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受欢迎的微积分 >

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

解答

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

解答

\frac{1}{4} (\frac{tan ^{2} ( \frac{x}{2} )}{2} + ln tan (\frac{x}{2})) + C

求解步骤

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

使用三角恒等式改写

= \int \frac{1}{2 cos ( x ) sin ( x ) + 2 sin ( x )} d x

使用换元积分法

= \int \frac{u ^{2} + 1}{4 u} d u

提出常数:

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

= \frac{1}{4} \cdot \int \frac{u ^{2} + 1}{u} d u

乘开

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

= \frac{1}{4} \cdot \int u + \frac{1}{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}{4} (\int u d u + \int \frac{1}{u} d u)

\int u d u = \frac{u ^{2}}{2}

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{4} (\frac{u ^{2}}{2} + ln ∣ u ∣)

u = tan (\frac{x}{2})

代回

= \frac{1}{4} (\frac{tan ^{2} ( \frac{x}{2} )}{2} + ln tan (\frac{x}{2}))

解答补常数

= \frac{1}{4} (\frac{tan ^{2} ( \frac{x}{2} )}{2} + ln tan (\frac{x}{2})) + C

作图

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\int (a x + 2) cos (x) d x

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

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

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integral of x/(4+25x^2)

\int \frac{x}{4 + 25 x ^{2}} d x