What is the integral of e^{-θ}cos(5θ) ?

The integral of e^{-θ}cos(5θ) is (5e^{-θ}sin(5θ))/(26)-(e^{-θ}cos(5θ))/(26)+C

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integral of e^{-θ}cos(5θ)

\int e^{- θ} cos (5 θ) d θ

\frac{5 e ^{- θ} sin ( 5 θ )}{26} - \frac{e ^{- θ} cos ( 5 θ )}{26} + C

Solution steps

\int e^{- θ} cos (5 θ) d θ

Apply Integration By Parts

= \frac{1}{5} e^{- θ} sin (5 θ) - \int - \frac{1}{5} e^{- θ} sin (5 θ) d θ

Take the constant out:

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

= \frac{1}{5} e^{- θ} sin (5 θ) - (- \frac{1}{5} \cdot \int e^{- θ} sin (5 θ) d θ)

Apply Integration By Parts

= \frac{1}{5} e^{- θ} sin (5 θ) - (- \frac{1}{5} (- \frac{1}{5} e^{- θ} cos (5 θ) - \int \frac{1}{5} e^{- θ} cos (5 θ) d θ))

Take the constant out:

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

= \frac{1}{5} e^{- θ} sin (5 θ) - (- \frac{1}{5} (- \frac{1}{5} e^{- θ} cos (5 θ) - \frac{1}{5} \cdot \int e^{- θ} cos (5 θ) d θ))

Therefore

\int e^{- θ} cos (5 θ) d θ = \frac{1}{5} e^{- θ} sin (5 θ) - (- \frac{1}{5} (- \frac{1}{5} e^{- θ} cos (5 θ) - \frac{1}{5} \cdot \int e^{- θ} cos (5 θ) d θ))

Isolate

\int e^{- θ} cos (5 θ) d θ

= \frac{5 e ^{- θ} sin ( 5 θ )}{26} - \frac{e ^{- θ} cos ( 5 θ )}{26}

Add a constant to the solution

= \frac{5 e ^{- θ} sin ( 5 θ )}{26} - \frac{e ^{- θ} cos ( 5 θ )}{26} + C

What is the integral of e^{-θ}cos(5θ) ?
The integral of e^{-θ}cos(5θ) is (5e^{-θ}sin(5θ))/(26)-(e^{-θ}cos(5θ))/(26)+C