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受欢迎的微积分 >

integral of 2tan^3(4x)sec^5(4x)

解答

\int 2 tan^{3} (4 x) sec^{5} (4 x) d x

解答

2 (\frac{sec ^{7} ( 4 x )}{28} - \frac{sec ^{5} ( 4 x )}{20}) + C

求解步骤

\int 2 tan^{3} (4 x) sec^{5} (4 x) d x

提出常数:

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

= 2 \cdot \int tan^{3} (4 x) sec^{5} (4 x) d x

使用三角恒等式改写

= 2 \cdot \int (- 1 + sec^{2} (4 x)) tan (4 x) sec^{5} (4 x) d x

使用换元积分法

= 2 \cdot \int \frac{u ^{4} ( u ^{2} - 1 )}{4} d u

乘开

\frac{u ^{4} ( u ^{2} - 1 )}{4} : \frac{u ^{6}}{4} - \frac{u ^{4}}{4}

= 2 \cdot \int \frac{u ^{6}}{4} - \frac{u ^{4}}{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

= 2 (\int \frac{u ^{6}}{4} d u - \int \frac{u ^{4}}{4} d u)

\int \frac{u ^{6}}{4} d u = \frac{u ^{7}}{28}

\int \frac{u ^{4}}{4} d u = \frac{u ^{5}}{20}

= 2 (\frac{u ^{7}}{28} - \frac{u ^{5}}{20})

u = sec (4 x)

代回

= 2 (\frac{sec ^{7} ( 4 x )}{28} - \frac{sec ^{5} ( 4 x )}{20})

解答补常数

= 2 (\frac{sec ^{7} ( 4 x )}{28} - \frac{sec ^{5} ( 4 x )}{20}) + C

作图

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