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受欢迎的微积分 >

integral of sqrt(1+4θ^2)

解答

\int 1 + 4 θ^{2} d θ

解答

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

求解步骤

\int 1 + 4 θ^{2} d θ

使用三角换元法

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

使用降次积分法

= \frac{1}{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

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

u = arctan (2 θ)

代回

= \frac{1}{2} (\frac{sec ^{2} ( arctan ( 2 θ ) ) sin ( arctan ( 2 θ ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (2 θ)) + sec (arctan (2 θ)) ∣)

化简

\frac{1}{2} (\frac{sec ^{2} ( arctan ( 2 θ ) ) sin ( arctan ( 2 θ ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (2 θ)) + sec (arctan (2 θ)) ∣) : \frac{1}{4} (2 θ 4 θ^{2} + 1 + ln 2 θ + 1 + 4 θ^{2})

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

解答补常数

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

作图

流行的例子

laplacetransform e^{2t}(t-1)^2

l a pl a ce t r an s f or m e^{2 t} (t - 1)^{2}

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\frac{d}{d x} ((1 - 3 x) arccoth (3 x))

y^{^{'}} + 4 x y^{3} = 0

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\frac{d}{d x} (\frac{4 x ^{3}}{3} + cos (x) + 1)

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\frac{d}{d x} (sin (n x))