integral of 1/(sqrt(1+a^2y^2))

解答

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

\frac{1}{a} ln a y + 1 + a^{2} y^{2} + C

求解步骤

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

使用三角换元法

= \int \frac{sec ( u )}{a} d u

提出常数:

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

= \frac{1}{a} \cdot \int sec (u) d u

使用积分常见定则:

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

= \frac{1}{a} ln ∣ tan (u) + sec (u) ∣

u = arctan (a y)

代回

= \frac{1}{a} ln ∣ tan (arctan (a y)) + sec (arctan (a y)) ∣

化简

\frac{1}{a} ln ∣ tan (arctan (a y)) + sec (arctan (a y)) ∣ : \frac{1}{a} ln a y + 1 + a^{2} y^{2}

= \frac{1}{a} ln a y + 1 + a^{2} y^{2}

解答补常数

= \frac{1}{a} ln a y + 1 + a^{2} y^{2} + C

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