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受欢迎的微积分 >

integral of (-sin(2x)tan(2x))/2

解答

\int \frac{- sin ( 2 x ) tan ( 2 x )}{2} d x

解答

- \frac{1}{4} (ln ∣ tan (2 x) + sec (2 x) ∣ - sin (2 x)) + C

求解步骤

\int \frac{- sin ( 2 x ) tan ( 2 x )}{2} d x

提出常数:

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

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

化简

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

使用三角恒等式改写

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

使用换元积分法

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

使用三角恒等式改写

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

u = 2 x

代回

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

化简

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

= - \frac{1}{4} (ln ∣ tan (2 x) + sec (2 x) ∣ - sin (2 x))

解答补常数

= - \frac{1}{4} (ln ∣ tan (2 x) + sec (2 x) ∣ - sin (2 x)) + C

作图

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