integral of (sec(m)x+tan(m)x)^2

解答

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

\frac{1}{3 cos ^{2} ( m )} (x^{3} + 2 x^{3} sin (m) + x^{3} sin^{2} (m)) + C

求解步骤

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

使用三角恒等式改写

= \int \frac{( x + x sin ( m ) ) ^{2}}{cos ^{2} ( m )} d x

提出常数:

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

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

乘开

(x + x sin (m))^{2} : x^{2} + 2 x^{2} sin (m) + x^{2} sin^{2} (m)

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

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int x^{2} d x = \frac{x ^{3}}{3}

\int 2 x^{2} sin (m) d x = \frac{2}{3} x^{3} sin (m)

\int x^{2} sin^{2} (m) d x = \frac{x ^{3} sin ^{2} ( m )}{3}

= \frac{1}{cos ^{2} ( m )} (\frac{x ^{3}}{3} + \frac{2}{3} x^{3} sin (m) + \frac{x ^{3} sin ^{2} ( m )}{3})

化简

\frac{1}{cos ^{2} ( m )} (\frac{x ^{3}}{3} + \frac{2}{3} x^{3} sin (m) + \frac{x ^{3} sin ^{2} ( m )}{3}) : \frac{1}{3 cos ^{2} ( m )} (x^{3} + 2 x^{3} sin (m) + x^{3} sin^{2} (m))

= \frac{1}{3 cos ^{2} ( m )} (x^{3} + 2 x^{3} sin (m) + x^{3} sin^{2} (m))

解答补常数

= \frac{1}{3 cos ^{2} ( m )} (x^{3} + 2 x^{3} sin (m) + x^{3} sin^{2} (m)) + C