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受欢迎的微积分 >

integral of (cos(x))/(1+cos^2(x))

解答

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

解答

\frac{1}{2 2} (ln \frac{sin ( x )}{2} + 1 - ln \frac{sin ( x )}{2} - 1) + C

求解步骤

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

使用三角换元法

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

使用换元积分法

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

提出常数:

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

= \frac{1}{2} \cdot \int \frac{1}{- v ^{2} + 1} d v

使用积分常见定则:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= \frac{1}{2} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

代回

= \frac{1}{2} \frac{ln \frac{s i n ( x )}{2} + 1}{2} - \frac{ln \frac{s i n ( x )}{2} - 1}{2}

化简

\frac{1}{2} \frac{ln \frac{s i n ( x )}{2} + 1}{2} - \frac{ln \frac{s i n ( x )}{2} - 1}{2} : \frac{1}{2 2} (ln \frac{sin ( x )}{2} + 1 - ln \frac{sin ( x )}{2} - 1)

= \frac{1}{2 2} (ln \frac{sin ( x )}{2} + 1 - ln \frac{sin ( x )}{2} - 1)

解答补常数

= \frac{1}{2 2} (ln \frac{sin ( x )}{2} + 1 - ln \frac{sin ( x )}{2} - 1) + C

作图

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\int (3 x^{2} + 1)^{2} d x

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