# A Surprising Integral

I had a fun encounter with an innocuous looking integral.

It all started with a simple directive:  evaluate $\int{cos(\sqrt{x}) \thinspace dx}$.

Integration is often tricky business.  Although there is a large body of integration techniques, there isn’t really one guaranteed procedure for evaluating an integral.  If you see what the answer is, you write it down; if you don’t, you try a technique in the hope that it makes you see what the answer is.  If that technique doesn’t work, you try another.

This particular problem is interesting in that it highlights a strange phenomenon that occasionally pops up in problem-solving:  sometimes making a problem look more complicated actually makes it easier to solve.

Let $u = \sqrt{x}$.  Thus, $du = \frac{1}{2\sqrt{x}} dx$, and so $dx = 2 \sqrt{x} du$.  But since $u = \sqrt{x}$, we have $dx = 2 \thinspace u \thinspace du$.

This gives us $\int{cos(\sqrt{x}) \thinspace dx} = \int{2 \thinspace u \thinspace cos(u) \thinspace du}$.

This actually looks a bit more difficult than the original problem, but now we can easily integrate using Integration by Parts!

After applying this technique, we’ll get $\int{2 \thinspace u \thinspace cos(u) \thinspace du} = 2 \thinspace u \thinspace sin(u) + 2 \thinspace cos(u) + C$.  And so, after un-substituting, we get $\int{cos(\sqrt{x})} = 2\sqrt{x} \thinspace sin(\sqrt{x}) + 2 \thinspace cos(\sqrt{x}) + C.$

I was surprised that this technique worked, so I actually differentiated to make sure I got the correct answer.  You can take my word for it, or you can verify with WolframAlpha.

One of the best parts of being a teacher is learning (or re-learning) something new every day!

Categories: Teaching

#### patrick honner

Math teacher in Brooklyn, New York #### Joshua Bowman (@Thalesdisciple) · June 8, 2012 at 10:34 pm

I like cos(sqrt(x)), because it looks like it should only be defined for positive x, but since cos(x) is even and analytic, you get a power series that works for all values of x. IIRC, the function defined by this series has a surprising graph for negative x. #### Timothy Gowers · February 27, 2013 at 8:59 am

A technique I like that works here is “successive approximation”. Suppose we start with a naive guess of sin(sqrt(x)), chosen because the sin will at least differentiate to the cos. Then the chain rule gives us an unwanted factor of (2x)^{-1/2}. So to compensate for that we go for (2x)^{1/2}sin(sqrt(x)). But now the product rule gives us an unwanted term of x^{-1/2}sin(sqrt(x)). At this point we can observe that the factor x^{-1/2} is exactly what we need to make the unwanted term easy to integrate. #### MrHonner · February 28, 2013 at 11:33 am

Wow. That’s really cool. It strikes me like integration by parts from another perspective: you are essentially replacing one integral with the sum of a function and another integral, hoping that the new integral is easier to compute, and you find it by playing around with the product rule and chain rule.

I just tried to use this technique to integrate f'(g(x)) in general, and noticed that the “second” approximation will work when g”(x) / (g'(x))^3 is a constant. This is precisely the situation we have here with g(x) = sqrt(x). #### Drew Armstrong · January 13, 2017 at 8:53 pm

I agree that the best part of teaching is learning.

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