integral of ((1+cos(x))^2)/2

解答

\int \frac{( 1 + cos ( x ) ) ^{2}}{2} d x

\frac{1}{8} (6 x + 8 sin (x) + sin (2 x)) + C

求解步骤

\int \frac{( 1 + cos ( x ) ) ^{2}}{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 + cos (x))^{2} d x

乘开

(1 + cos (x))^{2} : 1 + 2 cos (x) + cos^{2} (x)

= \frac{1}{2} \cdot \int 1 + 2 cos (x) + 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 d x + \int 2 cos (x) d x + \int cos^{2} (x) d x)

\int 1 d x = x

\int 2 cos (x) d x = 2 sin (x)

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

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

化简

\frac{1}{2} (x + 2 sin (x) + \frac{1}{2} (x + \frac{1}{2} sin (2 x))) : \frac{1}{8} (6 x + 8 sin (x) + sin (2 x))

= \frac{1}{8} (6 x + 8 sin (x) + sin (2 x))

解答补常数

= \frac{1}{8} (6 x + 8 sin (x) + sin (2 x)) + C

\int (1 + x^{2})^{3} d x

\int 49 - 4 x^{2} d x

\int \frac{1}{( x - 2 ) \cdot x + 2} d x

\int (6 x^{4} + 7 x + \frac{1}{x ^{3}}) d x

\int \frac{x ^{2} + 2 x + 2}{( x + 1 ) ( x ^{2} + 2 x + 4 )} d x