That's a great question! Let me take a stab at an explanation.

Generally you use trigonometric substitutions when you have a \(\displaystyle\sqrt{{{x}^{2}-{1}}}\)(or something similar) in your integral. If there is an x multiplied outside the square root, then you're fine, just use \(\displaystyle{u}={x}^{2}-{1}\) and then the x outside the square root is taken care of by the du=2xdx.However, when there's not that x there, you have to do something else. One thing that works would be to use one of the trigonometric identities. For example,

\(\displaystyle{{\sin}^{2}{\left({0}\right)}}+{{\cos}^{2}{\left({0}\right)}}={1}\Rightarrow{{\tan}^{2}{\left({0}\right)}}+{1}={{\sec}^{2}{\left({0}\right)}}\)

\(\displaystyle\Rightarrow{{\sec}^{2}{\left({0}\right)}}-{1}={{\tan}^{2}{\left({0}\right)}}\)

Thus you could substitute \(\displaystyle \sec{{\left({0}\right)}}={x}\text{instead of}\ {u}={x}^{2}-{1}\) and then the part under the square root becomes

\(\displaystyle\sqrt{{{{\sec}^{2}{\left({0}\right)}}}}=\sqrt{{{{\tan}^{2}{\left({0}\right)}}}}= \tan{{\left({0}\right)}}\)

which is much simpler to deal with. And then the differential part is much simpler to deal with as well because you just have \(\displaystyle \tan{{\left({0}\right)}} \sec{{\left({0}\right)}}{d}{0}={\left.{d}{x}\right.}\). Notice that there are no stray x's in there you have to find in your integral.

Let's look at an example.Consider the integral

\(\displaystyle\int\frac{{\sqrt{{{x}^{2}-{1}}}}}{{x}}{\left.{d}{x}\right.}\)

Firstly notice that the substitution \(\displaystyle{u}={x}^{2}-{1}\)

won't work because the extra x is on the denominator so it's not directly multiplied by dx. So let's try the substitution I mentioned above.

\(\displaystyle{L}{e}{t} \sec{{\left(\theta\right)}}={x}.\text{Then, taking derivatives of both sides, we see that}\ \tan{{\left(\theta\right)}} \sec{{\left(\theta\right)}}{d}\theta={\left.{d}{x}\right.}\). Substituting this in, we get.

Generally you use trigonometric substitutions when you have a \(\displaystyle\sqrt{{{x}^{2}-{1}}}\)(or something similar) in your integral. If there is an x multiplied outside the square root, then you're fine, just use \(\displaystyle{u}={x}^{2}-{1}\) and then the x outside the square root is taken care of by the du=2xdx.However, when there's not that x there, you have to do something else. One thing that works would be to use one of the trigonometric identities. For example,

\(\displaystyle{{\sin}^{2}{\left({0}\right)}}+{{\cos}^{2}{\left({0}\right)}}={1}\Rightarrow{{\tan}^{2}{\left({0}\right)}}+{1}={{\sec}^{2}{\left({0}\right)}}\)

\(\displaystyle\Rightarrow{{\sec}^{2}{\left({0}\right)}}-{1}={{\tan}^{2}{\left({0}\right)}}\)

Thus you could substitute \(\displaystyle \sec{{\left({0}\right)}}={x}\text{instead of}\ {u}={x}^{2}-{1}\) and then the part under the square root becomes

\(\displaystyle\sqrt{{{{\sec}^{2}{\left({0}\right)}}}}=\sqrt{{{{\tan}^{2}{\left({0}\right)}}}}= \tan{{\left({0}\right)}}\)

which is much simpler to deal with. And then the differential part is much simpler to deal with as well because you just have \(\displaystyle \tan{{\left({0}\right)}} \sec{{\left({0}\right)}}{d}{0}={\left.{d}{x}\right.}\). Notice that there are no stray x's in there you have to find in your integral.

Let's look at an example.Consider the integral

\(\displaystyle\int\frac{{\sqrt{{{x}^{2}-{1}}}}}{{x}}{\left.{d}{x}\right.}\)

Firstly notice that the substitution \(\displaystyle{u}={x}^{2}-{1}\)

won't work because the extra x is on the denominator so it's not directly multiplied by dx. So let's try the substitution I mentioned above.

\(\displaystyle{L}{e}{t} \sec{{\left(\theta\right)}}={x}.\text{Then, taking derivatives of both sides, we see that}\ \tan{{\left(\theta\right)}} \sec{{\left(\theta\right)}}{d}\theta={\left.{d}{x}\right.}\). Substituting this in, we get.