MHB 303 AP calculus int with initial value

[karush]

View attachment 9383
ok image to avoid typo... try to solve before looking at suggested solutions

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Spoiler

so then we have first $$\displaystyle y=\int 2\sin x \,dx=-2cos x +C$$ if $y(\pi)=1$
then
$$y(\pi)=-2cos(\pi) +C=1 $$
then
$$2cos(\pi)+1=C $$
and
$$2(-1)+1=C=-1$$
and finally
$$y=-2cos{x}-1$$
which is $\textbf{(E)}$

ok I think you could do this by observation if you are careful with signs

 

[Monoxdifly]

Spoiler

$$\frac{dy}{dx}=2sinx$$
$$y=-2cosx+C$$
$$y(\pi)=1$$
$$-2cos\pi+C=1$$
$$-2(-1)+C=1$$
$$2+C=1$$
$$C=-1$$
$$y=-cosx+C$$
$$y=-cosx-1$$
I got E as the answer.