What is the integral of e^{-θ}cos(6θ) ?

The integral of e^{-θ}cos(6θ) is (6e^{-θ}sin(6θ))/(37)-(e^{-θ}cos(6θ))/(37)+C

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integral of e^{-θ}cos(6θ)

\int e^{- θ} cos (6 θ) d θ

\frac{6 e ^{- θ} sin ( 6 θ )}{37} - \frac{e ^{- θ} cos ( 6 θ )}{37} + C

Solution steps

\int e^{- θ} cos (6 θ) d θ

Apply Integration By Parts

: \frac{1}{6} e^{- θ} sin (6 θ) - \int - \frac{1}{6} e^{- θ} sin (6 θ) d θ

= \frac{1}{6} e^{- θ} sin (6 θ) - \int - \frac{1}{6} e^{- θ} sin (6 θ) d θ

Take the constant out:

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

= \frac{1}{6} e^{- θ} sin (6 θ) - (- \frac{1}{6} \cdot \int e^{- θ} sin (6 θ) d θ)

Apply Integration By Parts

: - \frac{1}{6} e^{- θ} cos (6 θ) - \int \frac{1}{6} e^{- θ} cos (6 θ) d θ

= \frac{1}{6} e^{- θ} sin (6 θ) - (- \frac{1}{6} (- \frac{1}{6} e^{- θ} cos (6 θ) - \int \frac{1}{6} e^{- θ} cos (6 θ) d θ))

Take the constant out:

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

= \frac{1}{6} e^{- θ} sin (6 θ) - (- \frac{1}{6} (- \frac{1}{6} e^{- θ} cos (6 θ) - \frac{1}{6} \cdot \int e^{- θ} cos (6 θ) d θ))

Therefore

\int e^{- θ} cos (6 θ) d θ = \frac{1}{6} e^{- θ} sin (6 θ) - (- \frac{1}{6} (- \frac{1}{6} e^{- θ} cos (6 θ) - \frac{1}{6} \cdot \int e^{- θ} cos (6 θ) d θ))

Isolate

\int e^{- θ} cos (6 θ) d θ

= \frac{6 e ^{- θ} sin ( 6 θ )}{37} - \frac{e ^{- θ} cos ( 6 θ )}{37}

Add a constant to the solution

= \frac{6 e ^{- θ} sin ( 6 θ )}{37} - \frac{e ^{- θ} cos ( 6 θ )}{37} + C

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What is the integral of e^{-θ}cos(6θ) ?
The integral of e^{-θ}cos(6θ) is (6e^{-θ}sin(6θ))/(37)-(e^{-θ}cos(6θ))/(37)+C