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

解

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

\frac{1}{4} (- cot (x) - tan (x)) + C

解答ステップ

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

三角関数の公式を使用して書き換える

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

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

三角関数の公式を使用して書き換える

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

拡張

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

= \frac{1}{4} \cdot \int \frac{1}{sin ^{2} ( x )} - \frac{1}{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

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

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

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

= \frac{1}{4} (- cot (x) - tan (x))

定数を解答に追加する

= \frac{1}{4} (- cot (x) - tan (x)) + C