What is the integral of e^{-θ}cos(4θ) ?

The integral of e^{-θ}cos(4θ) is (4e^{-θ}sin(4θ))/(17)-(e^{-θ}cos(4θ))/(17)+C

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integral of e^{-θ}cos(4θ)

\int e^{- θ} cos (4 θ) d θ

\frac{4 e ^{- θ} sin ( 4 θ )}{17} - \frac{e ^{- θ} cos ( 4 θ )}{17} + C

Solution steps

\int e^{- θ} cos (4 θ) d θ

Apply Integration By Parts

: \frac{1}{4} e^{- θ} sin (4 θ) - \int - \frac{1}{4} e^{- θ} sin (4 θ) d θ

= \frac{1}{4} e^{- θ} sin (4 θ) - \int - \frac{1}{4} e^{- θ} sin (4 θ) d θ

Take the constant out:

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

= \frac{1}{4} e^{- θ} sin (4 θ) - (- \frac{1}{4} \cdot \int e^{- θ} sin (4 θ) d θ)

Apply Integration By Parts

: - \frac{1}{4} e^{- θ} cos (4 θ) - \int \frac{1}{4} e^{- θ} cos (4 θ) d θ

= \frac{1}{4} e^{- θ} sin (4 θ) - (- \frac{1}{4} (- \frac{1}{4} e^{- θ} cos (4 θ) - \int \frac{1}{4} e^{- θ} cos (4 θ) d θ))

Take the constant out:

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

= \frac{1}{4} e^{- θ} sin (4 θ) - (- \frac{1}{4} (- \frac{1}{4} e^{- θ} cos (4 θ) - \frac{1}{4} \cdot \int e^{- θ} cos (4 θ) d θ))

Therefore

\int e^{- θ} cos (4 θ) d θ = \frac{1}{4} e^{- θ} sin (4 θ) - (- \frac{1}{4} (- \frac{1}{4} e^{- θ} cos (4 θ) - \frac{1}{4} \cdot \int e^{- θ} cos (4 θ) d θ))

Isolate

\int e^{- θ} cos (4 θ) d θ

= \frac{4 e ^{- θ} sin ( 4 θ )}{17} - \frac{e ^{- θ} cos ( 4 θ )}{17}

Add a constant to the solution

= \frac{4 e ^{- θ} sin ( 4 θ )}{17} - \frac{e ^{- θ} cos ( 4 θ )}{17} + C

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What is the integral of e^{-θ}cos(4θ) ?
The integral of e^{-θ}cos(4θ) is (4e^{-θ}sin(4θ))/(17)-(e^{-θ}cos(4θ))/(17)+C