integral of (7sec(x)-cos(x))^2

解答

\int (7 sec (x) - cos (x))^{2} d x

49 tan (x) - \frac{27}{2} x + \frac{1}{4} sin (2 x) + C

求解步骤

\int (7 sec (x) - cos (x))^{2} d x

使用三角恒等式改写

= \int \frac{( 7 - cos ^{2} ( x ) ) ^{2}}{cos ^{2} ( x )} d x

乘开

\frac{( 7 - cos ^{2} ( x ) ) ^{2}}{cos ^{2} ( x )} : \frac{49}{cos ^{2} ( x )} - 14 + cos^{2} (x)

= \int \frac{49}{cos ^{2} ( x )} - 14 + cos^{2} (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{49}{cos ^{2} ( x )} d x - \int 14 d x + \int cos^{2} (x) d x

\int \frac{49}{cos ^{2} ( x )} d x = 49 tan (x)

\int 14 d x = 14 x

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

= 49 tan (x) - 14 x + \frac{1}{2} (x + \frac{1}{2} sin (2 x))

化简

49 tan (x) - 14 x + \frac{1}{2} (x + \frac{1}{2} sin (2 x)) : 49 tan (x) - \frac{27}{2} x + \frac{1}{4} sin (2 x)

= 49 tan (x) - \frac{27}{2} x + \frac{1}{4} sin (2 x)

解答补常数

= 49 tan (x) - \frac{27}{2} x + \frac{1}{4} sin (2 x) + C

\frac{d}{d x} (e^{- \frac{1}{x ^{4}}})

\frac{d}{d x} ((sin (cos (2 x)))^{2})

\frac{d}{d t} (20 - 25 sin^{2} (t))

\int 3 sin (x) d x

\frac{\partial}{\partial x} (x^{y} + ln (z))