Integrate cos (4x+3) w.r.t x

$\frac{s i n (4 x + 3)}{4} + c$

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$- \frac{s i n (4 x + 3)}{4} + c$

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sin(4x+3)+c

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-sin(4x+3)+c

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Solution

The correct option is A
$\frac{s i n (4 x + 3)}{4} + c$

Let $u$ =4x+3
$\Rightarrow \frac{d u}{d x} = 4$

$\Rightarrow d x = \frac{d u}{4}$

$I = \int c o s (4 x + 3) d x = \int \frac{c o s u}{4} d u$

= $\frac{1}{4} s i n u + c$

= $\frac{1}{4} s i n (4 x + 3) + c$

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