Question

integrate: 4dx/ (4-x^2)^(3/2)

Answer 1

u= 4-x^2
du =-2xdx
4xdx = -2du

integrate -2 (du/u^(3/2))

Answer 2

@Algebraic! it is 4dx not 4xdx, your substitution is not correct.

Answer 3

so it is...

Answer 4

Correct. its 4 dx.

Answer 5

@shawngomez36@gmail.com use trigonometric substitution method.

Answer 6

for forms\[\sqrt{a^2-b^2x^2}\]let\[x=\frac ab\sin\theta\]

Answer 7

i don't understand the u substitution method. can someone explain it?

Answer 8

\[\cos \theta = \frac{ \sqrt{4-x^2} }{ 2 }\]
\[\cos ^{2} \theta = \frac{ 4-x^2 }{ 4 }\]
\[\frac{ 1}{\cos ^{2} \theta} = \frac{ 4}{ 4-x^2 }\]
\[\frac{ 1}{\cos ^{3} \theta} = \frac{ 4*2}{ (4-x^2)^{3/2} }\]
\[\frac{ 1}{4*2\cos ^{3} \theta} = \frac{ 1}{ (4-x^2)^{3/2} }\]

\[\frac{ x }{ 2 } = \sin \theta \]
\[dx = 2 \cos \theta d \theta \]

\[\frac{ 4dx }{ (4-x^2)^{3/2}} = \frac{ 4*2 \cos \theta d \theta }{ 4*2\cos ^{3} \theta }\]
\[\frac{ d \theta }{ \cos ^{2} \theta } = \sec ^{2}\theta d\theta\]

Answer 9

integrate to get tan(theta) which is x/(sqrt(4-x^2))