integral of (sin(x)tan(x))/(9+sec^2(x))

解答

\int \frac{sin ( x ) tan ( x )}{9 + sec ^{2} ( x )} d x

- \frac{1}{9} sin (x) + \frac{5}{27 2} ln 3 sin (x) + 10 - \frac{5}{27 2} ln 3 sin (x) - 10 + C

求解步骤

\int \frac{sin ( x ) tan ( x )}{9 + sec ^{2} ( x )} d x

使用换元积分法

= \int \frac{u ^{2}}{- 9 u ^{2} + 10} d u

将

\frac{u ^{2}}{- 9 u ^{2} + 10}

用部份分式展开:

- \frac{1}{9} + \frac{5}{9 2 ( 3 u + 10 )} - \frac{5}{9 2 ( 3 u - 10 )}

= \int - \frac{1}{9} + \frac{5}{9 2 ( 3 u + 10 )} - \frac{5}{9 2 ( 3 u - 10 )} 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 \frac{1}{9} d u + \int \frac{5}{9 2 ( 3 u + 10 )} d u - \int \frac{5}{9 2 ( 3 u - 10 )} d u

\int \frac{1}{9} d u = \frac{1}{9} u

\int \frac{5}{9 2 ( 3 u + 10 )} d u = \frac{5}{27 2} ln 3 u + 10

\int \frac{5}{9 2 ( 3 u - 10 )} d u = \frac{5}{27 2} ln 3 u - 10

= - \frac{1}{9} u + \frac{5}{27 2} ln 3 u + 10 - \frac{5}{27 2} ln 3 u - 10

u = sin (x)

代回

= - \frac{1}{9} sin (x) + \frac{5}{27 2} ln 3 sin (x) + 10 - \frac{5}{27 2} ln 3 sin (x) - 10

解答补常数

= - \frac{1}{9} sin (x) + \frac{5}{27 2} ln 3 sin (x) + 10 - \frac{5}{27 2} ln 3 sin (x) - 10 + C

t^{2} y^{^{''}} + 21 t y^{^{'}} + 100 y = 0

x \to 2 lim (5 a + a x)

\frac{d y}{d x} = \frac{2 - e ^{x}}{3 + 2 y}

d er i v a t i v e (8 x + 7)^{2}

n = 1 \sum \infty \frac{ln ( n )}{8 n}