integral from 0 to pi of tcos^2(t)

解答

\int_{0}^{π} t cos^{2} (t) d t

\frac{π ^{2}}{4}

十进制

2.46740 \dots

求解步骤

\int_{0}^{π} t cos^{2} (t) d t

使用分布积分法

= [\frac{1}{4} (t (2 t + sin (2 t)) - 4 \cdot \int \frac{1}{2} (t + \frac{1}{2} sin (2 t)) d t)]_{0}^{π}

\int \frac{1}{2} (t + \frac{1}{2} sin (2 t)) d t = \frac{1}{2} (\frac{t ^{2}}{2} - \frac{1}{4} cos (2 t))

= [\frac{1}{4} (t (2 t + sin (2 t)) - 4 \cdot \frac{1}{2} (\frac{t ^{2}}{2} - \frac{1}{4} cos (2 t)))]_{0}^{π}

化简

[\frac{1}{4} (t (2 t + sin (2 t)) - 4 \cdot \frac{1}{2} (\frac{t ^{2}}{2} - \frac{1}{4} cos (2 t)))]_{0}^{π} : [\frac{1}{8} (2 t^{2} + 2 t sin (2 t) + cos (2 t))]_{0}^{π}

= [\frac{1}{8} (2 t^{2} + 2 t sin (2 t) + cos (2 t))]_{0}^{π}

带入上下限计算后得:

\frac{2 π ^{2} + 1}{8} - \frac{1}{8}

= \frac{2 π ^{2} + 1}{8} - \frac{1}{8}

化简

= \frac{π ^{2}}{4}

\int sin (\frac{nπ x}{1}) d x

x \to 1 lim (\frac{x ^{2} + 1}{x - 2})

\int \frac{cos ( θ )}{1 - sin ( θ )} d θ

t an g e n t y = 6 ln (x^{2}), (e, 12)

t an g e n t f (x) = e^{x} sin (x), (0, 0)