NZ Level 8 (NZC) Level 3 (NCEA) [In development]
Finding primitive functions, indefinite integrals, definite integrals (tan)
Lesson

The expressions indefinite integral, antiderivative and primitive function refer to the same kind of mathematical object. We find primitive functions by anti-differentiation.

Many functions involving the trigonometric functions have primitives that can be found through prior knowledge of a sufficient range of derivatives but there is no general rule for writing the primitive of an arbitrary function in terms of simple or elementary functions, even when it is known to exist.

In this chapter, we find antiderivatives based on the known derivative

$\frac{\mathrm{d}}{\mathrm{d}x}\sin x=\cos x$ddxsinx=cosx

From this, we have

 $\frac{\mathrm{d}}{\mathrm{d}x}\cos x$ddx​cosx $=$= $\frac{\mathrm{d}}{\mathrm{d}x}\sin(\frac{\pi}{2}-x)$ddx​sin(π2​−x) $=$= $\cos(\frac{\pi}{2}-x)\times(-1)$cos(π2​−x)×(−1) $=$= $-\sin x$−sinx

Knowing the derivatives of the sine and cosine functions, we can find the derivative of the tangent function.

 $\frac{\mathrm{d}}{\mathrm{d}x}\tan x$ddx​tanx $=$= $\frac{\mathrm{d}}{\mathrm{d}x}\frac{\sin x}{\cos x}$ddx​sinxcosx​ $=$= $\frac{\cos^2x+\sin^2x}{\cos^2x}$cos2x+sin2xcos2x​ $=$= $\sec^2x$sec2x

We may also need to recognise the derivatives of other functions, including polynomials, the logarithm function and the exponential function.

##### Example 1

Find the primitive $\int3\sec^2\frac{x}{2}\ \mathrm{d}x$3sec2x2 dx.

The $\sec^2$sec2 in the expression immediately suggests that the tangent function belongs in the solution. If $\tan\frac{x}{2}$tanx2 is differentiated, the result is $\frac{1}{2}\sec^2\frac{x}{2}$12sec2x2. So, in order to obtain $3\sec^2\frac{x}{2}$3sec2x2 on differentiation, we must start with $6\tan\frac{x}{2}$6tanx2.

The complete solution is $\int3\sec^2\frac{x}{2}\ \mathrm{d}x=6\tan\frac{x}{2}+C$3sec2x2 dx=6tanx2+C, where $C$C is any constant - called a constant of integration.

##### Example 2

Evaluate the definite integral, $\int_0^{\frac{\pi}{3}}\sec^2t+t^2\ \mathrm{d}t$π30sec2t+t2 dt.

We first find antiderivatives of $\sec^2t$sec2t and $t^2$t2 in the integrand. The simplest antiderivatives are $\tan t$tant and $\frac{t^3}{3}$t33 respectively. So, the required primitive is $\tan t+\frac{t^3}{3}$tant+t33. (In view of the next step, the constant of integration is not needed when evaluating definite integrals.)

We evaluate $\left[\tan t+\frac{t^3}{3}\right]_0^{\frac{\pi}{3}}$[tant+t33]π30.

This is, $\tan\frac{\pi}{3}+\frac{\left(\frac{\pi}{3}\right)^3}{3}-0$tanπ3+(π3)330$=\sqrt{3}+\frac{1}{3}\left(\frac{\pi}{3}\right)^3$=3+13(π3)3

##### Example 3

Evaluate $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\tan t\ \mathrm{d}t$π4π6tant dt.

We find an antiderivative for $\tan t$tant by first writing it as $\frac{\sin t}{\cos t}$sintcost. The occurrence of $\sin t$sint in the numerator and $\cos t$cost in the denominator comes about by differentiating $-\ln\left(\cos t\right)$ln(cost). (Another explanation of this is found here (Example 4}.)

So, we must evaluate

 $\left[-\ln\left(\cos t\right)\right]_{\frac{\pi}{6}}^{\frac{\pi}{4}}$[−ln(cost)]π4​π6​​ $=$= $-\ln\left(\cos\frac{\pi}{4}\right)-\left(-\ln\left(\cos\frac{\pi}{6}\right)\right)$−ln(cosπ4​)−(−ln(cosπ6​)) $=$= $\ln\left(\frac{\sqrt{3}}{2}\right)-\ln\left(\frac{1}{\sqrt{2}}\right)$ln(√32​)−ln(1√2​) $\approx$≈ $0.20273$0.20273

#### More Examples

##### Question 1

State the primitive function of $4\sec^2\left(x\right)$4sec2(x).

You may use $c$c as a constant.

### 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