integral of sin^2(x)sec(x)

解

\int sin^{2} (x) sec (x) d x

ln ∣ tan (x) + sec (x) ∣ - sin (x) + C

解答ステップ

\int sin^{2} (x) sec (x) d x

簡素化

sin^{2} (x) sec (x) : sin (x) tan (x)

= \int sin (x) tan (x) d x

三角関数の公式を使用して書き換える

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

拡張

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

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

\int \frac{1}{cos ( x )} d x = ln ∣ tan (x) + sec (x) ∣

\int cos (x) d x = sin (x)

= ln ∣ tan (x) + sec (x) ∣ - sin (x)

定数を解答に追加する

= ln ∣ tan (x) + sec (x) ∣ - sin (x) + C