integral of 7sin^5(x)

解答

\int 7 sin^{5} (x) d x

7 (- cos (x) + \frac{2 cos ^{3} ( x )}{3} - \frac{cos ^{5} ( x )}{5}) + C

求解步骤

\int 7 sin^{5} (x) d x

提出常数:

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

= 7 \cdot \int sin^{5} (x) d x

使用三角恒等式改写

= 7 \cdot \int (1 - cos^{2} (x))^{2} sin (x) d x

使用换元积分法

= 7 \cdot \int - 1 + 2 u^{2} - u^{4} 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

= 7 (- \int 1 d u + \int 2 u^{2} d u - \int u^{4} d u)

\int 1 d u = u

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

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

= 7 (- u + \frac{2 u ^{3}}{3} - \frac{u ^{5}}{5})

u = cos (x)

代回

= 7 (- cos (x) + \frac{2 cos ^{3} ( x )}{3} - \frac{cos ^{5} ( x )}{5})

解答补常数

= 7 (- cos (x) + \frac{2 cos ^{3} ( x )}{3} - \frac{cos ^{5} ( x )}{5}) + C