Question

what the integration of sqrt(16-x^2) dx. i try solving it but im not getting the answer.

Answer 1

\[I=\int\limits \sqrt{16-x^2}*1dx\]
integrate by parts.

Answer 2

you got it or should i proceed.

Answer 3

integration by parts will produce a bigger mess (unless your thinking of some other variable substitution, I dunno could be missing your strategy), the derivative of a square root function is another square root function but in the denominator. Use substitution. u= [sqrt(16-x^2)]^2, then du = 2x, that should clean it up to a point where you can evaluate it

Answer 4

x/sqrt(16-x^2) from int by parts actually will make substitution easier

Answer 5

i know what x already

Answer 6

x = 4*sin theta , dx=4*cos theta

Answer 7

\[I=\sqrt{16-x^2}.x-\int\limits \frac{ -2x*x }{2\sqrt{16-x^2} }dx\]
\[=x \sqrt{16-x^2}-\int\limits \frac{ 16-x^2-16 }{\sqrt{16-x^2} }dx\]
\[I=x \sqrt{16-x^2}-\int\limits \sqrt{16-x^2}dx+16\int\limits \frac{ dx }{\sqrt{4^2-x^2} }\]
\[2I=x \sqrt{16-x^2}+16\sin^{-1} \frac{ x }{4 }+c\]
find I

Answer 8

you need to use trigonometric substitution. you are doing more work

Answer 9

ok.if you are more comfortable with trigonometric substitution you can solve that way.
My aim was to tell another method.

Answer 10

you forgot to multiplied the 1/2 on the x*sqrt(16-x^2)

Answer 11

i have calculated 2I
so to find I divide by 2

Answer 12

\[\int\limits \sqrt{16-x^2}dx=I\]