integral of x(3sec^2(x)-tan^2(x))

解

\int x (3 sec^{2} (x) - tan^{2} (x)) d x

x^{2} - \frac{x ^{2}}{2} + 2 x tan (x) + 2 ln ∣ cos (x) ∣ + C

解答ステップ

\int x (3 sec^{2} (x) - tan^{2} (x)) d x

三角関数の公式を使用して書き換える

= \int \frac{x ( 3 - sin ^{2} ( x ) )}{cos ^{2} ( x )} d x

拡張

\frac{x ( 3 - sin ^{2} ( x ) )}{cos ^{2} ( x )} : \frac{3 x}{cos ^{2} ( x )} - \frac{x sin ^{2} ( x )}{cos ^{2} ( x )}

= \int \frac{3 x}{cos ^{2} ( x )} - \frac{x sin ^{2} ( x )}{cos ^{2} ( x )} d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{3 x}{cos ^{2} ( x )} d x - \int \frac{x sin ^{2} ( x )}{cos ^{2} ( x )} d x

\int \frac{3 x}{cos ^{2} ( x )} d x = 6 (\frac{1}{2} x tan (x) + \frac{1}{2} ln ∣ cos (x) ∣)

\int \frac{x sin ^{2} ( x )}{cos ^{2} ( x )} d x = - x^{2} + x tan (x) + \frac{x ^{2}}{2} + ln ∣ cos (x) ∣

= 6 (\frac{1}{2} x tan (x) + \frac{1}{2} ln ∣ cos (x) ∣) - (- x^{2} + x tan (x) + \frac{x ^{2}}{2} + ln ∣ cos (x) ∣)

簡素化

6 (\frac{1}{2} x tan (x) + \frac{1}{2} ln ∣ cos (x) ∣) - (- x^{2} + x tan (x) + \frac{x ^{2}}{2} + ln ∣ cos (x) ∣) : x^{2} - \frac{x ^{2}}{2} + 2 x tan (x) + 2 ln ∣ cos (x) ∣

= x^{2} - \frac{x ^{2}}{2} + 2 x tan (x) + 2 ln ∣ cos (x) ∣

定数を解答に追加する

= x^{2} - \frac{x ^{2}}{2} + 2 x tan (x) + 2 ln ∣ cos (x) ∣ + C