Question

So I did this math problem but I feel like I might've made a mistake because my answer is just weird

Answer 1

\[14\int\limits_{0}^{1}x^3\sqrt(1-x^2)dx\]

Answer 2

I used trig substitution:

Answer 3

\[\sin \theta =\frac{ x }{ 1 }\]
\[dx=\cos \theta dt \theta\]

Answer 4

\[14\int\limits_{0}^{1}\sin^3 \theta \sqrt(1-\sin^2)* \cos \theta d \theta\]

Answer 5

Simplified:

\[14\int\limits_{0}^{1}\sin^3 \theta \cos^2 \theta d \theta\]

Answer 6

split the sin^3 theta to be able to do U substitution

Answer 7

\[14\int\limits_{0}^{1}\sin^2 \theta \cos^2 \theta \sin \theta d \theta\]

Answer 8

\[u= \cos \theta\]
\[du= - \sin \theta d \theta\]

Answer 9

How's it looking so far?

Answer 10

Gimme a min

Answer 11

How about
\[u=x^2~~~du=2xdx\]

Answer 12

Then,
\[14 \int\limits^1_0 x^3 \sqrt{1-x^2} \, dx \\ \\ 2 \int\limits^1_0 u \sqrt{1-u} \, du\]

Answer 13

Get it?

Answer 14

I'm just splitting the \(x^3\) into \(x^2\) and \(x\), and from the derivation letting \(u=x^2\), \(du=2xdx\), Then \(\frac{du}{2}=xdx\)
\[14 \int\limits\limits^1_0 x^3 \sqrt{1-x^2} \, dx \\ \\ 14 \int\limits\limits^1_0 x^2 \sqrt{1-x^2} (x)\, dx \\ \\ 2 \int\limits\limits^1_0 u \sqrt{1-u} \, du \]

Answer 15

Your u-sub looks great :)
But you have too many sines floating around.
You'll need to apply your Pythagorean Identity,
\(\large\rm 14\int \color{orangered}{sin^2\theta}cos^2\theta~sin\theta~d \theta\)
\(\large\rm 14\int \color{orangered}{(1-cos^2\theta)}cos^2\theta~sin\theta~d \theta\)
And at this point, we should remove the limits of integration.
They are not limits of theta, they were applying to x.

Answer 16

The 2 is actually 7 btw, I made a typo

Answer 17

yeah that is what I did let me show you

Answer 18

\[14\int\limits_{0}^{1}(1-\cos^2 \theta)\cos^2 \theta \sin^2 \theta d \theta\]

Answer 19

You don't have to use trig-sub, from here
\[7 \int\limits\limits\limits^1_0 u \sqrt{1-u} \, du\]
Let \(h=1-u\), \(dh=-du\)
\[7 \int\limits\limits\limits^1_0 (h-1)\sqrt{h}\, dh\]
Integrate it now

Answer 20

\[u= \cos \theta\]
\[du= -\sin \theta \]

Answer 21

But that's the method he/she choose to use.

Answer 22

chose*

Answer 23

\[-14\int\limits_{1}^{\cos(1)}(1-u^2)u^2 du\]

Answer 24

Ooo your limits look strange..
let's just leave them off for now.

Answer 25

\[-14\int\limits_{1}^{\cos(1)}u^2-u^4du= -14[\frac{ u^3 }{ 3 }-\frac{ u^5 }{ 5 }]\]

Answer 26

ohh yeah that part confused me a bit

Answer 27

Let's factor a 15 out, should clean up the fractions,
and distribute the negative,\[\large\rm =\frac{14}{15}(3u^5-5u^3)\]Then we'll undo our u-substitution as a first step of unwinding things, ya?

Answer 28

Factored a 1/15 out* not 15, blah

Answer 29

oh ok so we have \[-\frac{ 14 }{ 15 }[3\cos^5 \theta - 5\cos^3 \theta]\]

Answer 30

No more negative in front,
if we have the 5th power term leading.

Answer 31

ohh right I see now

Answer 32

Ummm to get back to x is another thing...
we can do fancy triangle business...
but, since we were given limits for x, we probably can avoid doing that.

Answer 33

\(\large\rm x=sin\theta\)
x=0: \(\rm 0=sin\theta,\qquad\qquad\theta=?\)
x=1: \(\rm 1=sin\theta,\qquad\qquad\theta=?\)

Answer 34

x= 0, theta= arc sin(0)

x=1 theta= arc sin(1)

Answer 35

x= 0, theta= 0

x=1 theta= pi/2

Answer 36

\[\large\rm \frac{ 14 }{ 15 }[3\cos^5 \theta - 5\cos^3 \theta]_0^{\pi/2}\]Ok good.

Answer 37

I get 28/15

Answer 38

https://www.wolframalpha.com/input/?i=14(integral+from+0+to+1+of+x%5E3+sqrt(1-x%5E2)dx)
yayyy good job \c:/

Answer 39

Thanks! :D