integral of 6sin^4(x)

解答

\int 6 sin^{4} (x) d x

6 (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x))) + C

求解步骤

\int 6 sin^{4} (x) d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int sin^{4} (x) d x

使用降次积分法

= 6 (- \frac{cos ( x ) sin ^{3} ( x )}{4} + \frac{3}{4} \cdot \int sin^{2} (x) d x)

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

= 6 (- \frac{cos ( x ) sin ^{3} ( x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (x - \frac{1}{2} sin (2 x)))

化简

6 (- \frac{cos ( x ) sin ^{3} ( x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (x - \frac{1}{2} sin (2 x))) : 6 (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)))

= 6 (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)))

解答补常数

= 6 (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x))) + C

d er i v a t i v e y = ln (\frac{x ^{2} - 1}{x ^{2} + 1})

d er i v a t i v e - 10 e^{- 5 x^{2}} x

\int 6 x ln (5 x) d x

a re a y = x^{4} - 4 x^{2} + 4, y = x^{2}

in v erse l a pl a ce \frac{4}{s ^{2} ( s + 1 )}