Question

Help Trigonometric substitution
\[\int\limits_{0}^{5/2}\frac{ dx }{ \sqrt{25-x ^{2}} }\]

Answer 1

so which substitution u tried ?
hint : for a^2-x^2 try x = a sin t

Answer 2

ok so \[\frac{ dx }{ \sqrt{25-(5\sin \theta)^{2} }}\]

Answer 3

you need to find dx too!

Answer 4

x = 5 sin t
dx = .... ?

Answer 5

dx=5cost

Answer 6

so \[\frac{ 5\cos \theta }{ \sqrt{25\cos ^{2}\theta } }\] correct?

Answer 7

yes!

Answer 8

and now what?

Answer 9

simplify the denominator!
whats square root of 25 ?

Answer 10

so 5cos t in the denominator?

Answer 11

yes!

Answer 12

so wouldn't that simplify to 1?

Answer 13

absolutely.

Answer 14

but we r not done yet!

Answer 15

we need to change the limits of integration too, right ???

Answer 16

to?

Answer 17

pi/2, -pi/2?

Answer 18

how ?
when x= 0
what is t ?
x = 5 sin t

Answer 19

0

Answer 20

correct!
when x= 5/2
what is t ?
x = 5 sin t

Answer 21

pi/6

Answer 22

and there's the answer

Answer 23

yes,
as integral of 1 is just x
and then u put the limits
pi/6-0 = pi/6

Answer 24

oh I see I totally forgot that this was an integral question at its heart. Besides that my substitution skills were a little lacking that is. Ok this makes sense now

Answer 25

Thanks for always being so helpful.

Answer 26

you're welcome ^_^