integral from 0 to arctan(3) of sin(2x)

解答

\int_{0}^{a r c t a n (3)} sin (2 x) d x

\frac{9}{10}

十进制

0.9

求解步骤

\int_{0}^{a r c t a n (3)} sin (2 x) d x

使用换元积分法

= \int_{0}^{2 a r c t a n (3)} sin (u) \frac{1}{2} d u

提出常数:

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

= \frac{1}{2} \cdot \int_{0}^{2 a r c t a n (3)} sin (u) d u

使用积分常见定则:

\int sin (u) d u = - cos (u)

= \frac{1}{2} [- cos (u)]_{0}^{2 a r c t a n (3)}

带入上下限计算后得:

\frac{9}{5}

= \frac{1}{2} \cdot \frac{9}{5}

化简

= \frac{9}{10}

\int_{0}^{\infty} \frac{3}{x + 5} d x

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