integral from 1 to 1/x of cos^3(u)

解

\int_{1}^{\frac{1}{x}} cos^{3} (u) d u

sin (\frac{1}{x}) - sin (1) - \frac{sin ^{3} ( \frac{1}{x} ) - sin ^{3} ( 1 )}{3}

解答ステップ

\int_{1}^{\frac{1}{x}} cos^{3} (u) d u

三角関数の公式を使用して書き換える

= \int_{1}^{\frac{1}{x}} (1 - sin^{2} (u)) cos (u) d u

u置換積分法を適用する

= \int_{s i n (1)}^{s i n (\frac{1}{x})} 1 - v^{2} d v

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_{s i n (1)}^{s i n (\frac{1}{x})} 1 d v - \int_{s i n (1)}^{s i n (\frac{1}{x})} v^{2} d v

\int_{s i n (1)}^{s i n (\frac{1}{x})} 1 d v = sin (\frac{1}{x}) - sin (1)

\int_{s i n (1)}^{s i n (\frac{1}{x})} v^{2} d v = \frac{sin ^{3} ( \frac{1}{x} ) - sin ^{3} ( 1 )}{3}

= sin (\frac{1}{x}) - sin (1) - \frac{sin ^{3} ( \frac{1}{x} ) - sin ^{3} ( 1 )}{3}