integral from tan(2x) to 1 of (t^3+2t)^2

解

\int_{t a n (2 x)}^{1} (t^{3} + 2 t)^{2} d t

\frac{1 - tan ^{7} ( 2 x )}{7} + \frac{4 ( 1 - tan ^{5} ( 2 x ) )}{5} + \frac{4 ( 1 - tan ^{3} ( 2 x ) )}{3}

解答ステップ

\int_{t a n (2 x)}^{1} (t^{3} + 2 t)^{2} d t

拡張

(t^{3} + 2 t)^{2} : t^{6} + 4 t^{4} + 4 t^{2}

= \int_{t a n (2 x)}^{1} t^{6} + 4 t^{4} + 4 t^{2} d t

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_{t a n (2 x)}^{1} t^{6} d t + \int_{t a n (2 x)}^{1} 4 t^{4} d t + \int_{t a n (2 x)}^{1} 4 t^{2} d t

\int_{t a n (2 x)}^{1} t^{6} d t = \frac{1 - tan ^{7} ( 2 x )}{7}

\int_{t a n (2 x)}^{1} 4 t^{4} d t = \frac{4 ( 1 - tan ^{5} ( 2 x ) )}{5}

\int_{t a n (2 x)}^{1} 4 t^{2} d t = \frac{4 ( 1 - tan ^{3} ( 2 x ) )}{3}

= \frac{1 - tan ^{7} ( 2 x )}{7} + \frac{4 ( 1 - tan ^{5} ( 2 x ) )}{5} + \frac{4 ( 1 - tan ^{3} ( 2 x ) )}{3}