What is the integral of e^{-θ}cos(8θ) ?

The integral of e^{-θ}cos(8θ) is (8e^{-θ}sin(8θ))/(65)-(e^{-θ}cos(8θ))/(65)+C

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integral of e^{-θ}cos(8θ)

\int e^{- θ} cos (8 θ) d θ

\frac{8 e ^{- θ} sin ( 8 θ )}{65} - \frac{e ^{- θ} cos ( 8 θ )}{65} + C

Solution steps

\int e^{- θ} cos (8 θ) d θ

Apply Integration By Parts

: \frac{1}{8} e^{- θ} sin (8 θ) - \int - \frac{1}{8} e^{- θ} sin (8 θ) d θ

= \frac{1}{8} e^{- θ} sin (8 θ) - \int - \frac{1}{8} e^{- θ} sin (8 θ) d θ

Take the constant out:

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

= \frac{1}{8} e^{- θ} sin (8 θ) - (- \frac{1}{8} \cdot \int e^{- θ} sin (8 θ) d θ)

Apply Integration By Parts

: - \frac{1}{8} e^{- θ} cos (8 θ) - \int \frac{1}{8} e^{- θ} cos (8 θ) d θ

= \frac{1}{8} e^{- θ} sin (8 θ) - (- \frac{1}{8} (- \frac{1}{8} e^{- θ} cos (8 θ) - \int \frac{1}{8} e^{- θ} cos (8 θ) d θ))

Take the constant out:

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

= \frac{1}{8} e^{- θ} sin (8 θ) - (- \frac{1}{8} (- \frac{1}{8} e^{- θ} cos (8 θ) - \frac{1}{8} \cdot \int e^{- θ} cos (8 θ) d θ))

Therefore

\int e^{- θ} cos (8 θ) d θ = \frac{1}{8} e^{- θ} sin (8 θ) - (- \frac{1}{8} (- \frac{1}{8} e^{- θ} cos (8 θ) - \frac{1}{8} \cdot \int e^{- θ} cos (8 θ) d θ))

Isolate

\int e^{- θ} cos (8 θ) d θ

= \frac{8 e ^{- θ} sin ( 8 θ )}{65} - \frac{e ^{- θ} cos ( 8 θ )}{65}

Add a constant to the solution

= \frac{8 e ^{- θ} sin ( 8 θ )}{65} - \frac{e ^{- θ} cos ( 8 θ )}{65} + C

\frac{\partial}{\partial x} (\frac{1}{x y ^{2} - x ^{2} y})

x \to 0 lim ((e^{x} + x)^{\frac{1}{x}})

(x + 1) \frac{d y}{d x} + y = x ln^{2} (x)

x \to \infty lim (\frac{( ln ( e ^{3 x} + x ) )}{x})

t \frac{d y}{d t} + 2 y = t^{4}

What is the integral of e^{-θ}cos(8θ) ?
The integral of e^{-θ}cos(8θ) is (8e^{-θ}sin(8θ))/(65)-(e^{-θ}cos(8θ))/(65)+C