integral of 1/(sin(x)cos^5(x))

解答

\int \frac{1}{sin ( x ) cos ^{5} ( x )} d x

\frac{tan ^{4} ( x )}{4} + tan^{2} (x) + ln ∣ tan (x) ∣ + C

求解步骤

\int \frac{1}{sin ( x ) cos ^{5} ( x )} d x

使用三角换元法

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

(u^{2} + 1)^{\frac{3}{2}} = (u^{2} + 1) u^{2} + 1

= \int \frac{( u ^{2} + 1 ) u ^{2} + 1 u ^{2} + 1}{u} d u

(u^{2} + 1) u^{2} + 1 u^{2} + 1 = (u^{2} + 1)^{2}

= \int \frac{( u ^{2} + 1 ) ^{2}}{u} d u

乘开

\frac{( u ^{2} + 1 ) ^{2}}{u} : u^{3} + 2 u + \frac{1}{u}

= \int u^{3} + 2 u + \frac{1}{u} 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 2 u d u + \int \frac{1}{u} d u

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{4}}{4} + u^{2} + ln ∣ u ∣

u = tan (x)

代回

= \frac{tan ^{4} ( x )}{4} + tan^{2} (x) + ln ∣ tan (x) ∣

解答补常数

= \frac{tan ^{4} ( x )}{4} + tan^{2} (x) + ln ∣ tan (x) ∣ + C

\frac{d}{d x} (x (x - 7)^{2})

\frac{d}{d x} (\frac{7}{4 - 3 x})

d er i v a t i v e f (x) = \frac{x + x ^{3}}{x}

x \to 4 lim (\frac{5}{x - 3})

d er i v a t i v e y = \frac{x - 2}{x - 2}