integral from-1 to 1 of (sin(x))^2

解答

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

1 - \frac{1}{2} sin (2)

十进制

0.54535 \dots

求解步骤

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

化简

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

使用三角恒等式改写

= \int_{- 1}^{1} \frac{1 - cos ( 2 x )}{2} d x

提出常数:

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

= \frac{1}{2} \cdot \int_{- 1}^{1} 1 - cos (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

= \frac{1}{2} (\int_{- 1}^{1} 1 d x - \int_{- 1}^{1} cos (2 x) d x)

\int_{- 1}^{1} 1 d x = 2

\int_{- 1}^{1} cos (2 x) d x = sin (2)

= \frac{1}{2} (2 - sin (2))

化简

\frac{1}{2} (2 - sin (2)) : 1 - \frac{1}{2} sin (2)

= 1 - \frac{1}{2} sin (2)

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

\int_{0}^{l o g_{8} (3)} 8^{x} d x

\int_{0}^{1} y e^{2 y} d y

\int_{- 2}^{4} e^{∣ x ∣} d x

\int_{0}^{1} \frac{( e ^{x} + 3 )}{e ^{x}} d x