F(x)=integral from 0 to x of sin(2t)dt
Evaluate F(pi)

Question

F(x)=integral from 0 to x of sin(2t)dt
Evaluate F(pi)

anonymous · Answer

$$F(x)=\int\limits_{0}^{x}\sin(2t)dt$$
Evaluate $$F(\pi )$$

miracrown · Answer

the idea is we do the indefinate integral with respect to t 
then evaluate from 0 to pi

anonymous · Answer

so t=pi

$$\int\limits_{0}^{\pi} \sin (2t) dt$$

$$=\left[ -\cos(2\pi)-(-\cos (0)) \right]$$

$$=\left[ -\cos(2\pi)+ 1 \right]$$

anonymous · Answer

sorry forgot something

anonymous · Answer

you best use the rule of sin (2t)= 2 sin t  cos t

anonymous · Answer

then set u=sin t; du= cos t dt -> dt = du/cos t, substitute in the integral, and eliminate what you can. Then solve the integral using  u.du. Then you've only got sin² (t) left. Apply the substraction rule for limited integrals. [sin²(t2)-sin²(t1)] Plug in the limit values pi and 0. Calculate.