- #1

karush

Gold Member

MHB

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View attachment 9383

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

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

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

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 $\textbf{(E)}$

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