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受欢迎的微积分 >

integral of (sin(x))/(sin^3(x)+cos^3(x))

解答

\int \frac{sin ( x )}{sin ^{3} ( x ) + cos ^{3} ( x )} d x

解答

- \frac{1}{3} ln ∣ tan (x) + 1 ∣ + \frac{1}{6} (ln (4 - tan (x) + sec^{2} (x)) + 23 arctan (\frac{1}{3} (2 tan (x) - 1))) + C

求解步骤

\int \frac{sin ( x )}{sin ^{3} ( x ) + cos ^{3} ( x )} d x

使用三角换元法

= \int \frac{u}{u ^{3} + 1} d u

将

\frac{u}{u ^{3} + 1}

用部份分式展开:

- \frac{1}{3 ( u + 1 )} + \frac{u + 1}{3 ( u ^{2} - u + 1 )}

= \int - \frac{1}{3 ( u + 1 )} + \frac{u + 1}{3 ( u ^{2} - u + 1 )} 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

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

\int \frac{1}{3 ( u + 1 )} d u = \frac{1}{3} ln ∣ u + 1 ∣

\int \frac{u + 1}{3 ( u ^{2} - u + 1 )} d u = \frac{1}{6} (ln 4 u^{2} - 4 u + 4 + 23 arctan (\frac{2 u - 1}{3}))

= - \frac{1}{3} ln ∣ u + 1 ∣ + \frac{1}{6} (ln 4 u^{2} - 4 u + 4 + 23 arctan (\frac{2 u - 1}{3}))

u = tan (x)

代回

= - \frac{1}{3} ln ∣ tan (x) + 1 ∣ + \frac{1}{6} (ln 4 tan^{2} (x) - 4 tan (x) + 4 + 23 arctan (\frac{2 tan ( x ) - 1}{3}))

化简

- \frac{1}{3} ln ∣ tan (x) + 1 ∣ + \frac{1}{6} (ln 4 tan^{2} (x) - 4 tan (x) + 4 + 23 arctan (\frac{2 tan ( x ) - 1}{3})) : - \frac{1}{3} ln ∣ tan (x) + 1 ∣ + \frac{1}{6} (ln (4 - tan (x) + sec^{2} (x)) + 23 arctan (\frac{1}{3} (2 tan (x) - 1)))

= - \frac{1}{3} ln ∣ tan (x) + 1 ∣ + \frac{1}{6} (ln (4 - tan (x) + sec^{2} (x)) + 23 arctan (\frac{1}{3} (2 tan (x) - 1)))

解答补常数

= - \frac{1}{3} ln ∣ tan (x) + 1 ∣ + \frac{1}{6} (ln (4 - tan (x) + sec^{2} (x)) + 23 arctan (\frac{1}{3} (2 tan (x) - 1))) + C

作图

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