integral of sin(x)sec(x)tan(x)

解答

\int sin (x) sec (x) tan (x) d x

tan (x) - x + C

求解步骤

\int sin (x) sec (x) tan (x) d x

使用三角恒等式改写

= \int \frac{1 - cos ^{2} ( x )}{cos ^{2} ( x )} d x

乘开

\frac{1 - cos ^{2} ( x )}{cos ^{2} ( x )} : \frac{1}{cos ^{2} ( x )} - 1

= \int \frac{1}{cos ^{2} ( x )} - 1 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

= \int \frac{1}{cos ^{2} ( x )} d x - \int 1 d x

\int \frac{1}{cos ^{2} ( x )} d x = tan (x)

\int 1 d x = x

= tan (x) - x

解答补常数

= tan (x) - x + C

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

\int 4 x^{2} sin (7 x^{3}) d x

\int (12 x^{7} + 25 x^{6} - 22 x^{5} + 15 x^{4}) d x

\int \frac{x + 3}{2 x + 3} d x

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