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受欢迎的微积分 >

integral of (sin(26x))/(1+cos^2(26x))

解答

\int \frac{sin ( 26 x )}{1 + cos ^{2} ( 26 x )} d x

解答

\frac{1}{26} arctan (tan^{2} (13 x)) + C

求解步骤

\int \frac{sin ( 26 x )}{1 + cos ^{2} ( 26 x )} d x

使用换元积分法

= \int \frac{sin ( u )}{26 ( 1 + cos ^{2} ( u ) )} d u

提出常数:

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

= \frac{1}{26} \cdot \int \frac{sin ( u )}{1 + cos ^{2} ( u )} d u

使用三角恒等式改写

= \frac{1}{26} \cdot \int \frac{sin ( u )}{- sin ^{2} ( u ) + 2} d u

使用换元积分法

= \frac{1}{26} \cdot \int \frac{2 v}{v ^{4} + 1} d v

提出常数:

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

= \frac{1}{26} \cdot 2 \cdot \int \frac{v}{v ^{4} + 1} d v

使用换元积分法

= \frac{1}{26} \cdot 2 \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}{26} \cdot 2 \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 = arctan (w)

= \frac{1}{26} \cdot 2 \cdot \frac{1}{2} arctan (w)

代回

= \frac{1}{26} \cdot 2 \cdot \frac{1}{2} arctan (tan^{2} (\frac{26 x}{2}))

化简

\frac{1}{26} \cdot 2 \cdot \frac{1}{2} arctan (tan^{2} (\frac{26 x}{2})) : \frac{1}{26} arctan (tan^{2} (13 x))

= \frac{1}{26} arctan (tan^{2} (13 x))

解答补常数

= \frac{1}{26} arctan (tan^{2} (13 x)) + C

作图

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