Question

Use trigonometric substitution to find the anti-derivative x^3(4-x^2)^3/2 dx

Answer 1

I'm so tired to type everything...best is to use substitution x=2sint so dx=2costdt....plug this into your function and you'll get integral nice to solve :D

Answer 2

\[\int x^3(4-x^2)^{3/2}dx\]\[x=2\sin tdt\to dx=2\cos tdt\]\[\int2^3\sin^3t(4-4\sin^2t)^{3/2}2\cos tdt\]\[=2^7\int\sin^3(1-\sin^2t)^{3/2}\cos tdt\]\[=2^7\int\sin^3t(\cos^3t)\cos tdt\]\[=2^7\int\sin t(1-\cos^2t)\cos^4tdt\]\[=2^7\int\cos^4t\sin t-\cos^6\sin tdt\]now it's just a u-sub and recognizing how to get back to x using the triangle

Answer 3

here's your triangleyou should be able to take it from here

Answer 4

I got the triangle but this is where i'm stuck, I'm doing something wrong with the rest of the calculations

Answer 5

can you show me what you did?

Answer 6

I got \[8.16\int\limits_{?}^{?}\sin^3t \cos^4tdt\]

Answer 7

how did you get a decimal outside
the rest is right, and I showed how to integrate it above

Answer 8

you wind up pulling 2^3=8 out of the parentheses after the sub\[(4-(2\sin t)^2)^{3/2}=(4-4\sin^2t)^{3/2}=4^{3/2}(1-\sin^2t)^{3/2}\] \[=(2^2)^{3/2}(1-\sin^2t)^{3/2}=2^3(1-\sin^2t)^{3/2}=2^3\cos^3t\]so the whole integral is\[\int(2\sin t)^32^3\cos^3t(2\cos t)dt=2^7\int\sin^3t\cos^4tdt\]now strip out a sin t and convert the remaining sin^2t to (1-cos^2t)...

Answer 9

\[\dots=2^7\int\sin^2t\cos^4t\sin tdt\]\[=2^7\int(1-\cos^2t)\cos^4t\sin tdt\]\[=2^7\int\cos^4t\sin t-\cos^6t\sin tdt\]

Answer 10

Oh that makes sense now

Answer 11

What did you get for a final answer ?

Answer 12

\[u=\cos t\to du=-\sin tdt\]\[2^7\int\cos^4t\sin t-\cos^6t\sin tdt=2^7\int u^6-u^4du\]\[=\frac{2^7}7\cos^7t-\frac{2^7}5\cos^5t+C\]\[=\frac{2^7}7\frac{(4-x^2)^{7/2}}{2^7}-\frac{2^7}5\frac{(4-x^2)^{5/2}}{2^5}+C\]\[=(4-x^2)^{5/2}[\frac17(4-x^2)-\frac45]+C\]\[=-\frac1{35}(4-x^2)^{2/5}(8+5x^2)+C\]there we go!

Answer 13

I got \[(4-x^2)^{5/2}\] for the final answer, I'm guessing that's what you wanted to put right ?