integral of sec^2(x)(1-4x)

解

\int sec^{2} (x) (1 - 4 x) d x

tan (x) - 8 (\frac{1}{2} x tan (x) + \frac{1}{2} ln ∣ cos (x) ∣) + C

解答ステップ

\int sec^{2} (x) (1 - 4 x) d x

三角関数の公式を使用して書き換える

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

拡張

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

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

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

\int \frac{4 x}{cos ^{2} ( x )} d x = 8 (\frac{1}{2} x tan (x) + \frac{1}{2} ln ∣ cos (x) ∣)

= tan (x) - 8 (\frac{1}{2} x tan (x) + \frac{1}{2} ln ∣ cos (x) ∣)

定数を解答に追加する

= tan (x) - 8 (\frac{1}{2} x tan (x) + \frac{1}{2} ln ∣ cos (x) ∣) + C