integral of 2xsec^2(5x)

解答

\int 2 x sec^{2} (5 x) d x

x^{2} sec^{2} (5 x) - x^{2} tan^{2} (5 x) + \frac{2}{5} x tan (5 x) + \frac{2}{25} ln ∣ cos (5 x) ∣ - x^{2} + C

求解步骤

\int 2 x sec^{2} (5 x) d x

提出常数:

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

= 2 \cdot \int x sec^{2} (5 x) d x

使用三角恒等式改写

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

使用分布积分法

= 2 (\frac{1}{2} x^{2} sec^{2} (5 x) - \int 5 x^{2} sec^{2} (5 x) tan (5 x) d x)

\int 5 x^{2} sec^{2} (5 x) tan (5 x) d x = \frac{1}{50} (25 x^{2} tan^{2} (5 x) - 10 x tan (5 x) - 2 ln ∣ cos (5 x) ∣ + 25 x^{2})

= 2 (\frac{1}{2} x^{2} sec^{2} (5 x) - \frac{1}{50} (25 x^{2} tan^{2} (5 x) - 10 x tan (5 x) - 2 ln ∣ cos (5 x) ∣ + 25 x^{2}))

化简

2 (\frac{1}{2} x^{2} sec^{2} (5 x) - \frac{1}{50} (25 x^{2} tan^{2} (5 x) - 10 x tan (5 x) - 2 ln ∣ cos (5 x) ∣ + 25 x^{2})) : x^{2} sec^{2} (5 x) - x^{2} tan^{2} (5 x) + \frac{2}{5} x tan (5 x) + \frac{2}{25} ln ∣ cos (5 x) ∣ - x^{2}

= x^{2} sec^{2} (5 x) - x^{2} tan^{2} (5 x) + \frac{2}{5} x tan (5 x) + \frac{2}{25} ln ∣ cos (5 x) ∣ - x^{2}

解答补常数

= x^{2} sec^{2} (5 x) - x^{2} tan^{2} (5 x) + \frac{2}{5} x tan (5 x) + \frac{2}{25} ln ∣ cos (5 x) ∣ - x^{2} + C