integral of-sin(x)tan(x)

解答

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

- ln ∣ tan (x) + sec (x) ∣ + sin (x) + C

求解步骤

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

提出常数:

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

= - \int sin (x) tan (x) d x

使用三角恒等式改写

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

乘开

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

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

\int \frac{1}{cos ( x )} d x = ln ∣ tan (x) + sec (x) ∣

\int cos (x) d x = sin (x)

= - (ln ∣ tan (x) + sec (x) ∣ - sin (x))

化简

= - ln ∣ tan (x) + sec (x) ∣ + sin (x)

解答补常数

= - ln ∣ tan (x) + sec (x) ∣ + sin (x) + C