Question

http://i1084.photobucket.com/albums/j40 ... tempt3.png
What am I doing wrong?

Answer 1

One second

Answer 2

http://i1084.photobucket.com/albums/j409/QRAWarrior/MATA30/MATA30Y-WA-Attempt3.png

Answer 3

\[\int \frac{r^3}{\sqrt{16+r^2}}dr\]
right?

Answer 4

tan^2 + 1 = sec^2

r=4tan(t) ; dr = 4sec^2(t) dt

\[\int \frac{64tan^3(t)\ 4sec^2(t)}{4sec (t)}dt \]

\[\int 64tan^3(t)\ sec(t)\ dt \]

Answer 5

tan^2=sec^2-1

\[64\int (sec^2(t)-1)\ sec(t)tan(t)\ dt\]

\[64\int (sec^2(t))(sec(t)tan(t))-\frac{sin(t)}{cos^2(t)}dt\]

\[64 \frac{1}{3}sec^3(t)-64\int \frac{sin(t)}{cos^2(t)}dt\]