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人気のある微分積分 >

integral from-1 to 1 of 1/(1+sin(x))

解

\int_{- 1}^{1} \frac{1}{1 + sin ( x )} d x

解

\frac{4 tan ( \frac{1}{2} )}{( tan ( \frac{1}{2} ) + 1 ) ( - tan ( \frac{1}{2} ) + 1 )}

+1

十進法表記

3.11481 \dots

解答ステップ

\int_{- 1}^{1} \frac{1}{1 + sin ( x )} d x

u置換積分法を適用する

= \int_{- t a n (\frac{1}{2})}^{t a n (\frac{1}{2})} \frac{2}{1 + u ^{2} + 2 u} d u

定数を除く:

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

= 2 \cdot \int_{- t a n (\frac{1}{2})}^{t a n (\frac{1}{2})} \frac{1}{1 + u ^{2} + 2 u} d u

平方完成する

1 + u^{2} + 2 u : (u + 1)^{2}

= 2 \cdot \int_{- t a n (\frac{1}{2})}^{t a n (\frac{1}{2})} \frac{1}{( u + 1 ) ^{2}} d u

u置換積分法を適用する

= 2 \cdot \int_{- t a n (\frac{1}{2}) + 1}^{t a n (\frac{1}{2}) + 1} \frac{1}{v ^{2}} d v

べき乗則を適用する

= 2 [- \frac{1}{v}]_{- t a n (\frac{1}{2}) + 1}^{t a n (\frac{1}{2}) + 1}

限度を計算する:

\frac{2}{( tan ( \frac{1}{2} ) + 1 ) ( - tan ( \frac{1}{2} ) + 1 )} tan (\frac{1}{2})

= 2 \cdot \frac{2}{( tan ( \frac{1}{2} ) + 1 ) ( - tan ( \frac{1}{2} ) + 1 )} tan (\frac{1}{2})

簡素化

= \frac{4 tan ( \frac{1}{2} )}{( tan ( \frac{1}{2} ) + 1 ) ( - tan ( \frac{1}{2} ) + 1 )}

グラフ

人気の例

integral from 2 to 5 of 2^{2x}

\int_{2}^{5} 2^{2 x} d x

integral from 2 to 4 of (1-e)

\int_{2}^{4} (1 - e) d x

integral from 1 to sqrt(2 of) 1/(x^2)

\int_{1}^{2} \frac{1}{x ^{2}} d x

integral from 1 to 4 of 1/(2x(3-x))

\int_{1}^{4} \frac{1}{2 x ( 3 - x )} d x

integral from 0 to 2 of x^2sin(x^3)

\int_{0}^{2} x^{2} sin (x^{3}) d x