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受欢迎的微积分 >

integral of tan^3(2t)sec^4(2t)

解答

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

解答

\frac{sec ^{6} ( 2 t )}{12} - \frac{sec ^{4} ( 2 t )}{8} + C

求解步骤

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

使用三角恒等式改写

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

使用换元积分法

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

乘开

\frac{u ^{3} ( u ^{2} - 1 )}{2} : \frac{u ^{5}}{2} - \frac{u ^{3}}{2}

= \int \frac{u ^{5}}{2} - \frac{u ^{3}}{2} 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{u ^{5}}{2} d u - \int \frac{u ^{3}}{2} d u

\int \frac{u ^{5}}{2} d u = \frac{u ^{6}}{12}

\int \frac{u ^{3}}{2} d u = \frac{u ^{4}}{8}

= \frac{u ^{6}}{12} - \frac{u ^{4}}{8}

u = sec (2 t)

代回

= \frac{sec ^{6} ( 2 t )}{12} - \frac{sec ^{4} ( 2 t )}{8}

解答补常数

= \frac{sec ^{6} ( 2 t )}{12} - \frac{sec ^{4} ( 2 t )}{8} + C

作图

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\int cos (12 x) d x

integral of (sin^7(6x))/(cos^4(6x))

\int \frac{sin ^{7} ( 6 x )}{cos ^{4} ( 6 x )} d x

limit as x approaches 2.9 of x^2+3x

x \to 2.9 lim (x^{2} + 3 x)