Question

\[x=3sin\theta\]
\[\int x^3 \sqrt (9-x^2)\]

Answer 1

where are you stuck?

Answer 2

\[dx=3cos\theta d\theta\]
\[9\int \sin^ 3 \theta \cos\theta \sqrt(9-3\sin^ 2 \theta) d\theta\]

Answer 3

you forgot to square the whole sub\[x=3\sin\theta\implies x^2=9\sin\theta\]

Answer 4

what?

Answer 5

oh sorry

Answer 6

\[9\int \sin^ 3 \theta \cos\theta \sqrt(9-9\sin^ 2 \theta) d\theta\]

Answer 7

ok...i think I see it now

Answer 8

further\[x^3=3^3\sin\theta\]so this should be\[\int3^3\sin^3\theta\cos\theta\sqrt{9-9\sin^2\theta}d\theta=27\int\sin^3\theta\cos\theta\sqrt{9-9\sin^2\theta}d\theta\]now we need to get the 9 out of the radical

Answer 9

\[27\int \sin^ 3 \theta \cos\theta \sqrt(9-9\sin^ 2 \theta) d\theta\]

Answer 10

\[3^3\int\sin^3\theta\cos\theta\sqrt{9(1-\sin^2\theta)}d\theta=3^3\int\sin^3\theta\cos\theta(3)\sqrt{1-\sin^2\theta}d\theta\]\[=3^4\int\sin^3\theta\cos^2\theta d\theta\]so far so good?

Answer 11

yes sir

Answer 12

now it's just u substitution

Answer 13

I'm no "sir" lol

Answer 14

no way! LOL! my bad haha

Answer 15

no it's not because now our cos is to the second power :(
so more trouble ahead... not sure what the best move is actually

Answer 16

reduction formula perhaps

Answer 17

oh I see what to do :)

Answer 18

strip out a sin, and convert the sin^2 to 1-cos^2
then u-sub

Answer 19

sin ^3 = sin * sin ^ 2 = sin * (1 - cos ^ 2)

Answer 20

yeah thats the best way

Answer 21

\[=3^4\int\sin^3\theta\cos^2\theta d\theta\]
u=cos theta
du= sin theta dtheta
\[\int (1-cos\theta)cos^ 2 sin\theta d\theta\]

Answer 22

\[\int (1-u)u^ 2 du\]

Answer 23

\[3^4\int\sin^3\theta\cos^2\theta d\theta=3^4\int\sin^2\theta\cos^2\theta\cdot\sin\theta d\theta\]\[=3^4\int(1-\cos^2\theta)\cos^2\theta\sin\theta d\theta\]

Answer 24

somebody dropped a ^2

Answer 25

there I go again lol

Answer 26

it happens :)

Answer 27

\[\int (u^2 - u^4)du\]

Answer 28

=

Answer 29

lolz \[27(\frac 1 3 cos\theta- \frac 1 5 cos\theta) \]

Answer 30

careful with the signs\[u=\cos\theta\implies du=-\sin\theta d\theta\]

Answer 31

and where did your exponents go ?

Answer 32

\[27(\frac 1 3 cos^3\theta- \frac 1 5 cos^5\theta)\]
I found them...

Answer 33

wait my negative signs

Answer 34

yep, now that's your only problem

Answer 35

\[27(\frac 1 5 cos^5\theta-\frac 1 3 cos^3\theta)\]

Answer 36

right :)
can you put it back in terms of x ?

Answer 37

wait, it's supposed to be 3^4 on the outside

Answer 38

oh no wrong again
it should be 3^5=243

Answer 39

where's the 5th one from?

Answer 40

i think 3^4 is correct

Answer 41

\[\int x^3\sqrt{9-x^2}dx\]\[x=3\sin\theta\implies du=3\cos\theta d\theta\]\[\int(3\sin\theta)^3\sqrt{9-(3\sin\theta)^2}(3\cos\theta d\theta)\]\[=3^4\int\sin^3\theta\cos\theta\sqrt{9-9\sin^2\theta}d\theta\]\[=3^4\int\sin^3\theta\cos\theta(3)\sqrt{1-\sin^2\theta}d\theta\]\[=3^5\int\sin^3\theta\cos^2\theta d\theta\]

Answer 42

u=cos x
du= sinx then


right?

Answer 43

that would also make it easier to sub the x back in

Answer 44

i meant u= cos theta

Answer 45

yeah but you need to use the trig identity first

Answer 46

well of course
\[=3^5\int(1-cos^ 2) \cos^2\theta sin\theta d\theta\]

Answer 47

I'll be right back

Answer 48

I gotta go... thank for all of your help again!

Answer 49

\[\int x^3\sqrt{9-x^2}dx\]\[x=3\sin\theta\implies du=3\cos\theta d\theta\]\[\int(3\sin\theta)^3\sqrt{9-(3\sin\theta)^2}(3\cos\theta d\theta)\]\[=3^4\int\sin^3\theta\cos\theta\sqrt{9-9\sin^2\theta}d\theta\]\[=3^4\int\sin^3\theta\cos\theta(3)\sqrt{1-\sin^2\theta}d\theta\]\[=3^5\int\sin^3\theta\cos^2\theta d\theta\]\[=3^5\int\sin^2\theta\cos^2\theta\sin\theta d\theta\]\[=3^5\int(1-\cos^2\theta)\cos^2\theta\sin\theta d\theta\]\[=3^5\int(\cos^2\theta-\cos^4\theta )\sin\theta d\theta\]\[u=\cos\theta\implies du=-\sin\theta d\theta\implies\sin\theta d\theta=-du\]\[3^5\int(u^4-u^2)du=3^5(\frac15u^5-\frac13u^3)=3^5(\frac15\cos^5\theta-\frac13\cos^3\theta)\]from our initila substitution we get the triangle from which we can find cos in terms of x\[\int x^3\sqrt{9-x^2}dx=3^5\left(\frac15\cdot{(9-x^2)^{5/2}\over3^{5/2}}-\frac13\cdot{(9-x^2)^{3/2}\over3^{3/2}}\right)+C\]which we can simplify a bit