What's my mistake in this integration problem?

Here's the problem: ##\int_0^{2\pi} \cos^{-1}(\sin(x)) \mathrm{d}x##
If I do the substitution ##u = \sin(x)##, both the limits of integration become 0 and the integral would result in 0. But the graph of the function tells me the area isn't 0. Where am I going wrong?

murshid_islam said:

TL;DR Summary: My calculation of the integral leads to 0 when I do u-substitution. But from the graph of the function, I can see that the area is obviously not 0.

Here's the problem: ##\int_0^{2\pi} \cos^{-1}(\sin(x)) \mathrm{d}x##
If I do the substitution ##u = \sin(x)##, both the limits of integration become 0 and the integral would result in 0. But the graph of the function tells me the area isn't 0. Where am I going wrong?

The new variable in a substitution must be an interval. Otherwise, with an appropriate substitution you could make the limits equal on any integral.

Ok, I tried it with integration by parts with ##u = \cos^{−1}(\sin(x))## and ##\mathrm{d}v = \mathrm{d}x##, which gives us ##\mathrm{d}u = \frac{−\cos(x)}{\sqrt{1−\sin^2 x}}\mathrm{d}x = −\mathrm{d}x## and ##v=x##.
The integral becomes ##x\cos^{−1}(\sin(x)) + \int x \mathrm{d}x = x\cos^{-1}(\sin(x)) + \frac{x^2}{2} + C##
Evaluating that from 0 to ##2\pi##, we get ##\left(2\pi\cos^{−1}(0) − 0\right) + \frac{1}{2}\left(4\pi^2 − 0\right) = \pi^2 + 2\pi^2 = 3\pi^2##

But the actual answer seems to be ##\pi^2##. What am I missing?

That arccos is periodic, of period exactly ##2\pi## and that also sin is periodic. So you must be careful when computing the derivative of ##u## between the limits.

dextercioby said:

That arccos is periodic, of period exactly ##2\pi## and that also sin is periodic. So you must be careful when computing the derivative of ##u## between the limits.

The range of sin(x) is the interval [-1, 1], which is also the domain of arccos(x). I'm not being able to figure out how that will affect ##\mathrm{d}u##. What will ##\mathrm{d}u## be if ##u = \arccos(\sin x)##?

murshid_islam said:

The range of sin(x) is the interval [-1, 1], which is also the domain of arccos(x). I'm not being able to figure out how that will affect ##\mathrm{d}u##. What will ##\mathrm{d}u## be if ##u = \arccos(\sin x)##?

You have to split the integral, so that ##u## in each case takes values from a simple interval. We look at the values of ##\sin x##:

##x \in [0, \frac \pi 2]## gives ##\sin x \in [0, 1]##
##x \in [\frac \pi 2, \frac {3\pi} 2]## gives ##\sin x \in [1, -1]##
##x \in [\frac {3\pi} 2, 2\pi]## gives ##\sin x \in [-1, 0]##

To illustrate the point. Consider the integral: ##\int_0^1 x \ dx## and use the substitution ##u = x(1-x)##. This transforms the integeral to: ##\int_0^0 f(u) \ du## for some function ##f##, which you can work out. But, we've screwed the limits on the integral.

What about using ##\sin x = \cos(x - \frac \pi 2)##?

murshid_islam said:

But the actual answer seems to be ##\pi^2##. What am I missing?

I agree with the answer ##\pi^2##. As above, the integrand takes a different form across the interval of integration. So, you have to identify this and split the integral up.

Here's a quick way. Note that the function ##\cos^{-1}(\cos x)## has period ##2\pi##:

For ##x \in [0, \pi]## we have ##\cos^{-1}(\cos x) = x##

For ##x \in [\pi, 2\pi]## we have ##\cos^{-1}(\cos x) = 2\pi - x##

A graph of this function is a triangle of base ##2\pi## and height ##\pi##, hence area ##\pi^2##. Hence:
$$\int_0^{2\pi} \cos^{-1}(\cos x) \ dx = \pi^2$$As the function is periodic, for any ##a##, we have:
$$\int_a^{a + 2\pi} \cos^{-1}(\cos x) \ dx = \pi^2$$Finally, using ##\sin x = \cos (x - \frac \pi 2)##, we have:
$$\int_0^{2\pi} \cos^{-1}(\sin x) \ dx = \int_0^{2\pi} \cos^{-1}(\cos(x - \frac \pi 2)) \ dx$$$$= \int_{-\frac \pi 2}^{\frac {3\pi} 2} \cos^{-1}(\cos u) \ du = \pi^2$$