Question

Evaluate the integral

∫ (7dx)/(√16-49x^2)

Is it ok if I simplify it to ∫(7dx)/(4-7x) ??

Answer 1

No it cannot be simplified that way

Answer 2

\[\int\limits_{}^{}\frac{ 7 }{ \sqrt{16-49x^2} }dx\]

Answer 3

T.T ok

Answer 4

By doing that I don't know if the square root would go away though or would it?? :/

Answer 5

Ya the square root disappears\[(x^a)^b=x^{ab}\]\[\large (x^{\frac{ 1 }{ 2 }})^2=x^1=x\]

Answer 6

I don't think I can do that, hmmm...

Answer 7

oh ok :/ I'll try to google this...lets see

Answer 8

\[\Large \frac{7}{\sqrt{16 - 49x^2}}\]


\[\Large \frac{7}{\sqrt{16(1 - \frac{49}{16}x^2)}}\]


\[\Large \frac{7}{\sqrt{16}\sqrt{1 - \frac{49}{16}x^2}}\]


\[\Large \frac{7}{4\sqrt{1 - \frac{49}{16}x^2}}\]


\[\Large \frac{7}{4\sqrt{1 - \left(\frac{7x}{4}\right)^2}}\]

Answer 9

so

\[\Large \int\frac{7dx}{\sqrt{16 - 49x^2}}\]

is the same as

\[\Large \int\frac{7dx}{4\sqrt{1 - \left(\frac{7x}{4}\right)^2}}\]

\[\Large \frac{7}{4}\int\frac{dx}{\sqrt{1 - \left(\frac{7x}{4}\right)^2}}\]

\[\Large \frac{7}{4}\int\frac{4du}{7\sqrt{1 - u^2}}\]

\[\Large \frac{7}{4}*\frac{4}{7}\int\frac{du}{\sqrt{1 - u^2}}\]

\[\Large \int\frac{du}{\sqrt{1 - u^2}}\]

where u = 7x/4 ---> du = 7dx/4 ---> dx = 4du/7

Answer 10

Wow! Thank you so much! Both of you thanks! =)

Answer 11

from there, you use the idea that if y = arcsin(x), then dy/dx = 1/sqrt(1 - x^2)

Answer 12

but you do it in reverse

Answer 13

once you've integrated, don't forget to replace u with what it is in terms of x

Answer 14

oh and i'm sure you know this, but don't forget the +C (it gets me everytime)

Answer 15

yes thank you so much! I was doing it out right now, I'll post the answer once I'm done. :)

Answer 16

ok, yw

Answer 17

hmm...I got[ (1-u^2)^1.5]/1.5 + c but I think its wrong did I have to include the du=7dx/4?? It's probably all wrong...

Answer 18

QUESTION!! Is this correct?

\[7\int\limits \frac{ dx }{ \sqrt{49}\times \sqrt{\frac{ 16 }{ 49}-x^2}} \]

\[= \frac{ 7 }{ 7 }\int\limits \frac{ dx }{ \sqrt{\frac{ 9 }{ 7 }^2 -x^2} }\]

Answer 19

*BTW the numbers that look like 0 are 9

Answer 20

no you should have arcsine in your answer

Answer 21

\[\Large \int\frac{du}{\sqrt{1 - u^2}} = \arcsin(u) + C\]

Answer 22

\[\Large \int\frac{du}{\sqrt{1 - u^2}} = \arcsin(u) + C\]


\[\Large \int\frac{du}{\sqrt{1 - \left(\frac{7x}{4}\right)^2}} = \arcsin\left(\frac{7x}{4}\right) + C\]


\[\Large \int\frac{7dx}{\sqrt{16 - 49x^2}} = \arcsin\left(\frac{7x}{4}\right) + C\]

Answer 23

would the answer be:

\[\sin ^{-1} (\frac{ x }{ \frac{ 9 }{ 7 } }) +c\]

Answer 24

i agree with the sine inverse, but not sure how you got the 9/7

Answer 25

oops would the final answer be:
\[\sin ^{-1} (7/4 x)+c\]

Answer 26

yep, that's what I wrote above

arcsin = sin^(-1)

Answer 27

ohh lolol ^_^ ok TY!!

Answer 28

np