Question

Calculus - need work shown

Answer 1

$\frac{ - 1}{2} [\cos(8x(x + 3))]+ c$

Answer 2

Answer:

$\large -\dfrac{1}{2}\cos (8x^2+24x) + C$

Step-by-step explanation:

$\large \displaystyle\begin{aligned}\textsf{let }\:u & =8x^2+24x\\\\\implies \dfrac{du}{dx} & =16x+24\\ & =2(8x+12)\\\\\implies dx & =\dfrac{1}{2(8x+12)}\: du\\\end{aligned}$

$\large\displaystyle\begin{aligned}\int (8x+12) \sin (8x^2+24x)\:dx & = \int (8x+12) \sin (u) \cdot \dfrac{1}{2(8x+12)}\:du\\\\& = \int \dfrac{(8x+12)\sin (u)}{2(8x+12)}\:du\\\\& = \int \dfrac{1}{2}\sin (u)\:du\\\\& = -\dfrac{1}{2}\cos (u) + C\\\\& = -\dfrac{1}{2}\cos (8x^2+24x) + C\\\\\end{aligned}$