Question

intergral (r^3)/sqrt(4+r^2) from 0 to 1

Answer 1

\[\int_0^1\frac{r^3}{\sqrt{4+r^2}}dr\]
Have you tried the following substitution?
\[r=2\tan u\]

Answer 2

yes i did

Answer 3

\[dr=2\sec^2 u \;du\\
\text{The integral then becomes}\\
\int_0^{2\tan1}\dfrac{(2\tan u)^2(2\sec^2u)}{\sqrt{4+(2\tan u)^2}}du\]
What can you do with that? And if you have any questions about how you get the above, ask away.

Answer 4

Oh, I've made a slight mistake. In the numerator you should have
\[(2\tan u)^3 \text{ instead of }(2 \tan u)^2.\]

Answer 5

how did you get the bounds of the integral

Answer 6

Ah, that's the second time I've made that mistake. They should be . Sorry!

Since r = 2 tanu, we know that
u = arctan(r/2)
Upper u = arctan(1/2)
Lower u = arctan(0/2) = arctan(0) = 0
So the integral should be fixed now,
\[\int_0^{\arctan2}\frac{(2\tan u)^3 (2\sec^2u)}{\sqrt{4+(2\tan u)^2}}du\]

Answer 7

**They should be 0 and arctan(1/2)**

Answer 8

i get
2^3(1/3(sec^3u)-secu)

Answer 9

as my derivative of the intergral

Answer 10

What do you mean by "derivative of the integral"? You should be getting a number as your answer.

Answer 11

before i minus the lower bound
from the upper bound

Answer 12

my final answer i get
(1/3)((-7sqrt5) +16)

Answer 13

I don't see where the 1/3 is coming from. Here's what you should be getting, step by step.
\[\int_{0}^{\arctan2}\frac{2^4 \tan^3u\sec^2u}{\sqrt{4+4\tan^2u}}du\\
16\int_{0}^{\arctan2}\frac{\tan^3u\sec^2u}{\sqrt{4}\sqrt{1+\tan^2u}}du\\
8\int_{0}^{\arctan2}\frac{\tan^3u\sec^2u}{\sqrt{\sec^2u}}du\\
8\int_{0}^{\arctan2}\frac{\tan^3u\sec^2u}{\sec u}du\\
8\int_{0}^{\arctan2}\tan^3u\sec udu\\
8\int_{0}^{\arctan2}\tan^2u\sec u\tan udu\\
8\int_{0}^{\arctan2}(\sec^2u-1)\sec u\tan udu\\
8\int_{0}^{\arctan2}\left[\sec^2u\sec u\tan u-\sec u\tan u\right]du\\
8\left[\int_{0}^{\arctan2}\sec^2u\sec u\tan u\;du-\int_{0}^{\arctan2}\sec u\tan u\;du\right]\]
For the first integral,
let t = secu,
dt = secu tanu du
\[8\left[\int_{1}^{\sqrt5}t^2\;dt-\int_{0}^{\arctan2}\sec u\tan u\;du\right]\]
The rest is pretty easy.

Answer 14

i used 8*intergral of t^2-1 dt

Answer 15

when i intergrate that i get
1/3t^3-t

Answer 16

Okay. What about the 1, in (t² - 1)? Is that what secu tanu becomes?

Answer 17

doesnt 1 become t when you int. and t=secu

Answer 18

Not quite. The second integral,
\[\int_0^{\arctan2}\sec u\tan u \;du = \left[\sec u\right]_0^{\arctan2}\]
If you were to use the substitution t = secu, you'd have
\[[t]_1^{\sqrt5}\]
Besides, it's not clear how you get t²-1 in the first place.
You have the int(t² dt) - int(secu tanu du), and neither of these terms reduce to 1.