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受欢迎的微积分 >

integral of (tan^2(2x))/(sec(2x))

解答

\int \frac{tan ^{2} ( 2 x )}{sec ( 2 x )} d x

解答

\frac{1}{2} (ln ∣ tan (2 x) + sec (2 x) ∣ - sin (2 x)) + C

求解步骤

\int \frac{tan ^{2} ( 2 x )}{sec ( 2 x )} d x

使用三角恒等式改写

= \int \frac{1 - cos ^{2} ( 2 x )}{cos ( 2 x )} d x

使用换元积分法

= \int \frac{sin ( u ) tan ( u )}{2} d u

提出常数:

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

= \frac{1}{2} \cdot \int sin (u) tan (u) d u

使用三角恒等式改写

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

乘开

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

= \frac{1}{2} \cdot \int \frac{1}{cos ( u )} - cos (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

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

\int \frac{1}{cos ( u )} d u = ln ∣ tan (u) + sec (u) ∣

\int cos (u) d u = sin (u)

= \frac{1}{2} (ln ∣ tan (u) + sec (u) ∣ - sin (u))

u = 2 x

代回

= \frac{1}{2} (ln ∣ tan (2 x) + sec (2 x) ∣ - sin (2 x))

解答补常数

= \frac{1}{2} (ln ∣ tan (2 x) + sec (2 x) ∣ - sin (2 x)) + C

作图

流行的例子

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\int (7 x^{3} + 5 x^{2} + \frac{9}{x ^{2}} + \frac{2}{a ^{2}}) d x

integral of ((1+x^2))/x

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

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\int 60 (1 + x^{2}) (7 + 15 x + 5 x^{3})^{16} d x

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\int y e^{\frac{- y}{x}} d y

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\int \frac{1}{y ^{2} + y - 2} d y