integral of (sin^2(x))/(cos^3(x))

解

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

- x sec^{3} (x) cos^{3} (x) + x + \frac{1}{2} sec (x) tan (x) - \frac{1}{2} ln ∣ tan (x) + sec (x) ∣ + C

解答ステップ

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

拡張

\frac{1 - cos ^{2} ( x )}{cos ^{3} ( x )} : \frac{1}{cos ^{3} ( x )} - \frac{1}{cos ( x )}

= \int \frac{1}{cos ^{3} ( x )} - \frac{1}{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 ^{3} ( x )} d x - \int \frac{1}{cos ( x )} d x

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

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

= - x sec^{3} (x) cos^{3} (x) + x + \frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣ - ln ∣ tan (x) + sec (x) ∣

類似した元を足す:

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

= - x sec^{3} (x) cos^{3} (x) + x + \frac{1}{2} sec (x) tan (x) - \frac{1}{2} ln ∣ tan (x) + sec (x) ∣

定数を解答に追加する

= - x sec^{3} (x) cos^{3} (x) + x + \frac{1}{2} sec (x) tan (x) - \frac{1}{2} ln ∣ tan (x) + sec (x) ∣ + C