Question

Anyone know how to do this?

Answer 1

\[\int\limits_{?}^{?}\frac{ \sin \sqrt{x} }{ \sqrt{x} }\]

Answer 2

Why does it have question marks on top and bottom of the integral...

Answer 3

oh im not sure theres no numbers in the integral it is just suppose to be the integral by itself

Answer 4

\[\int\limits \frac{ \sin \sqrt{x} }{ \sqrt{x} }\]

Answer 5

let u = \(\sqrt x\)

Answer 6

du=1/2x^-1/2 but how would I substitute this into the integral?

Answer 7

\[\mathrm d u=\frac{1}{2}x^{-\frac{1}{2}}\mathrm d x=\frac{1}{2\sqrt{x}}\mathrm d x\]

Answer 8

just substitute , \(\int -1/2 sin u du\)

Answer 9

you can pull the 1/2 (constant) out in front of the integral

Answer 10

As for the lack of upper and lower limits of integration, it just means you need a general antiderivative (we aren't sure of the constants), so the typical convention is to add +C after you integrate

Answer 11

where C is any constant

Answer 12

would the answer be \[-\cos \sqrt{x}+C\]

Answer 13

\[\frac{-1}{2}\int\limits \sin{u}\mathrm d u\] I believe you forgot about the -1/2

Answer 14

er was it positive 1/2

Answer 15

Yeah it's positive; my mistake. Otherwise it's correct, so you should have\[-\frac{1}{2}\cos{\sqrt{x}}+C\]

Answer 16

wait, i may have made another mistake -_-

Answer 17

found it.\[\mathrm d u=\frac{1}{2\sqrt{x}}\mathrm d x\] \[2\mathrm d u=\frac{1}{\sqrt{x}}\mathrm d x\] Sorry, sloppy maths from me tonight ;)
It's a 2 we took in front, not a half.

Answer 18

so its \[2\cos \sqrt{x}+C\]

Answer 19

\[-2\cos{\sqrt{x}}+C\] since the antiderivative of sin(u) is -cos(u)

Answer 20

thank you so much!!! it was correct!!!

Answer 21

My pleasure. Sorry about all the mistakes, hope I didn't confuse you ;)