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受欢迎的微积分 >

integral of (cos(4x))/(4-sin^2(4x))

解答

\int \frac{cos ( 4 x )}{4 - sin ^{2} ( 4 x )} d x

解答

\frac{1}{16} (ln \frac{sin ( 4 x )}{2} + 1 - ln \frac{sin ( 4 x )}{2} - 1) + C

求解步骤

\int \frac{cos ( 4 x )}{4 - sin ^{2} ( 4 x )} d x

使用三角恒等式改写

= \int \frac{cos ( 4 x )}{cos ^{2} ( 4 x ) + 3} d x

使用换元积分法

= \int \frac{cos ( u )}{4 ( cos ^{2} ( u ) + 3 )} d u

提出常数:

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

= \frac{1}{4} \cdot \int \frac{cos ( u )}{cos ^{2} ( u ) + 3} d u

使用三角恒等式改写

= \frac{1}{4} \cdot \int \frac{cos ( u )}{- sin ^{2} ( u ) + 4} d u

使用换元积分法

= \frac{1}{4} \cdot \int \frac{1}{- v ^{2} + 4} d v

使用换元积分法

= \frac{1}{4} \cdot \int \frac{1}{2 ( - w ^{2} + 1 )} d w

提出常数:

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

= \frac{1}{4} \cdot \frac{1}{2} \cdot \int \frac{1}{- w ^{2} + 1} d w

使用积分常见定则:

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

= \frac{1}{4} \cdot \frac{1}{2} (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2})

代回

= \frac{1}{4} \cdot \frac{1}{2} \frac{ln \frac{s i n ( 4 x )}{2} + 1}{2} - \frac{ln \frac{s i n ( 4 x )}{2} - 1}{2}

化简

\frac{1}{4} \cdot \frac{1}{2} \frac{ln \frac{s i n ( 4 x )}{2} + 1}{2} - \frac{ln \frac{s i n ( 4 x )}{2} - 1}{2} : \frac{1}{16} (ln \frac{sin ( 4 x )}{2} + 1 - ln \frac{sin ( 4 x )}{2} - 1)

= \frac{1}{16} (ln \frac{sin ( 4 x )}{2} + 1 - ln \frac{sin ( 4 x )}{2} - 1)

解答补常数

= \frac{1}{16} (ln \frac{sin ( 4 x )}{2} + 1 - ln \frac{sin ( 4 x )}{2} - 1) + C

作图

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