integral of sin^7(θ)cos^5(θ)

解答

\int sin^{7} (θ) cos^{5} (θ) d θ

- \frac{cos ^{6} ( θ )}{6} + \frac{3 cos ^{8} ( θ )}{8} - \frac{3 cos ^{10} ( θ )}{10} + \frac{cos ^{12} ( θ )}{12} + C

求解步骤

\int sin^{7} (θ) cos^{5} (θ) d θ

使用三角恒等式改写

= \int (1 - cos^{2} (θ))^{3} sin (θ) cos^{5} (θ) d θ

使用换元积分法

= \int - u^{5} (1 - u^{2})^{3} d u

乘开

- u^{5} (1 - u^{2})^{3} : - u^{5} + 3 u^{7} - 3 u^{9} + u^{11}

= \int - u^{5} + 3 u^{7} - 3 u^{9} + u^{11} 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^{5} d u + \int 3 u^{7} d u - \int 3 u^{9} d u + \int u^{11} d u

\int u^{5} d u = \frac{u ^{6}}{6}

\int 3 u^{7} d u = \frac{3 u ^{8}}{8}

\int 3 u^{9} d u = \frac{3 u ^{10}}{10}

\int u^{11} d u = \frac{u ^{12}}{12}

= - \frac{u ^{6}}{6} + \frac{3 u ^{8}}{8} - \frac{3 u ^{10}}{10} + \frac{u ^{12}}{12}

u = cos (θ)

代回

= - \frac{cos ^{6} ( θ )}{6} + \frac{3 cos ^{8} ( θ )}{8} - \frac{3 cos ^{10} ( θ )}{10} + \frac{cos ^{12} ( θ )}{12}

解答补常数

= - \frac{cos ^{6} ( θ )}{6} + \frac{3 cos ^{8} ( θ )}{8} - \frac{3 cos ^{10} ( θ )}{10} + \frac{cos ^{12} ( θ )}{12} + C

t an g e n t f (x) = 5 x e^{x}, at x = 0

\frac{d}{d x} (\frac{3 x ^{5} - 7 x ^{2} - 4}{x ^{2}})

d er i v a t i v e sin (\frac{1}{x ^{2} + 1})

d er i v a t i v e \frac{sin ( x ) - x}{x ^{2}}

(- 2 x y 2 - 1 y) d x + (- 2 x 2 y + 1 x) d y = 0