∫sin2x+3dxPut 2x + 3 = t2.dx = dtdx=12dt=12∫sint dt=12−cos t +c=−12cos2x+3+c

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$\int \sin (2 x + 3) d x =$

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$- \frac{1}{2} \cos (2 x + 3) + c$

$\sin (2 x + 3) + c$

$\frac{1}{2} \sin (2 x + 3) + c$

$\frac{1}{2} \cos (2 x + 3) + c$

answer is D.

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Detailed Solution

$\int \sin (2 x + 3) d x$

Put 2x + 3 = t

2.dx = dt

$d x = \frac{1}{2} d t$

$\begin{array}{l} = \frac{1}{2} \int \sin t d t \\ = \frac{1}{2} (- \cos t) + c \\ = - \frac{1}{2} \cos (2 x + 3) + c \end{array}$

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