What is the integral of e^θcos(θ) ?

The integral of e^θcos(θ) is (e^θsin(θ))/2+(e^θcos(θ))/2+C

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integral of e^θcos(θ)

\int e^{θ} cos (θ) d θ

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

Solution steps

\int e^{θ} cos (θ) d θ

Apply Integration By Parts

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

\int e^{θ} cos (θ) d θ

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (\frac{ln ( x + 2 y )}{y ^{2}})

\int \frac{1}{e ^{x} - 3} d x

\frac{d}{d x} (ln (cos (x^{2} + 1)))

f (v) = e^{2 s i n (3 v)}

\int sin (x) - cos (x) d x

What is the integral of e^θcos(θ) ?
The integral of e^θcos(θ) is (e^θsin(θ))/2+(e^θcos(θ))/2+C