integral of (csc(φ))/(acsc(φ)-asin(φ))

解答

\int \frac{csc ( φ )}{a csc ( φ ) - a sin ( φ )} d φ

\frac{1}{a} tan (φ) + C

求解步骤

\int \frac{csc ( φ )}{a csc ( φ ) - a sin ( φ )} d φ

\frac{csc ( φ )}{csc ( φ ) a - sin ( φ ) a} = \frac{1}{a} \frac{csc ( φ )}{csc ( φ ) - sin ( φ )}

= \int \frac{1}{a} \cdot \frac{csc ( φ )}{csc ( φ ) - sin ( φ )} d φ

提出常数:

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

= \frac{1}{a} \cdot \int \frac{csc ( φ )}{csc ( φ ) - sin ( φ )} d φ

化简

\frac{csc ( φ )}{csc ( φ ) - sin ( φ )} : sec^{2} (φ)

= \frac{1}{a} \cdot \int sec^{2} (φ) d φ

使用积分常见定则:

\int sec^{2} (φ) d φ = tan (φ)

= \frac{1}{a} tan (φ)

解答补常数

= \frac{1}{a} tan (φ) + C

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