What is the integral of e^θsin(θ) ?

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

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integral of e^θsin(θ)

\int e^{θ} sin (θ) d θ

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

Solution steps

\int e^{θ} sin (θ) d θ

Apply Integration By Parts

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

Apply Integration By Parts

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

Therefore

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

Isolate

\int e^{θ} sin (θ) d θ

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{( ( - 3 ) ^{n - 1} )}{5 ^{n}}

\int 2 sin^{3} (x) d x

x \to \infty lim (x a)

\frac{d}{d x} (\frac{x ^{3} - 6}{x ^{2}})

\int \frac{1}{x ^{\frac{1}{2}} + x ^{\frac{1}{3}}} d x

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