Undergrad What's my mistake in this integration problem?

[murshid_islam]

TL;DR

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?

 

[PeroK]

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.

 

[murshid_islam]

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?

 

[dextercioby]

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.

 

[murshid_islam]

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 $u = \arccos(\sin x)$?

 

[PeroK]

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.

 

[PeroK]

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

 

[PeroK]

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.

 

[PeroK]

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:
$$\int_0^{2\pi} \cos^{-1}(\sin x) \ dx = \int_0^{2\pi} \cos^{-1}(\cos(x - \frac \pi 2)) \ dx$$$$= \int_{-\frac \pi 2}^{\frac {3\pi} 2} \cos^{-1}(\cos u) \ du = \pi^2$$