 New Zealand
Level 8 - NCEA Level 3

Integration of sin/cos (non-linear functions)

Lesson

The integrations considered in this chapter make use of a pattern that arises when a function of a function is differentiated.

According to the function-of-a-function rule, the derivative of a function $h(x)=f\left(g(x)\right)$h(x)=f(g(x)) is

$h'(x)=f'(g(x).g'(x)$h(x)=f(g(x).g(x)

This is also called the chain rule because of the way it looks when expressed using Leibniz's notation. If $w=g(x)$w=g(x) and $y=f(w)$y=f(w), we have

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}w}\cdot\frac{\mathrm{d}w}{\mathrm{d}x}$dydx=dydw·dwdx.

By reversing this process, we can obtain antiderivatives of the form $\int\ f'(g(x).g'(x)\ \mathrm{d}x$ f(g(x).g(x) dx or, equivalently, $\int\ \frac{\mathrm{d}y}{\mathrm{d}w}\cdot\frac{\mathrm{d}w}{\mathrm{d}x}\ \mathrm{d}x$ dydw·dwdx dx

Example 1

Let $y=2x.\cos(x^2)$y=2x.cos(x2). The antiderivative of this function can be found because $2x$2x is the derivative of $x^2$x2.

The $\cos$cos function is the derivative of the $\sin$sin function and we observe that differentiating $\sin(x^2)$sin(x2) gives $2x\cos(x^2)$2xcos(x2). So,

$\int\ 2x.\cos(x^2)\ \mathrm{d}x=\sin(x^2)+C$ 2x.cos(x2) dx=sin(x2)+C

We can use Leibniz notation to find $\int\ 2x.\cos(x^2)\ \mathrm{d}x$ 2x.cos(x2) dx by first putting $u=x^2$u=x2 and then $\frac{\mathrm{d}u}{\mathrm{d}x}=2x$dudx=2x. On making the substitutions, the integral becomes $\int\ \cos u.\frac{\mathrm{d}u}{\mathrm{d}x}\ \mathrm{d}x$ cosu.dudx dx.

It can be shown rigorously that this is the same as $\int\ \cos u\ \mathrm{d}u$ cosu du. We recognise this as $\sin u+C$sinu+C which is just $\sin(x^2)+C$sin(x2)+C.

Example 2

Find the definite integral $I=\int_0^{\pi}\ \left(3x^2+1\right)\sin(x^3+x)\ \mathrm{d}x$I=π0 (3x2+1)sin(x3+x) dx.

Since $3x^2+1$3x2+1 is the derivative of $x^3+x$x3+x we can write $I=\left[-\cos x^3+x\right]_0^{\pi}$I=[cosx3+x]π0

This is $I=-\cos\left(\pi^3+\pi\right)+\cos0\approx1.917$I=cos(π3+π)+cos01.917.

Example 3

Find the antiderivative $\int\ \tan u\ \mathrm{d}u$ tanu du.

Since $\tan u\equiv\frac{\sin u}{\cos u}$tanusinucosu and $-\sin u$sinu is the derivative of $\cos u$cosu, we try putting $w=\cos u$w=cosu and then $\frac{\mathrm{d}w}{\mathrm{d}u}=-\sin u$dwdu=sinu.

This means $\int\ \tan u\ \mathrm{d}u=\int\ \frac{1}{w}\cdot-\frac{\mathrm{d}w}{\mathrm{d}u}\mathrm{d}u=-\int\ \frac{\mathrm{d}w}{w}$ tanu du= 1w·dwdudu= dww.

So, $\int\ \tan u\ \mathrm{d}u=-\ln w+C=-\ln\cos u+C$ tanu du=lnw+C=lncosu+C.

Worked Examples

Question 1

Consider the function $y=\sin\left(x^5\right)$y=sin(x5).

1. Find the derivative $\frac{dy}{dx}$dydx.

2. Hence find the value of $\int x^4\cos\left(x^5\right)dx$x4cos(x5)dx.

You may use $C$C to represent the constant of integration.

Question 2

Consider the function $y=\cos\left(6x^2-\frac{\pi}{3}\right)$y=cos(6x2π3).

1. Find the derivative $\frac{dy}{dx}$dydx.

2. Hence find the exact value of $\int_{\sqrt{\frac{\pi}{18}}}^{\sqrt{\frac{\pi}{6}}}\left(-12x\sin\left(6x^2-\frac{\pi}{3}\right)\right)dx$π6π18(12xsin(6x2π3))dx.

Question 3

Find the exact value of $I$I, where $I=\int_0^{\frac{\pi^2}{9}}\frac{\sin\sqrt{x}}{\sqrt{x}}dx$I=π290sinxxdx.

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