Question

Evaluate the integral.

Answer 1

\[\int\limits_{}^{}\frac{ x^2 }{ \sqrt{16-25x^2} }dx\]

Answer 2

Ooo a fun one :O

Answer 3

\[x = \frac{ 4 }{ 5 }\sin \theta, dx = \frac{ 4 }{ 5 }\cos \theta d \theta \]\[\int\limits_{}^{}\frac{ (\frac{ 4 }{ 5 }\sin \theta )^2 }{ \sqrt{16-25(\frac{ 4 }{ 5 }\sin \theta)^2} }*(4/5\cos \theta d \theta )\]\[\int\limits_{}^{}\frac{ \frac{ 16 }{ 25 }\sin^2 \theta }{ {16-16\sin^2 \theta} }*\frac{ 4 }{ 5 }\cos \theta d \theta \]\[\int\limits_{}^{}\frac{ 64 }{ 125 }*\frac{ \sin^2 \theta \cos \theta }{ \sqrt{16-16\sin^2 \theta } }d \theta \]\[\frac{ 64 }{ 125 } \int\limits_{}^{}\frac{ \sin^2 \theta \cos \theta }{ \cos \theta }d \theta \]\[\frac{ 64 }{ 125 }*\frac{ 1 }{ 4 } \int\limits_{}^{}\sin^2 \theta d \theta \]\[\frac{ 16 }{ 125 } \int\limits_{}^{}\frac{ 1 }{ 2 }(1-\cos2 \theta)d \theta \]\[\frac{ 8 }{ 125 }[ \theta - \frac{ 1 }{ 2 }\sin2 \theta]_{0}^{0.8}\]\[x = \frac{ 4 }{ 5 }\sin \theta, \frac{ 5 }{ 4 }x = \sin \theta, \sin^{-1}(\frac{ 5 }{ 4 }x) = theta \]\[[0\frac{ 8 }{ 125 }[\sin^{-1}(\frac{ 5 }{ 4 }x)-\frac{ 1 }{ 2 }\sin(2\sin^{-1}(\frac{ 5 }{ 4 }x)]]_{0}^{0.8}\]

I forgot to put that the equation is integrating from 0 to 0.8

Answer 4

I plug in stuff, the answer is: 0.10053 -- but how do I get the answer to be in the form of \[\frac{ 4\pi }{ 125 }\]?

Answer 5

elegant solution!!

Answer 6

opps at the last line, erase the 0 by the 8/125

Answer 7

I just focus on method, not pay attention at the calculation. It is soooooooooooo good to me. I learn a new method. And thanks for that.

Answer 8

3rd line, I forgot to put a square root for 16-16sin^2x at the denominator.

Answer 9

You need to apply your Sine Double Angle formula from this point,\[\frac{ 8 }{ 125 }[ \theta - \frac{ 1 }{ 2 }\sin2 \theta]_{0}^{0.8}\]
\[\frac{ 8 }{ 125 }[ \theta - \sin \theta \cos \theta]_{0}^{0.8}\]And then draw a triangle to get these exact sine and cosine values.

Answer 10

How about you find what theta is equal when x=0 and x=0.8
x=0 => theta=0
x=0.8 => theta=pi/2

Answer 11

I am with @math&ing001 about the limits

Answer 12

Oh yes, I suppose if you get boundaries for theta,
then you can skip the triangle business.

Forgot we have a definite integral :p

Answer 13

So while changing the limits of integration,
Upper limit: \[\frac{ 5 }{ 4 }(0.8) = \theta \]

Bottom limit:
\[\frac{ 5 }{ 4 }(0) = \theta\]

?

Answer 14

\(x= \dfrac{4}{5}sin \theta\)
when x = 0 , \(sin \theta =0\rightarrow \theta =0\)

Answer 15

Yea, so b = 1 and a = 0

Answer 16

when \(x= 0.8 = \dfrac{4}{5} sin \theta\), \(sin (\theta ) = \dfrac{0.8*5}{4}=1\)

Answer 17

Hence \(\theta = \pi/2\)

Answer 18

\[\frac{ 8 }{ 125 }[\theta - \sin \theta \cos \theta]_{0}^{\pi/2}\]
Like that?

Answer 19

Yes

Answer 20

And just a note:
You probably should have written x= for the other boundaries so it's clear that you're not plugging those values in for theta.\[\large\rm\frac{8}{125}\left[ \theta - \frac{ 1 }{ 2 }\sin2 \theta\right]_{x=0}^{0.8}\]

Answer 21

Whenever there are definite integrals, we don't use the triangle to find theta?

Answer 22

I still like to do it that way,
but yes, I suppose you can skip it since you can easily change your limits from x to theta.

Answer 23

Okay, lets say I do the triangle way,
\[\frac{ 8 }{ 125 }[\sin^{-1}(\frac{ 5 }{ 4 }x)-(\frac{ 5 }{ 4 }x)(\frac{ \sqrt{15-25x^2} }{ 4 })]\]
What would be the integrations for a and b? From 0 to 0.8?

Answer 24

Thanks all

Answer 25

Yes