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人気のある微分積分 >

integral of (sin(2x))/(1+e^{cos^2(x))}

解

\int \frac{sin ( 2 x )}{1 + e ^{c o s^{2} (x)}} d x

解

- ln (e^{c o s^{2} (x)}) + ln e^{c o s^{2} (x)} + 1 + C

解答ステップ

\int \frac{sin ( 2 x )}{1 + e ^{c o s^{2} (x)}} d x

u置換積分法を適用する

= \int - \frac{1}{u ( 1 + u )} d u

定数を除く:

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

= - \int \frac{1}{u ( 1 + u )} d u

以下の部分分数を得る:

\frac{1}{u ( 1 + u )} : \frac{1}{u} - \frac{1}{u + 1}

= - \int \frac{1}{u} - \frac{1}{u + 1} d u

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}{u} d u - \int \frac{1}{u + 1} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u + 1} d u = ln ∣ u + 1 ∣

= - (ln ∣ u ∣ - ln ∣ u + 1 ∣)

代用を戻す

u = e^{c o s^{2} (x)}

= - (ln e^{c o s^{2} (x)} - ln e^{c o s^{2} (x)} + 1)

簡素化

- (ln e^{c o s^{2} (x)} - ln e^{c o s^{2} (x)} + 1) : - ln (e^{c o s^{2} (x)}) + ln e^{c o s^{2} (x)} + 1

= - ln (e^{c o s^{2} (x)}) + ln e^{c o s^{2} (x)} + 1

定数を解答に追加する

= - ln (e^{c o s^{2} (x)}) + ln e^{c o s^{2} (x)} + 1 + C

グラフ

人気の例

integral of (-sec^2(x))/5

\int \frac{- sec ^{2} ( x )}{5} d x

integral of (8x)/(sqrt(x^2+4))

\int \frac{8 x}{x ^{2} + 4} d x

integral of (6x)/(x^2+3)

\int \frac{6 x}{x ^{2} + 3} d x

integral of 1/(sqrt(5-16x^2))

\int \frac{1}{5 - 16 x ^{2}} d x

integral of 7x(2x+1)^{-3}

\int 7 x (2 x + 1)^{- 3} d x