What is the integral from 0 to pi of e^xcox(x) ?

The integral from 0 to pi of e^xcox(x) is co(e^pipi^2-2e^pipi+2e^pi-2)

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integral from 0 to pi of e^xcox(x)

\int_{0}^{π} e^{x} co x (x) d x

co (e^{π} π^{2} - 2 e^{π} π + 2 e^{π} - 2)

Solution steps

\int_{0}^{π} e^{x} co xx d x

Take the constant out:

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

= co \cdot \int_{0}^{π} e^{x} xx d x

Apply Integration By Parts

: [x (e^{x} x - e^{x}) - \int e^{x} x - e^{x} d x]_{0}^{π}

= co [x (e^{x} x - e^{x}) - \int e^{x} x - e^{x} d x]_{0}^{π}

\int e^{x} x - e^{x} d x = e^{x} x - 2 e^{x}

= co [x (e^{x} x - e^{x}) - (e^{x} x - 2 e^{x})]_{0}^{π}

Expand

x (e^{x} x - e^{x}) - (e^{x} x - 2 e^{x}) : e^{x} x^{2} - 2 e^{x} x + 2 e^{x}

= co [e^{x} x^{2} - 2 e^{x} x + 2 e^{x}]_{0}^{π}

Compute the boundaries:

e^{π} π^{2} - 2 e^{π} π + 2 e^{π} - 2

= co (e^{π} π^{2} - 2 e^{π} π + 2 e^{π} - 2)

\int_{1}^{2} (x^{- 3}) d x

\int_{0.1}^{1} \frac{1}{x} d x

\int_{0}^{π} sin (3 x) sin (3 x) d x

\int_{2}^{4} (16 - x^{2}) d x

\int_{0}^{1} e^{3 x^{3}} d x

What is the integral from 0 to pi of e^xcox(x) ?
The integral from 0 to pi of e^xcox(x) is co(e^pipi^2-2e^pipi+2e^pi-2)