Question

INTEGRATION of: -98x/sqrt of 16 -49x^2

Answer 1

for\[\sqrt{a^2-b^2x}\]use the sub\[x=\frac ab\sin\theta\]

Answer 2

oh never mind you can so this with a u-sub

Answer 3

\[u=16-49x^2\]

Answer 4

du=-98x dx ??

Answer 5

yes...

Answer 6

it's too hard 4me

Answer 7

Does this help?
Confused about any step here? :D