What is the integral of e^{-θ}cos(3θ) ?

The integral of e^{-θ}cos(3θ) is (3e^{-θ}sin(3θ))/(10)-(e^{-θ}cos(3θ))/(10)+C

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integral of e^{-θ}cos(3θ)

\int e^{- θ} cos (3 θ) d θ

\frac{3 e ^{- θ} sin ( 3 θ )}{10} - \frac{e ^{- θ} cos ( 3 θ )}{10} + C

Solution steps

\int e^{- θ} cos (3 θ) d θ

Apply Integration By Parts

: \frac{1}{3} e^{- θ} sin (3 θ) - \int - \frac{1}{3} e^{- θ} sin (3 θ) d θ

= \frac{1}{3} e^{- θ} sin (3 θ) - \int - \frac{1}{3} e^{- θ} sin (3 θ) d θ

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} e^{- θ} sin (3 θ) - (- \frac{1}{3} \cdot \int e^{- θ} sin (3 θ) d θ)

Apply Integration By Parts

: - \frac{1}{3} e^{- θ} cos (3 θ) - \int \frac{1}{3} e^{- θ} cos (3 θ) d θ

= \frac{1}{3} e^{- θ} sin (3 θ) - (- \frac{1}{3} (- \frac{1}{3} e^{- θ} cos (3 θ) - \int \frac{1}{3} e^{- θ} cos (3 θ) d θ))

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} e^{- θ} sin (3 θ) - (- \frac{1}{3} (- \frac{1}{3} e^{- θ} cos (3 θ) - \frac{1}{3} \cdot \int e^{- θ} cos (3 θ) d θ))

Therefore

\int e^{- θ} cos (3 θ) d θ = \frac{1}{3} e^{- θ} sin (3 θ) - (- \frac{1}{3} (- \frac{1}{3} e^{- θ} cos (3 θ) - \frac{1}{3} \cdot \int e^{- θ} cos (3 θ) d θ))

Isolate

\int e^{- θ} cos (3 θ) d θ

= \frac{3 e ^{- θ} sin ( 3 θ )}{10} - \frac{e ^{- θ} cos ( 3 θ )}{10}

Add a constant to the solution

= \frac{3 e ^{- θ} sin ( 3 θ )}{10} - \frac{e ^{- θ} cos ( 3 θ )}{10} + C

\int \frac{x}{x - 4} d x

\frac{\partial}{\partial x} (r^{2})

\int e^{- s t} \cdot e^{a t} d t

f (x) = \frac{2}{3 x + 5}, (- 1, - 1)

f (x) = x^{2} - 2 x, at x = - 1

What is the integral of e^{-θ}cos(3θ) ?
The integral of e^{-θ}cos(3θ) is (3e^{-θ}sin(3θ))/(10)-(e^{-θ}cos(3θ))/(10)+C