I have an equation and want to determine the area under the curve, but my calculus days are MANY years behind me, and this seems a more complex problem than I am able to easily figure by referencing my old text books.

What would the integral for this be? $$y=12+12\sin\left(\frac\pi2\cdot\frac{x-12}{12}\right)$$


Hint: $$\int{\sin x \,\mathrm d x} = -\cos x + C$$

substitute $u = \dfrac\pi {24}\left(x-12\right)$, then $\mathrm dx = \dfrac{24}\pi\,\mathrm du$.

EDIT: steps -

$\int [12 + 12\sin(\frac{\pi}{24}(x-12) )] dx$ = $12\int{dx}$ + 12$\int \sin(\frac{\pi}{24}(x-12) ) dx$

substitute in u and du:

12 $\int{dx}$ + 12$\int \frac{24}{\pi}\sin(u) du $ = $12x + \frac{288}{\pi}\int \sin (u) du$ + $C_1$

Since, $\int \sin (u) du$ = -$\cos u$ + $C_2$ = -$\cos(\frac{\pi}{24}(x-12))$ + $C_2$, you can substitute this to get the answer

12x -$\frac{288}{\pi}\cos(\frac{\pi}{24}(x-12))$ + C, where C = $C_1 + C_2$

  • $\begingroup$ correct, not a line, not sure what I was thinking when I typed that $\endgroup$ – GRW Oct 17 '16 at 19:58
  • $\begingroup$ Just figured I would mention that. Not sure, if you wanted to know how to solve the problem or needed the solution. Hope the answer helped. $\endgroup$ – RJM Oct 17 '16 at 20:03
  • $\begingroup$ Really just need the solution for this problem, but would like to know how to solve for future problems $\endgroup$ – GRW Oct 17 '16 at 20:29
  • $\begingroup$ Is there enough there? I can edit and provide more, if you'd like. $\endgroup$ – RJM Oct 17 '16 at 20:30
  • $\begingroup$ Think when I have time to sit down and work through it I could figure it out, but if you could provide more details to simplify the process that would be appreciated. $\endgroup$ – GRW Oct 17 '16 at 20:36

$$ \int 12 + 12\sin(\frac{\pi}{24}(x-12) ) dx $$

$$ \Longrightarrow 12x - \frac{288}{\pi}\cos(\frac{\pi}{24}(x-12) ) + C $$


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