integral of sin^4(θ)cos^5(θ)

解答

\int sin^{4} (θ) cos^{5} (θ) d θ

\frac{sin ^{5} ( θ )}{5} - \frac{2 sin ^{7} ( θ )}{7} + \frac{sin ^{9} ( θ )}{9} + C

求解步骤

\int sin^{4} (θ) cos^{5} (θ) d θ

使用三角恒等式改写

= \int (1 - sin^{2} (θ))^{2} cos (θ) sin^{4} (θ) d θ

使用换元积分法

= \int u^{4} (1 - u^{2})^{2} d u

乘开

u^{4} (1 - u^{2})^{2} : u^{4} - 2 u^{6} + u^{8}

= \int u^{4} - 2 u^{6} + u^{8} d u

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 u^{4} d u - \int 2 u^{6} d u + \int u^{8} d u

\int u^{4} d u = \frac{u ^{5}}{5}

\int 2 u^{6} d u = \frac{2 u ^{7}}{7}

\int u^{8} d u = \frac{u ^{9}}{9}

= \frac{u ^{5}}{5} - \frac{2 u ^{7}}{7} + \frac{u ^{9}}{9}

u = sin (θ)

代回

= \frac{sin ^{5} ( θ )}{5} - \frac{2 sin ^{7} ( θ )}{7} + \frac{sin ^{9} ( θ )}{9}

解答补常数

= \frac{sin ^{5} ( θ )}{5} - \frac{2 sin ^{7} ( θ )}{7} + \frac{sin ^{9} ( θ )}{9} + C

a re a - x, \frac{1}{2} x - \frac{3}{2}, - 1, 3

\frac{d}{d t} (1 + t^{2})

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

d er i v a t i v e f (x) = 9

\frac{d}{d x} (4 x^{2} cos (x) + 9 x)