Question

check my work, integrals

Answer 1

\[\int\limits_{0}^{\pi/8}\sin2xdx\]

Answer 2

=-cos2x
-cos(2(pi/8-0))
is that right

Answer 3

No, that's incorrect. o.o

Answer 4

for some reason not getting right answer

Answer 5

There's something missing..

Answer 6

ok wha

Answer 7

Try u = 2x before doing the integral.

Answer 8

@Vandreigan havent gotten that far yet

Answer 9

Well the general should be \(\LARGE -\frac{1}{2} cos~2x+C\)

Answer 10

and you drop the c, correct?

Answer 11

Yea, drop it, plug in your limits and use the fundamental theorem

Answer 12

so -1/2 cos(2(pi/8-0))

Answer 13

Do you know the theorem?

Answer 14

u = 2x , du= 2dx

Answer 15

\[\int\limits_0^{\pi/8} \sin(2x)dx = \frac{1}{2} \int\limits_0^{\pi/4} \sin(u)du\]

if u = 2x, du = 2dx

Now try the integral.

Answer 16

fb-fa

Answer 17

2+2=4

Answer 18

\[-\sqrt{2}/4\]

Answer 19

that not right either looking for

Answer 20

\[(2-\sqrt{2})/4\]

Answer 21

?

Answer 22

Cos(0) = 1, not 0