integral of (sin(x)+1)^2

解答

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

\frac{3}{2} x - \frac{1}{4} sin (2 x) - 2 cos (x) + C

求解步骤

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

乘开

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

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

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

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

\int 1 d x = x

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

化简

\frac{1}{2} (x - \frac{1}{2} sin (2 x)) - 2 cos (x) + x : \frac{3}{2} x - \frac{1}{4} sin (2 x) - 2 cos (x)

= \frac{3}{2} x - \frac{1}{4} sin (2 x) - 2 cos (x)

解答补常数

= \frac{3}{2} x - \frac{1}{4} sin (2 x) - 2 cos (x) + C

\int 8 t e^{- \frac{t}{5}} d t

\int i x d x

\int 8 x^{\frac{1}{3}} - 8 x^{- \frac{1}{3}} d x

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

\int 5 e^{3 x + 1} d x