integral of 64sin^2(x)

解答

\int 64 sin^{2} (x) d x

32 (x - \frac{1}{2} sin (2 x)) + C

求解步骤

\int 64 sin^{2} (x) d x

提出常数:

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

= 64 \cdot \int sin^{2} (x) d x

使用三角恒等式改写

= 64 \cdot \int \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

= 64 \cdot \frac{1}{2} \cdot \int 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

= 64 \cdot \frac{1}{2} (\int 1 d x - \int cos (2 x) d x)

\int 1 d x = x

\int cos (2 x) d x = \frac{1}{2} sin (2 x)

= 64 \cdot \frac{1}{2} (x - \frac{1}{2} sin (2 x))

化简

64 \cdot \frac{1}{2} (x - \frac{1}{2} sin (2 x)) : 32 (x - \frac{1}{2} sin (2 x))

= 32 (x - \frac{1}{2} sin (2 x))

解答补常数

= 32 (x - \frac{1}{2} sin (2 x)) + C

t an g e n t f (x) = x^{3} - 11 x, at x = 3

n = 2 \sum \infty (\frac{3}{4})^{n}

\frac{d}{d x} ((4 x^{5} - 2 x) (3 x + e^{x}))

d er i v a t i v e \frac{7}{t}

x \to \infty lim (2 x^{3} - 5 x - 4 x^{3})