integral of 8sec(x)(sec(x)-4tan(x))

解答

\int 8 sec (x) (sec (x) - 4 tan (x)) d x

8 (tan (x) - 4 sec (x)) + C

求解步骤

\int 8 sec (x) (sec (x) - 4 tan (x)) d x

提出常数:

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

= 8 \cdot \int sec (x) (sec (x) - 4 tan (x)) d x

使用三角恒等式改写

= 8 \cdot \int \frac{1 - 4 sin ( x )}{cos ^{2} ( x )} d x

乘开

\frac{1 - 4 sin ( x )}{cos ^{2} ( x )} : \frac{1}{cos ^{2} ( x )} - \frac{4 sin ( x )}{cos ^{2} ( x )}

= 8 \cdot \int \frac{1}{cos ^{2} ( x )} - \frac{4 sin ( x )}{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

= 8 (\int \frac{1}{cos ^{2} ( x )} d x - \int \frac{4 sin ( x )}{cos ^{2} ( x )} d x)

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

\int \frac{4 sin ( x )}{cos ^{2} ( x )} d x = 4 sec (x)

= 8 (tan (x) - 4 sec (x))

解答补常数

= 8 (tan (x) - 4 sec (x)) + C

y^{^{''}} + 8 y^{^{'}} + 4 y = t

\frac{\partial}{\partial y} (sin (z))

a re a 4 x + 9, 144 - x, - 9, 144

d er i v a t i v e \frac{2}{x ^{3}} - \frac{8}{x}

d er i v a t i v e y = (\frac{f ( x )}{g ( x )})^{2}