In how many different ways can you find answers to the following integrals?

Below are some suggested methods for each of the integrals. Did you use any of these ideas? Did you come up with different approaches?

\(\displaystyle {\int_{0}^{\pi} \cos 2x \, dx}\)

When looking at the graph of \(y = \cos 2x\) we see that \(\displaystyle {\int_{0}^{\pi} \cos 2x \, dx} = 0,\) as the shaded areas above and below the \(x\)-axis cancel each other out.

Would your answer differ if the question had asked for the *area* bounded by the curve and the \(x\)-axis between \(0\) and \(\pi\)?

If we know how to find \(\int\cos x \, dx\) then we might consider transforming \(\cos 2x\) into \(\cos x\).

The transformation is a stretch, scale factor 2, parallel to the \(x\)-axis. If we think about the integral \(\int_{0}^{\pi} \cos 2x \, dx\) being stretched by a scale factor of 2, then we see that the shaded area will double and the limits of the integral will change. Therefore we can write

\[\displaystyle {\int_{0}^{\pi} \cos 2x \, dx} = \frac{1}{2}\displaystyle {\int_{0}^{2\pi} \cos x \, dx}.\]

Of course in this case the answer is still zero, but would this approach would work no matter what the limits were?

This is a simple example of integration by substitution. Have a look at Slippery areas for a resource based on these ideas.

\(\displaystyle {\int_{0}}^{\frac{\pi}{2}} \sin^2 x \, dx\)

Sketching the graph of \(y = \sin^2 x\) helps us to see the symmetry in the function. It is an even function and it also has rotational symmetry around points such as \((\frac{\pi}{4}, \frac{1}{2})\). If we consider the rectangle drawn above, which has an area of \(\frac{\pi}{2} \times 1\) we can say that the area beneath the function is half this value. Therefore \[\displaystyle{\int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx = \frac{\pi}{4}}.\]

How can we convince ourselves that \(y = \sin^2 x\) has rotational symmetry?

Trigonometric identities give us a way to transform certain functions into others more easily. Identities that involve \(\sin^2 x\) include \(\sin ^2 x + \cos ^2 x = 1\) and \(\cos 2x = 1 - 2\sin ^2 x.\) Since we already know how to integrate \(\cos 2x\) from above, we can write:

\[\displaystyle {\int_{0}}^{\frac{\pi}{2}} \sin^2 x \, dx = \int_{0}^{\frac{\pi}{2}} \frac{1}{2} - \frac{1}{2}\cos 2x \, dx.\]

Since \(\displaystyle {\int_{0}^{\frac{\pi}{2}} \cos 2x \, dx = 0}\), we are left with \(\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{1}{2} \, dx = \frac{\pi}{4}.\)

\(\displaystyle {\int_{-1}^{1}} \arcsin x \, dx\)

Drawing the graph reminds us that \(\arcsin x\) is an odd function and so \(\displaystyle {\int_{-1}^{1}} \arcsin x \, dx =0.\)

You might also think about the inverse function of \(\arcsin x\) and how that could help you. To explore this idea further, have a look at Inverse integrals.

\(\displaystyle {\int_{-1}^{1}} \arccos x \, dx\)

By drawing \(y = \arcsin x\) and \(y = \arccos x\) on the same graph, we can see that one can be transformed into the other.

From the graph we can write down \(\arccos x = -\arcsin x + \dfrac{\pi}{2},\) so \[\begin{align*} \displaystyle {\int_{-1}^{1}} \arccos x \, dx &= \displaystyle {\int_{-1}^{1}} -\arcsin x + \dfrac{\pi}{2} \, dx \\ &= -\displaystyle {\int_{-1}^{1}} \arcsin x \, dx + \displaystyle{\int_{-1}^{1}} \dfrac{\pi}{2} \, dx \\ &= 0 + \pi \\ &= \pi. \end{align*}\]The rectangle drawn in diagram above has an area of \(\pi \times 2.\) Since the shaded area is half of the rectangle, we can say that \[\displaystyle {\int_{-1}^{1}} \arccos x \, dx = \pi.\]