Question

Help with Trigonometric Substitution? Equation in comments.

Answer 1

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

Answer 2

I'm guessing first step is \[x = 2 \sin \theta, dx = 2 \cos \theta d \theta\]
Then

Thus\[2 \cos \theta = \sqrt{4-x^2}\]

What do I do next?

Answer 3

put it all together

Answer 4

\[\int\limits_{}^{}\frac{ 2\cos \theta }{ (2\sin \theta)^2(2\cos \theta)}d \theta\]

Answer 5

yes

Answer 6

\[\int\limits_{}^{}\frac{ 1 }{ (2\sin \theta)^2}d \theta\]

Answer 7

\[\int\limits_{}^{}\frac{ 1 }{ 4 \sin^2 \theta }d \theta\]

Answer 8

\[(\cot \theta )' = - \csc^2 \theta\]

Answer 9

\[\frac{ 1 }{ 4 }\int\limits_{}^{} \frac{ 1 }{ \sin^2 \theta} d \theta = -\frac{ 1 }{ 4 }\cot \theta + C\]

Answer 10

that's what i make it too

Answer 11

\[\cot \theta = \frac{ \sqrt{4-x^2} }{ x } \rightarrow -\frac{ \sqrt{4-x^2} }{ 4x } + C\]

Answer 12

crumbs, that was quick.

but yes, i got that too....from \(\tan x = \dfrac{x}{\sqrt{4 - x^2}}\)

Answer 13

ooop. \(\theta \), not \(x\) ....again

Answer 14

Aight thanks. :D