Question

Find the indefinite integral using trigonometric substitution.

Answer 1

\[\int\limits_{}^{} \frac{ 1 }{ x^2 \sqrt{4-x^2} } dx\]

Answer 2

let x = sin u
then 4-x^2 = cos^2 u

Answer 3

*sqrt(4-x^2)

Answer 4

Yes, I got that, can someone please do this out? :)

Answer 5

When you have something in the form of
\[\sqrt{a^{2}-u^{2}}\]where a is a constant and u is a variable or variable term, you let u = asin(theta) and the entire radical be acos(theta), In your case, a is 2 and u is simply x. So let's rewrite:
\[\int\limits_{}^{}\frac{ 1 }{ (2\sin \theta)^{2}2\cos \theta }du\]I left the du there because we always need to get du as well. And since u = 2sin(theta), du = 2cos(theta). So Ill add that in there now, too.

\[\int\limits_{}^{}\frac{ 1 }{ (2\sin \theta)^{2}2\cos \theta }*2\cos \theta = \frac{ 1 }{ 4 } \int\limits_{}^{}\csc^{2} \theta d \theta\]
Now csc^2 has a basic integration, which is -cot(theta), meaning we're left with after integration:

\[-\frac{ 1 }{ 4 }\cot \theta\] Now we need to convert from theta back into terms of x. Now cot(theta) is cos or sin, and we do have defined values for cos and sin from earlier

u = x = 2sin(theta), therefore sin(theta) = x/2
sqrt(4 - x^2) = 2cos(theta), so cos(theta) = sqrt(4-x^2)/2. Putting all of it together I have:

\[-\frac{ 1 }{ 4 }\frac{ \frac{ \sqrt{4-x^{2}} }{ 2 } }{ \frac{ x }{ 2 } } = -\frac{ \sqrt{4-x^{2}} }{ 4x } + C\]