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受欢迎的微积分 >

integral of 2(sin(2x))^2

解答

\int 2 (sin (2 x))^{2} d x

解答

x - \frac{1}{4} sin (4 x) + C

求解步骤

\int 2 (sin (2 x))^{2} d x

提出常数:

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

= 2 \cdot \int (sin (2 x))^{2} d x

使用三角恒等式改写

= 2 \cdot \int 4 cos^{2} (x) sin^{2} (x) d x

提出常数:

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

= 2 \cdot 4 \cdot \int cos^{2} (x) sin^{2} (x) d x

使用三角恒等式改写

= 2 \cdot 4 \cdot \int \frac{1 - cos ( 4 x )}{8} d x

提出常数:

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

= 2 \cdot 4 \cdot \frac{1}{8} \cdot \int 1 - cos (4 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

= 2 \cdot 4 \cdot \frac{1}{8} (\int 1 d x - \int cos (4 x) d x)

\int 1 d x = x

\int cos (4 x) d x = \frac{1}{4} sin (4 x)

= 2 \cdot 4 \cdot \frac{1}{8} (x - \frac{1}{4} sin (4 x))

化简

2 \cdot 4 \cdot \frac{1}{8} (x - \frac{1}{4} sin (4 x)) : x - \frac{1}{4} sin (4 x)

= x - \frac{1}{4} sin (4 x)

解答补常数

= x - \frac{1}{4} sin (4 x) + C

作图

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