Question

how do i integrate....

Answer 1

\[\int\limits_{0}^{\pi/6}(cosx-2x)dx\]

Answer 2

oops
i wrote it wrong

Answer 3

hold on

Answer 4

\[\int\limits_{0}^{\pi/6}(cosx-\sin2x)\]

Answer 5

alright

Answer 6

gah, forgot dx ..

Answer 7

\[\left[ sinx+0.5\cos2x \right]\]

Answer 8

how did u get 1/2cos2x

Answer 9

idk how to integrate sin2x

Answer 10

\[\int\limits_{0}^{\pi/6}(\cos x-\sin2x)\text dx\]

Answer 11

Derivate -cos2x is 2sin2x
so u have to think of how to get rid of the 2 before sin2x

Answer 12

so multiply 0.5 on -cos2x

Answer 13

how did you know what's the integrate if theres a 2 inside tho, i know all the basic integral and derivatives but idk what to do when its a (2x)

Answer 14

well let 2x=t so that dx =2 dt
just make this substitution and you should have the ans @gedtajia

Answer 15

sorry,,i meant dx= dt/2

Answer 16

there is a fomular derivative f(g(x))=g'(x)*f'(g(x))

Answer 17

if i derive sin2x, wouldnt i get 2cosx

Answer 18

if i use that formula

Answer 19

nonono, it's 2cos2x

Answer 20

oh so i keep the inside

Answer 21

yes

Answer 22

ooo i have to review derivatives >_< thank you

Answer 23

:)