integral from 0 to pi/4 of 1-4sin^2(x)

解答

\int_{0}^{\frac{π}{4}} 1 - 4 sin^{2} (x) d x

- \frac{π}{4} + 1

十进制

0.21460 \dots

求解步骤

\int_{0}^{\frac{π}{4}} 1 - 4 sin^{2} (x) d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{\frac{π}{4}} 1 d x - \int_{0}^{\frac{π}{4}} 4 sin^{2} (x) d x

\int_{0}^{\frac{π}{4}} 1 d x = \frac{π}{4}

\int_{0}^{\frac{π}{4}} 4 sin^{2} (x) d x = \frac{π}{2} - 1

= \frac{π}{4} - (\frac{π}{2} - 1)

化简

\frac{π}{4} - (\frac{π}{2} - 1) : - \frac{π}{4} + 1

= - \frac{π}{4} + 1

\int 2 u sin (u) d u

\frac{d y}{d x} = \frac{x + y}{x}

x \to 5 lim (x^{5} - 9 x + 1)

\int - 4 sin (x) d x

t \to 2 lim (t^{3} i + t^{4} j + t^{5} k)