integral of (msec(θ)-tan(θ))^2

解答

\int (m sec (θ) - tan (θ))^{2} d θ

tan (θ) - θ - 2 m sec (θ) + m^{2} tan (θ) + C

求解步骤

\int (m sec (θ) - tan (θ))^{2} d θ

使用三角恒等式改写

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

乘开

\frac{( - sin ( θ ) + m ) ^{2}}{cos ^{2} ( θ )} : \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )} - \frac{2 m sin ( θ )}{cos ^{2} ( θ )} + \frac{m ^{2}}{cos ^{2} ( θ )}

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

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{sin ^{2} ( θ )}{cos ^{2} ( θ )} d θ - \int \frac{2 m sin ( θ )}{cos ^{2} ( θ )} d θ + \int \frac{m ^{2}}{cos ^{2} ( θ )} d θ

\int \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )} d θ = tan (θ) - θ

\int \frac{2 m sin ( θ )}{cos ^{2} ( θ )} d θ = 2 m sec (θ)

\int \frac{m ^{2}}{cos ^{2} ( θ )} d θ = m^{2} tan (θ)

= tan (θ) - θ - 2 m sec (θ) + m^{2} tan (θ)

解答补常数

= tan (θ) - θ - 2 m sec (θ) + m^{2} tan (θ) + C