Question

How to integrate cost dt for the interval 0, Pi/6?

Answer 1

\[\int\limits_{0}^{\pi/6}\cos(t)dt=[\sin(t)]\]
let t=pi/6 and let t=0, then you find the different
ans.\[\sin(\pi/6)-\sin(0)=\frac{ 1 }{ 2 }\]

Answer 2

\[\int\limits_{0}^{\pi/6} cost dt =sint \]

pi/6 and 0 are t

\[\sin \pi/6 - \sin 0 = 1/2\]

Answer 3

first do integration of cost dt that is sint t...them do sin(pi/6)-sin0=1/2
anybody tell am i right??