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

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

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integral of e^{-2θ}cos(θ)

\int e^{- 2 θ} cos (θ) d θ

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

Solution steps

\int e^{- 2 θ} cos (θ) d θ

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

\int e^{- 2 θ} cos (θ) d θ

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

Add a constant to the solution

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

x \to 0 lim (x cos (\frac{2}{x}))

x \to 1 + lim (\frac{8}{1 - x})

x = y

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

3^{x^{5}}

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