$6sin2xcos2x dx

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 $\log_e\cos2x$logecos2x, we would obtain $-\frac{-2\sin2x}{\cos2x}$2sin2xcos2x

It follows that $\int\frac{6\sin2x}{\cos2x}\ \mathrm{d}x=-3\log_e\cos2x+c$6sin2xcos2x dx=3logecos2x+c and this can be confirmed by differentiating the expression on the right.


Example 2

Find the general primitive for $f(x)=\sin x\cos x$f(x)=sinxcosx.

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

$F(x)=\int u\frac{\mathrm{d}u}{\mathrm{d}x}\ \mathrm{d}x$F(x)=ududx dx

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

Thus, $F(x)=\frac{u^2}{2}+c=\frac{1}{2}\sin^2x+c$F(x)=u22+c=12sin2x+c which, again, can be confirmed by differentiation.

Worked Examples

Question 1

Find the primitive of $\frac{\cos x}{\sin x}$cosxsinx.

You may use $c$c as a constant.

Question 2

Question 3

Find the primitive of $\frac{\sin x}{1-\cos x}$sinx1cosx.



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

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