This antiderivative has arguable the highest ratio of “really hard to compute” to “really easy to write”:

As we’ve seen in this series, the answer is

It turns out that this can be simplified somewhat as long as and . I’ll use the trig identity

When I apply this trig identity for and , I obtain

.

So we can conclude that

for some integer that depends on . The is important, as a cursory look reveals that and have different graphs. (The vertical lines in the orange graph indicate where the right-hand side is undefined when or .

The two graphs coincide when but differ otherwise. However, it appears that the two graphs differ by a constant. Indeed, if I subtract from the orange graph if and add to the orange graph if , then they match:

So, evidently

if ,

if ,

if .

So as long as and , this constant , , or can be absorbed into the constant :

.

However, a picture may be persuasive but is not a proof, and there are some subtle issues with this simplification. I’ll discuss these further details in tomorrow’s post.

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