Question

1/x^2sqrt(4-x^2)

Answer 1

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

Answer 2

first let x=2tan u dx= 2sec^2 u

Answer 3

= \[\frac{ \sec^2 u }{ 4\tan^2u \sqrt{4(1-\tan^2 u)} }\]

Answer 4

\[\frac{ 1 }{ 4 }\int\limits_{}^{}\frac{ \sec^2u }{ \tan^2u \sec u }\]

Answer 5

\[\frac{ 1 }{ 4 }\int\limits_{}^{}\frac{ \sec^2u }{ \sec u - \sec^3 u }\]

Answer 6

\[\frac{ 1 }{ 4 }\int\limits_{}^{}\frac{ \sec^2u }{ \sec^2u }\]

Answer 7

\[\frac{ 1 }{ 4 }\int\limits_{}^{}1du\]

Answer 8

\[\frac{ u }{ 4 }+c\]

Answer 9

am i correct?

Answer 10

Your substitution won't be helpful. Instead, use \(x=2\sin u\), you have \(dx=2\cos u~du\).
\[\int\frac{dx}{x^2\sqrt{4-x^2}}=\int\frac{2\cos u}{4\cos^2u\sqrt{4-4\sin^2u}}~du\]
Simplifying:
\[\frac{1}{4}\int\frac{\cos u}{\cos^2u\sqrt{1-\sin^2u}}~du=\frac{1}{4}\int\frac{du}{\cos^2u}=\frac{1}{4}\int\sec^2u~du\]

Answer 11

i made a mistake when i lablelled the question, its actualy\[\frac{ 1 }{ x^2 \sqrt{4+x^2}}\]

Answer 12

thats why i used tan instead of sin

Answer 13

Okay so \(x=2\tan u\), you have \(dx=2\sec^2 u~du\).
\[\int\frac{dx}{x^2\sqrt{4+x^2}}=\int\frac{2\sec^2 u}{4\tan^2u\sqrt{4+4\tan^2u}}~du\]
Simplifying:
\[\int\frac{2\sec^2 u}{4\tan^2u\sqrt{4+4\tan^2u}}~du=\frac{1}{4}\int\frac{\sec u}{\tan^2u}~du\]
No need to change tangent to secant. Instead, use the definition of secant and tangent, you get
\[\frac{\sec u}{\tan^2u}=\frac{\dfrac{1}{\cos u}}{\dfrac{\sin^2u}{\cos^2u}}=\frac{\cos u}{\sin^2u}\]
Finally a simple substitution, \(t=\sin u\) gives \(dt=\cos u~du\), and you have
\[\frac{1}{4}\int\frac{dt}{t^2}\]