integral of ((1+tan(t))^2)/(sec^2(t))

解

\int \frac{( 1 + tan ( t ) ) ^{2}}{sec ^{2} ( t )} d t

t + sin^{2} (t) + C

解答ステップ

\int \frac{( 1 + tan ( t ) ) ^{2}}{sec ^{2} ( t )} d t

拡張

\frac{( 1 + tan ( t ) ) ^{2}}{sec ^{2} ( t )} : \frac{1}{sec ^{2} ( t )} + \frac{2 tan ( t )}{sec ^{2} ( t )} + \frac{tan ^{2} ( t )}{sec ^{2} ( t )}

= \int \frac{1}{sec ^{2} ( t )} + \frac{2 tan ( t )}{sec ^{2} ( t )} + \frac{tan ^{2} ( t )}{sec ^{2} ( t )} 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 \frac{1}{sec ^{2} ( t )} d t + \int \frac{2 tan ( t )}{sec ^{2} ( t )} d t + \int \frac{tan ^{2} ( t )}{sec ^{2} ( t )} d t

\int \frac{1}{sec ^{2} ( t )} d t = \frac{1}{2} (t + \frac{1}{2} sin (2 t))

\int \frac{2 tan ( t )}{sec ^{2} ( t )} d t = sin^{2} (t)

\int \frac{tan ^{2} ( t )}{sec ^{2} ( t )} d t = \frac{1}{2} (t - \frac{1}{2} sin (2 t))

= \frac{1}{2} (t + \frac{1}{2} sin (2 t)) + sin^{2} (t) + \frac{1}{2} (t - \frac{1}{2} sin (2 t))

簡素化

\frac{1}{2} (t + \frac{1}{2} sin (2 t)) + sin^{2} (t) + \frac{1}{2} (t - \frac{1}{2} sin (2 t)) : t + sin^{2} (t)

= t + sin^{2} (t)

定数を解答に追加する

= t + sin^{2} (t) + C