- Infinite series
Each integral is a little project of its own, and some are quite complicated to solve. It's also worth remembering that while every function you can write down has an analytical derivative (you can write it down with pencil and paper), Not every function can be integrated analytically. Some have to be done numerically, by techniques we'll learn in another section.
For our first example, let's tackle this integral:
The first think I like to try with any integral is u-substitution, but it just won't work here. The power of both x-terms is 2, so taking the derivative will give us x1, which doesn't help as we substitute for dx.
Now we multiply both sides of this equation by the denominator on the left and cancel appropriately,
to obtain the "basic equation,"
Now the basic equation is good for any value of x, so we can substitute in any way that makes it easy to find our coefficients, A, B, C and D. Letting x = 0 makes it easy to find B:
Substituting x = 1 and x = -1, and using some of the tricks we know from solving linear systems lets us find C:
Now we'll use x = 1 again and x = -2 to make some decisions about A and D:
Now this is tricky, but we see that D has to be 23 times A, but that also works if both A and B are zero, and that's the simplest decomposition we could find. It looks like this:
In blue above, I've found the common denominator and added those two terms just to convince you that the decomposition is really identical to the original integrand. So now we've simplified our integrand into two simpler ones:
The first integral was easy, but the second will be a bit more of a challenge. Remember, it might be more difficult, but at least it's do-able. It wasn't before we decomposed the original integrand. Now (x2 + 25)-1 looks a lot like the derivative of the inverse tangent (see implicit differentiation). It's just that pesky 25 that's a problem. Here's how to work around it. We'll take advantage of the 1/25 out in front.
The u-substitution (see – it's really useful) allows us to easily evaluate the integral. The solution is:
Now let's re-substitute x/5 for u and put the whole integral back together:
Finally, let's just check and see if we get the integrand back when we differentiate our result:
Yup. That's it. What a ride!
Here's an example that uses a combination of u-substitution and trig. substitution
We'll first do a u-substitution here, not for the purpose of directly solving the integral, but for changing it into a form that's more amenable to trig-substitution:
Now we can substitute the appropriate trig. expression for u using the dummy variable t.
Putting the integral together gives:
Now we employ a trick that you'd have to think long and hard about to come up with originally, but give a few days to ponder it, I imagine that you eventually would try something like this:
Now we can do a second u-substitution (though we'll use the letter a here):
That substitution dramatically reduces the integral, and we can solve it.
Here's the solution:
Now we can draw a triangle that reflects our original trig.-substitution, tan(t) = u / 21/2 (below):
... to get the next-to-last version of our solution, the version with u's:
And now our answer is apparent by re-substituting u = 5x:
Here's an example of an integral that needs lots of u-substitution and some trig. identities. Integrate:
We begin by recognizing that the combination of ln(x) and 1/x in any integral makes it a target for u-substitution. There are no guarantees that it will work, but it's worth a try. In this case, it's great:
Here's the substitution:
Now we use two trig. identities to rewrite this integral. The first is tan2(x) = sec2(x) - 1,
... and the second is tan(x) = sin(x)/cos(x):
Now we use substitution once again on each of these integrals. I'll use a and b instead of u to avoid confusion:
So those integrals reduce to pretty simple ones:
Now re-substitute the u's for the a's and b's:
... to get the final result. Lots of functions inside functions here.
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