What is the integral from 0 to pi of (1+x^2)cos(x) ?

The integral from 0 to pi of (1+x^2)cos(x) is -2pi

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integral from 0 to pi of (1+x^2)cos(x)

\int_{0}^{π} (1 + x^{2}) cos (x) d x

- 2 π

Decimal Notation

- 6.28318 \dots

Solution steps

\int_{0}^{π} (1 + x^{2}) cos (x) d x

Apply Integration By Parts

: [(1 + x^{2}) sin (x) - \int 2 x sin (x) d x]_{0}^{π}

= [(1 + x^{2}) sin (x) - \int 2 x sin (x) d x]_{0}^{π}

\int 2 x sin (x) d x = 2 (- x cos (x) + sin (x))

= [(1 + x^{2}) sin (x) - 2 (- x cos (x) + sin (x))]_{0}^{π}

Expand

(1 + x^{2}) sin (x) - 2 (- x cos (x) + sin (x)) : x^{2} sin (x) + 2 x cos (x) - sin (x)

= [x^{2} sin (x) + 2 x cos (x) - sin (x)]_{0}^{π}

Compute the boundaries:

- 2 π

= - 2 π

\int_{- 2}^{2} 7 x + 5 d x

\int_{4}^{5} x e^{- x} d x

\int_{- 1}^{4} ∣ x - 2 ∣ d x

\int_{0}^{\frac{π}{4}} \frac{1}{cos ( θ )} d θ

\int_{0}^{2} (2 t^{3} + t) d t

What is the integral from 0 to pi of (1+x^2)cos(x) ?
The integral from 0 to pi of (1+x^2)cos(x) is -2pi