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Integral Problems Discussion

In this post, I would like to write a solution to an integral problem (possibly more problems in the future) as well as give insights that are usually not taught, in spite of its importance. Without any longer, let us delve into the problem.

Problem 1

Find the value of the integral:
[ \int \frac{\sqrt{4x-x^2}}{x} dx ]

Insight 1

First of all, in dealing with integrals, we should usually try to avoid forms with square roots. Moreover, it is usually harder to manipulate the equation if we have multiple $x$ terms in a square root.
Thus, in this problem, the most troublesome part is $\sqrt{4x-x^2}$.
After knowing this, several ideas to simplify this form might come up. 
The first most common ways to deal with $x$ and $x^2$ terms is by completing the square. In this case, $4x-x^2$ can be written as $-(x-2)^2+4$. However, the problem with this approach is that the denominator is $x$, so if we want to do a substitution of $u=x-2$, the denominator would be in the form of $u+2$, which would not help us in solving this problem. If the denominator were $x-2$, this approach might be an appropriate way to solve the problem. So, this approach does not simplify much the problem.
Since completing the square does not work here, we should try another method. Notice that we can factor out $4x-x^2$ into $x$ and $4-x$. Conveniently, we can cancel out the $x$ terms from the numerator and denominator as the following.
[ \int \frac{\sqrt{4x-x^2}}{x} dx ]
[ = \int \frac{\sqrt{x} \cdot \sqrt{4-x}}{(\sqrt{x})^2} dx ]
[ = \int \frac{\sqrt{4-x}}{\sqrt{x}} dx ]
At first, this might not look simpler. Nonetheless, we have achieved our first goal, which is to remove the two $x$ terms from the square root.

Insight 2

Now, our next goal is to remove the square root entirely. We can do this by using substitution. (In many cases, square roots correlates to trigonometric substitutions.)
The hard part is now is the substitution that we want. Remember, our goal is to remove the square root entirely, so we should make both $\sqrt{4-x}$ and $\sqrt{x}$ an integer. As such, $4-x$ and $x$ should both be a perfect square. Looking at this, you might recall a somewhat similar form $sin^2u$ and $1-sin^2u$, which are both perfect squares, because $1-sin^2x=cos^2x$. Adjusting to the given $4$, we also get that $4sin^2u$ and $4-4sin^2u$ are also perfect squares. Therefore, $4sin^2u$ is the perfect substitution for this. By considering $x=4sin^2u$, we can get:
[ x=4sin^2u]
[ dx = 8 \cdot sinu \cdot cosu \cdot du]
[ \int \frac{\sqrt{4-4sin^2u}}{\sqrt{4sin^2u}} dx ]
[ = \int \frac{2cosu}{2sinu} \cdot 8 \cdot sinu \cdot cosu \cdot du]
[ = 8 \int \cos^2{u} \cdot du]

Final Wrap Up

This is a common form which can be easily solved by considering the identity $cos^2u=\frac{cos2u+1}{2}$. Thus, we finish off and get our answer.
[8 \int \cos^2{u} du]
[ = 8 \int \frac{cos2u+1}{2} du]
[ = 4 \int (\cos2u+1) \cdot du]
[ = 2sin2u + 4u + C]

Dont forget to substitute back to $x$. Recall that $x=4sin^2u$. 
Several important substitutions:
[sinu = \frac{\sqrt{x}}{2}]
[u=\sin^{-1}(\sqrt{\frac{x}{4}})]
[cos^2u = 1-sin^2u = 1-\frac{x}{4}]
[cosu = \frac{\sqrt{4-x}}{2}]
[sin2u = 2 \cdot sinu \cdot cosu =\frac{\sqrt{4x-x^2}}{2}]
So, we get that
[2sin2u + 4u + C]
[ =  \sqrt{4x-x^2} + 4 \cdot sin^{-1}(\sqrt{\frac{x}{4}}) + C]

Problem 2

Find the value of the integral:
[ \int sec^3{x} \ dx ]
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