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受欢迎的微积分 >

integral of (sec^4(θ))/(tan^5(θ))

解答

\int \frac{sec ^{4} ( θ )}{tan ^{5} ( θ )} d θ

解答

- \frac{1}{4} cot^{4} (θ) - \frac{1}{2} cot^{2} (θ) + C

求解步骤

\int \frac{sec ^{4} ( θ )}{tan ^{5} ( θ )} d θ

使用三角恒等式改写

= \int (1 + tan^{2} (θ)) sec^{2} (θ) \frac{1}{tan ^{5} ( θ )} d θ

使用换元积分法

= \int \frac{1 + u ^{2}}{u ^{5}} d u

乘开

\frac{1 + u ^{2}}{u ^{5}} : \frac{1}{u ^{5}} + \frac{1}{u ^{3}}

= \int \frac{1}{u ^{5}} + \frac{1}{u ^{3}} 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 \frac{1}{u ^{5}} d u + \int \frac{1}{u ^{3}} d u

\int \frac{1}{u ^{5}} d u = - \frac{1}{4 u ^{4}}

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

= - \frac{1}{4 u ^{4}} - \frac{1}{2 u ^{2}}

u = tan (θ)

代回

= - \frac{1}{4 tan ^{4} ( θ )} - \frac{1}{2 tan ^{2} ( θ )}

化简

- \frac{1}{4 tan ^{4} ( θ )} - \frac{1}{2 tan ^{2} ( θ )} : - \frac{1}{4} cot^{4} (θ) - \frac{1}{2} cot^{2} (θ)

= - \frac{1}{4} cot^{4} (θ) - \frac{1}{2} cot^{2} (θ)

解答补常数

= - \frac{1}{4} cot^{4} (θ) - \frac{1}{2} cot^{2} (θ) + C

作图

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