What is the integral of e^θcos(2θ) ?

The integral of e^θcos(2θ) is (2e^θsin(2θ))/5+(e^θcos(2θ))/5+C

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integral of e^θcos(2θ)

\int e^{θ} cos (2 θ) d θ

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

Solution steps

\int e^{θ} cos (2 θ) d θ

Apply Integration By Parts

: e^{θ} \frac{1}{2} sin (2 θ) - \int e^{θ} \frac{1}{2} sin (2 θ) d θ

= e^{θ} \frac{1}{2} sin (2 θ) - \int e^{θ} \frac{1}{2} sin (2 θ) d θ

Take the constant out:

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

= e^{θ} \frac{1}{2} sin (2 θ) - \frac{1}{2} \cdot \int e^{θ} sin (2 θ) d θ

Apply Integration By Parts

: - \frac{1}{2} e^{θ} cos (2 θ) - \int - \frac{1}{2} e^{θ} cos (2 θ) d θ

= e^{θ} \frac{1}{2} sin (2 θ) - \frac{1}{2} (- \frac{1}{2} e^{θ} cos (2 θ) - \int - \frac{1}{2} e^{θ} cos (2 θ) d θ)

Take the constant out:

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

= e^{θ} \frac{1}{2} sin (2 θ) - \frac{1}{2} (- \frac{1}{2} e^{θ} cos (2 θ) - (- \frac{1}{2} \cdot \int e^{θ} cos (2 θ) d θ))

Therefore

\int e^{θ} cos (2 θ) d θ = e^{θ} \frac{1}{2} sin (2 θ) - \frac{1}{2} (- \frac{1}{2} e^{θ} cos (2 θ) - (- \frac{1}{2} \cdot \int e^{θ} cos (2 θ) d θ))

Isolate

\int e^{θ} cos (2 θ) d θ

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

Add a constant to the solution

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

\frac{d ^{2}}{d x ^{2}} ((1 + \frac{1}{x})^{3})

f (x) = x^{2} + 2, (- 3, 11)

\frac{\partial}{\partial x} (x e^{y} + y e^{z} + z e^{x})

y (1 + x^{2}) y^{^{'}} - x (1 + y^{2}) = 0

y = 12

What is the integral of e^θcos(2θ) ?
The integral of e^θcos(2θ) is (2e^θsin(2θ))/5+(e^θcos(2θ))/5+C