integral of tan(5x)sec^2(5x)

解答

\int tan (5 x) sec^{2} (5 x) d x

\frac{1}{10} tan^{2} (5 x) + C

求解步骤

\int tan (5 x) sec^{2} (5 x) d x

使用换元积分法

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

提出常数:

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

= \frac{1}{5} \cdot \int u d u

使用幂法则

= \frac{1}{5} \cdot \frac{u ^{2}}{2}

u = tan (5 x)

代回

= \frac{1}{5} \cdot \frac{tan ^{2} ( 5 x )}{2}

化简

\frac{1}{5} \cdot \frac{tan ^{2} ( 5 x )}{2} : \frac{1}{10} tan^{2} (5 x)

= \frac{1}{10} tan^{2} (5 x)

解答补常数

= \frac{1}{10} tan^{2} (5 x) + C

a re a x - 2, x, [0, 4]

d er i v a t i v e [x + (x + sin^{2} (x))^{7}]^{5}

\int x 1 + 2 x^{2} d x

\int \frac{3}{x ^{7}} d x

x \to \infty lim (e^{- 5})