integral of sin^7(x)cos^3(x)

解答

\int sin^{7} (x) cos^{3} (x) d x

- \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{2} - \frac{3 cos ^{8} ( x )}{8} + \frac{cos ^{10} ( x )}{10} + C

求解步骤

\int sin^{7} (x) cos^{3} (x) d x

使用三角恒等式改写

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

使用换元积分法

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

乘开

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

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

\int u^{3} d u = \frac{u ^{4}}{4}

\int 3 u^{5} d u = \frac{u ^{6}}{2}

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

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

= - \frac{u ^{4}}{4} + \frac{u ^{6}}{2} - \frac{3 u ^{8}}{8} + \frac{u ^{10}}{10}

u = cos (x)

代回

= - \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{2} - \frac{3 cos ^{8} ( x )}{8} + \frac{cos ^{10} ( x )}{10}

解答补常数

= - \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{2} - \frac{3 cos ^{8} ( x )}{8} + \frac{cos ^{10} ( x )}{10} + C

x \to 0 lim (x^{19 x})

\int_{1}^{\infty} f (x) d x

(19 x y^{2} - 15 y) + (19 x^{2} y - 15 x) y^{^{'}} = 0

\int \frac{sec ( x ) + cos ( x )}{7 cos ( x )} d x

\frac{d}{d x} (x^{4} \frac{1}{8} + \frac{1}{4 x ^{2}})