What is the integral of 1 (e^xcos(x)) ?

The integral of 1 (e^xcos(x)) is ((e^xsin(x))/2+(e^xcos(x))/2)d+C

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integral of 1 (e^xcos(x))

\int \frac{d}{1} (e^{x} cos (x)) d x

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

Solution steps

\int \frac{d}{1} e^{x} cos (x) d x

Take the constant out:

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

= d \cdot \int e^{x} cos (x) d x

Apply Integration By Parts

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

\int e^{x} cos (x) d x

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{1}{35} x^{5} + \frac{1}{4} x^{4} + x + 1)

f (x) = 8 x 16 + 0.5 x

\int_{2}^{5} \frac{x}{3} d x

n = 0 \sum \infty (1 + \frac{1}{n})^{n}

8 cot (\frac{π x}{16})

What is the integral of 1 (e^xcos(x)) ?
The integral of 1 (e^xcos(x)) is ((e^xsin(x))/2+(e^xcos(x))/2)d+C