What is the integral of e^{-θ}cos(2θ) ?

The integral of e^{-θ}cos(2θ) is (2e^{-θ}sin(2θ))/5-(e^{-θ}cos(2θ))/5+C

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integral of e^{-θ}cos(2θ)

\int e^{- θ} cos (2 θ) d θ

\frac{2 e ^{- θ} sin ( 2 θ )}{5} - \frac{e ^{- θ} cos ( 2 θ )}{5} + C

Solution steps

\int e^{- θ} cos (2 θ) d θ

Apply Integration By Parts

: \frac{1}{2} e^{- θ} sin (2 θ) - \int - \frac{1}{2} e^{- θ} sin (2 θ) d θ

= \frac{1}{2} e^{- θ} sin (2 θ) - \int - \frac{1}{2} e^{- θ} sin (2 θ) d θ

Take the constant out:

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

= \frac{1}{2} e^{- θ} sin (2 θ) - (- \frac{1}{2} \cdot \int e^{- θ} sin (2 θ) d θ)

Apply Integration By Parts

: - \frac{1}{2} e^{- θ} cos (2 θ) - \int \frac{1}{2} e^{- θ} cos (2 θ) d θ

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

Take the constant out:

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

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

Therefore

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

Isolate

\int e^{- θ} cos (2 θ) d θ

= \frac{2 e ^{- θ} sin ( 2 θ )}{5} - \frac{e ^{- θ} cos ( 2 θ )}{5}

Add a constant to the solution

= \frac{2 e ^{- θ} sin ( 2 θ )}{5} - \frac{e ^{- θ} cos ( 2 θ )}{5} + C

\int (2 x - 9) e^{2 x} d x

\frac{d}{d x} (- \frac{3 y}{3 x + 2 y})

\frac{d}{d x} (\frac{1}{( 3 - x ) ^{2}})

cos (k \cdot t)

\frac{d y}{d x} = y^{\frac{1}{3}}, y (1) = 0

What is the integral of e^{-θ}cos(2θ) ?
The integral of e^{-θ}cos(2θ) is (2e^{-θ}sin(2θ))/5-(e^{-θ}cos(2θ))/5+C