Question

Integrate sin(2x + pi/6)dx from 0 to pi/6

Answer 1

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

Answer 2

\[-\sqrt{3}/4\]

Answer 3

explain.... :/

Answer 4

again check it...it can't be negative :/

Answer 5

...? I'm lost...how did u arrive at that answer?

Answer 6

\[\sin(2x+\pi) = - \sin(2x)\]

so, we can write it as -sin2x /6

then integrate

Answer 7

i thinks thats wrong

Answer 8

kishan:i think it is sin(2x+(pi/6))dx
so let 2x+pi/6=t;
differnentiaite
2dx=dt
dx=dt/2
so it becomes integral of sin t dt/2
=-(cos t)/2
putting limits we get (cos 0 -cos pi/6)/2
=(1-1/2)/2
=1/4

Answer 9

but cos(\[\pi\]/6)==\[\sqrt{3}\]/2

Answer 10

oh sorry it will be (1-sqrt(3)/2)/2.