New Zealand
Level 8 - NCEA Level 3

# Finding primitive functions, indefinite integrals, definite integrals (sin/cos)

Lesson

Integration is typically about finding lengths, areas or volumes by summing a great many small pieces of the whole. Remarkably, where continuous and smooth functions are concerned, it turns out to be the inverse operation of differentiation.

Therefore, integration often begins with the search for an antiderivative.

We know, for example, that $\frac{\mathrm{d}}{\mathrm{d}x}\sin x=\cos x$ddxsinx=cosx and so, $\sin x$sinx is an antiderivative for $\cos x$cosx. An antiderivative is not unique since the derivative of $\sin x+k$sinx+k, for any constant $k$k, is also $\cos x$cosx. The general antiderivative in this case, $\sin x+k$sinx+k, is often called a primitive.

For the trigonometric functions, we look for antiderivatives by tracing back from known derivatives. The notation used involves the integral sign, which was introduced by Leibniz in the $17$17th century.

##### Example 1

If $f(x)=\sec^2x$f(x)=sec2x, find an antiderivative $F(x)=\int\sec^2x\ dx$F(x)=sec2x dx

We know from having previously differentiated the tangent function, that $F(x)=\tan x+C$F(x)=tanx+C for any constant $C$C, is an antiderivative for $\sec^2x$sec2x.

To find a definite integral, we first find an antiderivative. Then, we subtract its value at the lower terminal of integration from its value at the upper terminal.

##### Example 2

Evaluate $\int_0^{\frac{\pi}{2}}\cos t\ dt$π20cost dt.

The antiderivative is $\sin t$sint. So, we calculate $[\sin x]_0^{\frac{\pi}{2}}=\sin\frac{\pi}{2}-\sin0=1$[sinx]π20=sinπ2sin0=1

Antiderivatives can be difficult to find and may not exist in a form that can be written in terms of simple functions. Various substitutions and transformations are available that can be helpful.

##### Example 3

Find an antiderivative for $f(x)=\sin x\ \cos x$f(x)=sinx cosx.

We make use of the identity $\sin2x=2\sin x\ \cos x$sin2x=2sinx cosx. So, we require $F(x)=\int\frac{1}{2}\sin2x\ dx$F(x)=12sin2x dx.  You should check by differentiating that $F(x)=-\frac{1}{4}\cos2x$F(x)=14cos2x.

##### Example 4

Evaluate $\int_0^{\frac{\pi}{4}}\tan t\ dt$π40tant dt.

We write $\tan t$tant as $\frac{\sin t}{\cos t}$sintcost. Observing that $-\sin t$sint is the derivative of $\cos t$cost, we make the substitution $u=\cos t$u=cost. Therefore, $\frac{\mathrm{d}u}{\mathrm{d}t}=-\sin t$dudt=sint. and we can substitute for $\sin t$sint and $\cos t$cost to write the antiderivative $-\int\frac{1}{u}\frac{\mathrm{d}u}{\mathrm{d}t}\ dt$1ududt dt

It is established elsewhere that this is the same as $-\int\frac{1}{u}\ du$1u du and we call upon our knowledge of the logarithm function to recognise that this is $-\ln u$lnu.

Therefore, the antiderivative is $-\ln\cos t$lncost (which should be checked by differentiating) and we calculate

$-\left[\ln\cos t\right]_0^{\frac{\pi}{4}}=-\left[\ln\frac{1}{\sqrt{2}}-\ln1\right]=-\ln\frac{1}{\sqrt{2}}\approx0.3466$[lncost]π40=[ln12ln1]=ln120.3466

#### More Examples

##### Example 5

Integrate $\cos6x$cos6x.

You may use $C$C as the constant of integration.

##### Example 6

Integrate $-5\cos\left(\frac{x}{4}\right)$5cos(x4).

You may use $C$C as the constant of integration.

##### Example 7

Calculate $\int_{-\frac{\pi}{6}}^{\frac{7\pi}{6}}\cos3xdx$7π6π6cos3xdx.

### Outcomes

#### M8-11

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

#### 91579

Apply integration methods in solving problems