What is the integral of 1/(e^x)cos(pi)x ?

The integral of 1/(e^x)cos(pi)x is x/(e^x)+1/(e^x)+C

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integral of 1/(e^x)cos(pi)x

\int \frac{1}{e ^{x}} cos (π) x d x

\frac{x}{e ^{x}} + \frac{1}{e ^{x}} + C

Solution steps

\int \frac{1}{e ^{x}} cos (π) x d x

Take the constant out:

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

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

cos (π) = - 1

= (- 1) \cdot \int \frac{1}{e ^{x}} x d x

Simplify

\frac{1}{e ^{x}} x : \frac{x}{e ^{x}}

= (- 1) \cdot \int \frac{x}{e ^{x}} d x

Apply Integration By Parts

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

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

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

= (- 1) (- \frac{x}{e ^{x}} - \frac{1}{e ^{x}})

Simplify

= \frac{x}{e ^{x}} + \frac{1}{e ^{x}}

Add a constant to the solution

= \frac{x}{e ^{x}} + \frac{1}{e ^{x}} + C

y^{^{'}} = 10 x y

\int \frac{1}{20} (x + 1)^{2} (x - 3) (x + 4) d x

y^{^{''}} + 4 y^{^{'}} - 4 y = 0

f (x) = \frac{e ^{x}}{x}

5 cos^{2} (x) + 5 sin (x)

What is the integral of 1/(e^x)cos(pi)x ?
The integral of 1/(e^x)cos(pi)x is x/(e^x)+1/(e^x)+C