Question

Integral Help...

Answer 1

Prove that if \[U{n}=\int\limits_{0}^{2a}x ^{n}\sqrt{2ax-x ^{2}}dx=\left( \frac{ 2n+1 }{ n+2 } \right)U _{n-1}\]

Answer 2

put x= 2asin^2 thetha

Answer 3

let \(x= 2a\sin^2 \theta\)
\[\mathrm dx = \]

Answer 4

what do you get for dx in-terms of d(theta)

Answer 5

\[4a \sin \theta \cos \theta d\]

Answer 6

@UnkleRhaukus

Answer 7

so the integral becomes

\[U_{n}=\int\limits_{0}^{2a}x ^{n}\sqrt{2ax-x ^{2}}\,\mathrm dx\\
\qquad =\int\limits_{x=0}^{x=2a}(2a\sin^2\theta)^{n}\sqrt{2a(2a\sin^2\theta)-(2a\sin^2\theta)^{2}}\cdot 4a\sin\theta\cos\theta\,\mathrm d\theta\]

Answer 8

lets simplify the terms under the square root first

\[\sqrt{2a(2a\sin^2\theta)-(2a\sin^2\theta)^{2}}\]

Answer 9

don't forget to change the limits of integration

Answer 10

after factoring out common terms under the sqrt
use
\[\sin^2+\cos^2=1\\\cos=\sqrt{1-\sin^2}\]

Answer 11

are you still here @JOYAL

Answer 12

yes...@UnkleRhaukus

Answer 13

have you been able so simplify that surd?

Answer 14

no...

Answer 15

\[2a(2a\sin^2\theta)\]What does this simplify to be

Answer 16

no ideas...@UnkleRhaukus

Answer 17

\(2a(2a\sin^2\theta)\) simplifies to become \(4a^2\sin^2\theta\)

Answer 18

what does \[(2a\sin^2\theta)^2\] simplify to become ?

Answer 19

remember , its is just \[(2a\sin^2\theta)\times(2a\sin^2\theta)\]

Answer 20

no idea...

Answer 21

?

Answer 22

no...