What is the integral of e^{-4θ}cos(2θ) ?

The integral of e^{-4θ}cos(2θ) is (e^{-4θ}sin(2θ)}{10}-\frac{e^{-4θ}cos(2θ))/5+C

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integral of e^{-4θ}cos(2θ)

\int e^{- 4 θ} cos (2 θ) d θ

\frac{e ^{- 4 θ} sin ( 2 θ )}{10} - \frac{e ^{- 4 θ} cos ( 2 θ )}{5} + C

Solution steps

\int e^{- 4 θ} cos (2 θ) d θ

Apply Integration By Parts

: \frac{1}{2} e^{- 4 θ} sin (2 θ) - \int - 2 e^{- 4 θ} sin (2 θ) d θ

= \frac{1}{2} e^{- 4 θ} sin (2 θ) - \int - 2 e^{- 4 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^{- 4 θ} sin (2 θ) - (- 2 \cdot \int e^{- 4 θ} sin (2 θ) d θ)

Apply Integration By Parts

: - \frac{1}{2} e^{- 4 θ} cos (2 θ) - \int 2 e^{- 4 θ} cos (2 θ) d θ

= \frac{1}{2} e^{- 4 θ} sin (2 θ) - (- 2 (- \frac{1}{2} e^{- 4 θ} cos (2 θ) - \int 2 e^{- 4 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^{- 4 θ} sin (2 θ) - (- 2 (- \frac{1}{2} e^{- 4 θ} cos (2 θ) - 2 \cdot \int e^{- 4 θ} cos (2 θ) d θ))

Therefore

\int e^{- 4 θ} cos (2 θ) d θ = \frac{1}{2} e^{- 4 θ} sin (2 θ) - (- 2 (- \frac{1}{2} e^{- 4 θ} cos (2 θ) - 2 \cdot \int e^{- 4 θ} cos (2 θ) d θ))

Isolate

\int e^{- 4 θ} cos (2 θ) d θ

= \frac{e ^{- 4 θ} sin ( 2 θ )}{10} - \frac{e ^{- 4 θ} cos ( 2 θ )}{5}

Add a constant to the solution

= \frac{e ^{- 4 θ} sin ( 2 θ )}{10} - \frac{e ^{- 4 θ} cos ( 2 θ )}{5} + C

\int tan (X) d X

\frac{d y}{d x} = 4 y

y = \frac{6}{x}

4 t^{2}

\int 8 x^{4} - 5 x^{3} + 5 x^{2} + C d x

What is the integral of e^{-4θ}cos(2θ) ?
The integral of e^{-4θ}cos(2θ) is (e^{-4θ}sin(2θ)}{10}-\frac{e^{-4θ}cos(2θ))/5+C