The value of
\[\int e^{\sin x}\sin 2x\ dx\]is
\( I=Ae^{\sin(x)}\sin (x)+ Be^{\sin(x)}\cos (x)dx\)
Then derivatve \(I= A(cos(x)e^{\sin(x)} sin x+(cosx dx e^{\sin x)}) + ... I=Ae^{\sin(x)}\sin (2x)dx \).
Consider the option A: derivative is \(2 (cos(x) e^{\sin x} sin x + e^{sin(x)} (- cos(x) dy/dx )\) \(2e^{\sin(x)) sinxcosx}\) \( + C \)
The derivative of (A) is \(2e^{\sin x} cosx( sin x-1)dx = sin (2 x dx )\) ...
Therefore, the correct option is (A) 2 \(e^{\sin x} (sin x - 1) + C\)
We want to find $\int e^{\sin x} \sin 2x dx$.
$\int e^{\sin x} \sin 2x dx = \int e^{\sin x} 2 \sin x \cos x dx = 2 \int e^{\sin x} \sin x \cos x dx$.
Let $u = \sin x$, then $du = \cos x dx$. $2 \int e^u u du$.
Integration by parts: $\int u dv = uv - \int v du$.
Let $w = u$, $dv = e^u du$, then $dw = du$ and $v = e^u$.
$2 \int u e^u du = 2 (u e^u - \int e^u du) = 2 (u e^u - e^u) + C = 2 e^u (u-1) + C$
$= 2 e^{\sin x} (\sin x - 1) + C$.