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受欢迎的微积分 >

integral of sqrt(tan(4x))(sec(4x))^2

解答

\int tan (4 x) (sec (4 x))^{2} d x

解答

\frac{1}{6} tan^{\frac{3}{2}} (4 x) + C

求解步骤

\int tan (4 x) (sec (4 x))^{2} d x

使用三角恒等式改写

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

使用基本三角恒等式:

\frac{sin ( x )}{cos ( x )} = tan (x)

= \int sec^{2} (4 x) tan (4 x) d x

使用换元积分法

= \int \frac{u ^{2}}{2} d u

提出常数:

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

= \frac{1}{2} \cdot \int u^{2} d u

使用幂法则

= \frac{1}{2} \cdot \frac{u ^{3}}{3}

u = tan (4 x)

代回

= \frac{1}{2} \cdot \frac{( tan ( 4 x ) ) ^{3}}{3}

化简

\frac{1}{2} \cdot \frac{( tan ( 4 x ) ) ^{3}}{3} : \frac{1}{6} tan^{\frac{3}{2}} (4 x)

= \frac{1}{6} tan^{\frac{3}{2}} (4 x)

解答补常数

= \frac{1}{6} tan^{\frac{3}{2}} (4 x) + C

作图

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