Question

∫(x^3)/(16-x^2)^(1/2) from 0 to 4sin(pi/13). I keep getting to 64cos(theta)-(64/3)(cos(theta))^3. I don't even know if that is right, but the bounds are throwing me off.

Answer 1

\[\int\limits_{}^{}\frac{x^{3}}{\sqrt{16-x^{2}}}dx\]
\[u = 16-x^{2} \rightarrow du = -2xdx \rightarrow dx = -\frac{du}{2x}\]
\[=\frac{1}{2}\int\limits_{}^{}\frac{u-16}{\sqrt{u}}du\]
\[=\frac{1}{2}(\frac{2}{3}u^{3/2} - 32\sqrt{u}) = \frac{1}{3}u^{3/2} - 16\sqrt{u}\]
\[u = 16-x^{2}\]
\[\rightarrow \frac{1}{3}(16-x^{2})^{3/2} - 16(16-x^{2})^{1/2}\]
evaluate from 0 to 4sin(pi/13)
\[16 - (4\sin \pi/13)^{2} = 16(1 - (\sin \pi/13)^{2}) = 16(\cos \pi/13)^{2}\]
\[\frac{1}{3}(16(\cos \pi/13)^{2})^{3/2} - 16(16(\cos \pi/13)^{2})^{1/2} = \frac{64}{3}(\cos \pi/13)^{3} -64(\cos \pi/13)^{3} \]
\[\frac{1}{3}(16-0)^{3/2} - 16(16-0)^{1/2} = -\frac{128}{3}\]
\[=-\frac{128}{3}(\cos \pi/13)^{3} + \frac{128}{3} = \frac{128}{3}(1 -(\cos \pi/13)^{3})\]
\[\approx 3.61\]

Answer 2

cow there might be a mistake up there my calculator is saying approximately .05351675

Answer 3

yeah you are right, i'll check my work...

Answer 4

ok i found my mistake.. i made (cos^2)^(1/2) = cos^3 instead of cos
\[\rightarrow \frac{64}{3}(\cos \pi/13)^{3} - 64\cos \pi/13 + \frac{128}{3} \approx 0.05\]

Answer 5

good because i got the wrong answer too lol and i was looking for my mistake

Answer 6

bmlthex did you know you can check your definite integrals using a calculator?

Answer 7

if its a graphing calculator

Answer 8

well yeah if it is a graphing cal.

Answer 9

Thank you so much, guys! I feel like I tried this about 10 times and just couldn't get it. Yes, I did know you could check them. Thanks. I knew what the answer was, but I couldn't get my work to match it, haha. Now I know!

Answer 10

your welcome :)