Find $$\int_0^t A(s)ds$$ if $$A(t)=\begin{pmatrix}\sin(t),\cos(t)\\ -\sin(t),\cos(t)\end{pmatrix}$$ I'm a little confused with the format of the question because it asks me to integrate with respect to s, but the function is of t. Do I just integrate A(t) as an indefinite integral, so the answer would be $$A(t)=\begin{pmatrix}-\cos(t),\sin(t)\\ \cos(t),\sin(t)\end{pmatrix}$$ or do I integrate it a definite integral from 0 to t, so the answer would be $$A(t)=\begin{pmatrix}-\cos(t)+1,\ \sin(t)\\ \cos(t)-1,\ \sin(t)\end{pmatrix}$$ Or do I need to do something completely different?

  • $\begingroup$ The integral is with respect to $s$ because $t$ is a limit of integration. $\int_0^t A(t)dt$ doesn't make a whole lot of sense. Typically, we like to do the integral in its own little contextual world, and if we want to define a function based on the result of the integral (ie fundamental theorem of calculus), we tend to make that variable a limit of integration. $\endgroup$ – Emily Jan 20 '14 at 18:50


$$A(t)=\begin{bmatrix}~~~\sin(t) & ~~\cos(t)\\ -\sin(t) & ~~\cos(t)\end{bmatrix}$$


$$\int_0^t A(s)~ds = \int_0^t \begin{bmatrix}~~~\sin(s)& ~~\cos(s)\\ - \sin(s)&~~ \cos(s)\end{bmatrix}~ds = \begin{bmatrix} 1 - \cos t & ~~\sin t\\ \cos t - 1 & ~~ \sin t\end{bmatrix}$$

You are integrating each matrix element, one at a time as a function of $s$ and then applying the limits of integration on that result.

  • $\begingroup$ Thank for the answer, but I meant to have the matrix have 4 separate values, rather than 2. I edited my question to make it clearer. $\endgroup$ – john Jan 20 '14 at 18:51
  • $\begingroup$ I have updated my answer based on the correction to the problem statement. $\endgroup$ – Amzoti Jan 20 '14 at 18:57

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