What is the integral of (e^{-t}sin(t)i+cot(t)j) ?

The integral of (e^{-t}sin(t)i+cot(t)j) is i 1/2 (-e^{-t}cos(t)-e^{-t}sin(t))+jln|sin(t)|+C

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integral of (e^{-t}sin(t)i+cot(t)j)

\int (e^{- t} sin (t) i + cot (t) j) d t

i \frac{1}{2} (- e^{- t} cos (t) - e^{- t} sin (t)) + j ln ∣ sin (t) ∣ + C

Solution steps

\int e^{- t} sin (t) i + cot (t) j 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 e^{- t} sin (t) i d t + \int cot (t) j d t

\int e^{- t} sin (t) i d t = i \frac{1}{2} (- e^{- t} cos (t) - e^{- t} sin (t))

\int cot (t) j d t = j ln ∣ sin (t) ∣

= i \frac{1}{2} (- e^{- t} cos (t) - e^{- t} sin (t)) + j ln ∣ sin (t) ∣

Add a constant to the solution

= i \frac{1}{2} (- e^{- t} cos (t) - e^{- t} sin (t)) + j ln ∣ sin (t) ∣ + C

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What is the integral of (e^{-t}sin(t)i+cot(t)j) ?
The integral of (e^{-t}sin(t)i+cot(t)j) is i 1/2 (-e^{-t}cos(t)-e^{-t}sin(t))+jln|sin(t)|+C