zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的微积分 >

integral of 5xtan(4x^2-1)

解答

\int 5 x tan (4 x^{2} - 1) d x

解答

- \frac{5}{8} ln cos (4 x^{2} - 1) + C

求解步骤

\int 5 x tan (4 x^{2} - 1) d x

提出常数:

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

= 5 \cdot \int x tan (4 x^{2} - 1) d x

使用换元积分法

= 5 \cdot \int \frac{tan ( 4 u - 1 )}{2} d u

提出常数:

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

= 5 \cdot \frac{1}{2} \cdot \int tan (4 u - 1) d u

使用三角恒等式改写

= 5 \cdot \frac{1}{2} \cdot \int \frac{sin ( 4 u - 1 )}{cos ( 4 u - 1 )} d u

使用换元积分法

= 5 \cdot \frac{1}{2} \cdot \int - \frac{1}{4 v} d v

提出常数:

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

= 5 \cdot \frac{1}{2} (- \frac{1}{4} \cdot \int \frac{1}{v} d v)

使用积分常见定则:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 5 \cdot \frac{1}{2} (- \frac{1}{4} ln ∣ v ∣)

代回

= 5 \cdot \frac{1}{2} (- \frac{1}{4} ln cos (4 x^{2} - 1))

化简

5 \cdot \frac{1}{2} (- \frac{1}{4} ln cos (4 x^{2} - 1)) : - \frac{5}{8} ln cos (4 x^{2} - 1)

= - \frac{5}{8} ln cos (4 x^{2} - 1)

解答补常数

= - \frac{5}{8} ln cos (4 x^{2} - 1) + C

作图

流行的例子

integral of (3\sqrt[5]{x^2}+3)

\int (3 5 x^{2} + 3) d x

integral of (6x-4)/(sqrt(x^2-10x-119))

\int \frac{6 x - 4}{x ^{2} - 10 x - 119} d x

integral of z/(e^{3z)}

\int \frac{z}{e ^{3 z}} d z

integral of 2xsqrt(4x-5)

\int 2 x 4 x - 5 d x

integral of 1/((z^2-2z+5)^2)

\int \frac{1}{( z ^{2} - 2 z + 5 ) ^{2}} d z