What is the integral of (cos(pi)ti+sin(pi)tj+tk) ?

The integral of (cos(pi)ti+sin(pi)tj+tk) is (kt^2)/2-(it^2)/2+C

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integral of (cos(pi)ti+sin(pi)tj+tk)

\int (cos (π) t i + sin (π) t j + t k) d t

\frac{k t ^{2}}{2} - \frac{i t ^{2}}{2} + C

Solution steps

\int cos (π) t i + sin (π) t j + t k d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int cos (π) t i d t + \int sin (π) t j d t + \int t k d t

\int cos (π) t i d t = - \frac{i t ^{2}}{2}

\int sin (π) t j d t = 0

\int t k d t = \frac{k t ^{2}}{2}

= - \frac{i t ^{2}}{2} + 0 + \frac{k t ^{2}}{2}

Simplify

= \frac{k t ^{2}}{2} - \frac{i t ^{2}}{2}

Add a constant to the solution

= \frac{k t ^{2}}{2} - \frac{i t ^{2}}{2} + C

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What is the integral of (cos(pi)ti+sin(pi)tj+tk) ?
The integral of (cos(pi)ti+sin(pi)tj+tk) is (kt^2)/2-(it^2)/2+C