integral of 4cos^4(x)

解答

\int 4 cos^{4} (x) d x

4 (cos^{3} (x) sin (x) + \frac{3}{8} (x - \frac{1}{4} sin (4 x))) + C

求解步骤

\int 4 cos^{4} (x) d x

提出常数:

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

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

化简

cos^{4} (x) : cos^{3} (x) cos (x)

= 4 \cdot \int cos^{3} (x) cos (x) d x

使用分布积分法

= 4 (cos^{3} (x) sin (x) - \int - 3 cos^{2} (x) sin^{2} (x) d x)

\int - 3 cos^{2} (x) sin^{2} (x) d x = - \frac{3}{8} (x - \frac{1}{4} sin (4 x))

= 4 (cos^{3} (x) sin (x) - (- \frac{3}{8} (x - \frac{1}{4} sin (4 x))))

化简

= 4 (cos^{3} (x) sin (x) + \frac{3}{8} (x - \frac{1}{4} sin (4 x)))

解答补常数

= 4 (cos^{3} (x) sin (x) + \frac{3}{8} (x - \frac{1}{4} sin (4 x))) + C

\frac{d}{d x} ((\frac{x}{4})^{2})

\frac{d}{d x} (3 \frac{x - 1}{x + 1})

\frac{d}{d x} (x^{\frac{3}{5}} x^{3} + 5)

\frac{d}{d θ} (4 - θ^{2} sin (θ))

\frac{\partial}{\partial x} (\frac{1}{2 x + 3 y})