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

解答

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

\frac{x ^{3}}{3} + 2 (- x cos (x) + sin (x)) + \frac{1}{2} (x - \frac{1}{2} sin (2 x)) + C

求解步骤

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

乘开

(x + sin (x))^{2} : x^{2} + 2 x sin (x) + sin^{2} (x)

= \int x^{2} + 2 x sin (x) + sin^{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

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

\int x^{2} d x = \frac{x ^{3}}{3}

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

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

= \frac{x ^{3}}{3} + 2 (- x cos (x) + sin (x)) + \frac{1}{2} (x - \frac{1}{2} sin (2 x))

解答补常数

= \frac{x ^{3}}{3} + 2 (- x cos (x) + sin (x)) + \frac{1}{2} (x - \frac{1}{2} sin (2 x)) + C

l a pl a ce t r an s f or m t sin^{2} (t)

\int sec^{2} (θ) tan^{4} (θ) d θ

2 y^{^{'}} = y^{2}, y (0) = 7

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

\int sin^{2} (3 t) cos^{2} (3 t) d t