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人気のある微分積分 >

integral of (sin(19x))/(1+cos^2(19x))

解

\int \frac{sin ( 19 x )}{1 + cos ^{2} ( 19 x )} d x

解

\frac{1}{19} arctan (tan^{2} (\frac{19 x}{2})) + C

解答ステップ

\int \frac{sin ( 19 x )}{1 + cos ^{2} ( 19 x )} d x

u置換積分法を適用する

= \int \frac{sin ( u )}{19 ( 1 + cos ^{2} ( u ) )} d u

定数を除く:

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

= \frac{1}{19} \cdot \int \frac{sin ( u )}{1 + cos ^{2} ( u )} d u

三角関数の公式を使用して書き換える

= \frac{1}{19} \cdot \int \frac{sin ( u )}{- sin ^{2} ( u ) + 2} d u

u置換積分法を適用する

= \frac{1}{19} \cdot \int \frac{2 v}{v ^{4} + 1} d v

定数を除く:

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

= \frac{1}{19} \cdot 2 \cdot \int \frac{v}{v ^{4} + 1} d v

u置換積分法を適用する

= \frac{1}{19} \cdot 2 \cdot \int \frac{1}{2 ( w ^{2} + 1 )} d w

定数を除く:

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

= \frac{1}{19} \cdot 2 \cdot \frac{1}{2} \cdot \int \frac{1}{w ^{2} + 1} d w

共通積分を使用する:

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

= \frac{1}{19} \cdot 2 \cdot \frac{1}{2} arctan (w)

代用を戻す

= \frac{1}{19} \cdot 2 \cdot \frac{1}{2} arctan (tan^{2} (\frac{19 x}{2}))

簡素化

\frac{1}{19} \cdot 2 \cdot \frac{1}{2} arctan (tan^{2} (\frac{19 x}{2})) : \frac{1}{19} arctan (tan^{2} (\frac{19 x}{2}))

= \frac{1}{19} arctan (tan^{2} (\frac{19 x}{2}))

定数を解答に追加する

= \frac{1}{19} arctan (tan^{2} (\frac{19 x}{2})) + C

グラフ

人気の例

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