integral of (tan^3(x)+tan^5(x))

解答

\int (tan^{3} (x) + tan^{5} (x)) d x

\frac{1}{2} sec^{2} (x) - sec^{2} (x) + \frac{1}{4} sec^{4} (x) + C

求解步骤

\int tan^{3} (x) + tan^{5} (x) d x

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 tan^{3} (x) d x + \int tan^{5} (x) d x

\int tan^{3} (x) d x = - ln ∣ sec (x) ∣ + \frac{sec ^{2} ( x )}{2}

\int tan^{5} (x) d x = ln ∣ sec (x) ∣ - sec^{2} (x) + \frac{sec ^{4} ( x )}{4}

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

化简

- ln ∣ sec (x) ∣ + \frac{sec ^{2} ( x )}{2} + ln ∣ sec (x) ∣ - sec^{2} (x) + \frac{sec ^{4} ( x )}{4} : \frac{1}{2} sec^{2} (x) - sec^{2} (x) + \frac{1}{4} sec^{4} (x)

= \frac{1}{2} sec^{2} (x) - sec^{2} (x) + \frac{1}{4} sec^{4} (x)

解答补常数

= \frac{1}{2} sec^{2} (x) - sec^{2} (x) + \frac{1}{4} sec^{4} (x) + C

\int sec (x) sec (x) + tan (x) d x

\int θ^{2} + 1 d θ

\int 3 x^{2} - 2 x + 7 e^{x} d x

\int 5 x + 1 d x

\int \frac{2}{( x + 1 ) ( x - 4 )} d x