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 \frac{19 arctan ( x )}{x ^{2}} d x

x \to 0 lim (\frac{( x + 1 ) ^{3} - 1}{x})

\frac{1}{x - 3}

\frac{\partial}{\partial x} (\frac{s - 2 r}{t})

\int 2 t + 3 d t

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