Constant consistency

Constant consistency

Excerpt: Three different ways to work out an integral give results that appear to be very different. What is going on?

Problem

In how many different ways can you find this? $\displaystyle{\int {\cos (x) \sin (x)} \, dx}$

Some suggested approaches

Could you use a trigonometric identity?

Could you use a substitution?

Could you integrate by parts?

Anything else?

Some solutions

Using a trigonometric identity

The integrand is familiar from the identity $\sin(2x) \equiv 2\sin(x) \cos(x)$ so we could use this to help us.

$$\displaystyle{\int {\cos (x) \sin (x)} \, dx} = \displaystyle{\int \dfrac{1}{2} {\sin (2x)} \, dx} = -\dfrac{1}{4} \cos(2x) + C$$

Using a substitution

If we use the substitution $u=\cos(x)$, then $\dfrac{du}{dx} = -\sin(x)$.

Therefore, $\displaystyle{\int {\cos (x) \sin (x)} \, dx} = \displaystyle{\int {-u} \, du} = -\dfrac{1}{2} \cos^2 (x) + C$

Is this the only possible substitution we could use?

Integration by parts

We have a product of two different functions, so integration by parts may be useful. $\int uv' \, dx = uv - \int u'v \, dx$

We let $u=\sin(x)$ and $v'=\cos(x)$, so that $u'=\cos x$ and $v=\sin x$.

$$\displaystyle{\int {\cos (x) \sin (x)} \, dx}= \sin^2(x) - \displaystyle{\int {\cos (x) \sin (x)} \, dx}$$

The integral on the right hand side is the same as our original. Has this helped?

Let’s define $\displaystyle{\int {\cos (x) \sin (x)} \ dx} = I.$

Then we have $I = \sin^2(x) -I +c.$

Making

$I$

the subject, we have

$$\begin{align*} 2I &= sin^2(x) +c\\ \implies\quad I&= \dfrac{1}{2} \sin^2(x) + C. \end{align*}$$

Would the answer have been different if we had swapped the $u$ and $v'$?

We’ve integrated the function in three different ways and seem to have three different answers. What has happened?

• Would thinking about these solutions graphically help?

• Can we make sense of it algebraically?

Are these the same?

Graphical Approach

From the graphs we can see that each is a translation parallel to the $y$-axis of the same curve. Therefore we can add a constant value to each function and get the other solutions.

We used the same symbol, $C$ for the constant of integration in each case, but in fact they must be different constants.

Algebraic Approach

If the constants are different, let’s call them $P$, $Q$ and $R$ rather than $C$.

So we have the three answers as:

Are there any trigonometric identities that connect the trigonometric functions?

We know that $\cos(2x) \equiv 2\cos^2(x)-1 \equiv 1-2\sin^2(x)$ and $\cos^2(x) \equiv 1-\sin^2(x)$.

Therefore, $-\dfrac{1}{4} \cos(2x) + P = -\dfrac{1}{4} (2\cos^2(x)-1) +P = -\dfrac{1}{2} \cos^2(x) + \dfrac{1}{4} + P.$

Comparing $-\dfrac{1}{2} \cos^2(x) + \dfrac{1}{4} + P$ to $-\dfrac{1}{2} \cos^2 (x) + Q$ we can see they are the same except for the constants. If we equate them then $Q=\dfrac{1}{4} +P$.

Similarly, $-\dfrac{1}{4} \cos(2x) + P = -\dfrac{1}{4} (1-2\sin^2(x)) +P = \dfrac{1}{2} \sin^2(x) - \dfrac{1}{4} + P.$

Comparing again, we can see $R=-\dfrac{1}{4} +P$

So we can see the constant is not the same but the functions are equivalent.

How are the differences in the constants related to the translations of the graphs?

We worked out the integral in three different ways and got what at first seemed to be different answers. What would happen if we used the different methods to evaluate a definite integral such as $\displaystyle{\int_0^{\pi/2} {\cos (x) \sin (x)} \, dx}\text{ ?}$