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人気のある微分積分 >

integral of tan^5(x)sec^{-3/2}(x)

解

\int tan^{5} (x) sec^{- \frac{3}{2}} (x) d x

解

- \frac{2}{3 sec ^{\frac{3}{2}} ( x )} - 4 sec^{\frac{1}{2}} (x) + \frac{2}{5} sec^{\frac{5}{2}} (x) + C

解答ステップ

\int tan^{5} (x) sec^{- \frac{3}{2}} (x) d x

三角関数の公式を使用して書き換える

= \int (- 1 + sec^{2} (x))^{2} tan (x) sec^{- \frac{3}{2}} (x) d x

u置換積分法を適用する

= \int \frac{( - 1 + u ^{2} ) ^{2}}{u ^{\frac{5}{2}}} d u

拡張

\frac{( - 1 + u ^{2} ) ^{2}}{u ^{\frac{5}{2}}} : \frac{1}{u ^{\frac{5}{2}}} - \frac{2}{u ^{\frac{1}{2}}} + u^{\frac{3}{2}}

= \int \frac{1}{u ^{\frac{5}{2}}} - \frac{2}{u ^{\frac{1}{2}}} + u^{\frac{3}{2}} 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 ^{\frac{5}{2}}} d u - \int \frac{2}{u ^{\frac{1}{2}}} d u + \int u^{\frac{3}{2}} d u

\int \frac{1}{u ^{\frac{5}{2}}} d u = - \frac{2}{3 u ^{\frac{3}{2}}}

\int \frac{2}{u ^{\frac{1}{2}}} d u = 4 u^{\frac{1}{2}}

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

= - \frac{2}{3 u ^{\frac{3}{2}}} - 4 u^{\frac{1}{2}} + \frac{2}{5} u^{\frac{5}{2}}

代用を戻す

u = sec (x)

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

定数を解答に追加する

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

グラフ

人気の例

integral of x^2sqrt(x^2+9)

\int x^{2} x^{2} + 9 d x

integral of 1/((5x)^2)

\int \frac{1}{( 5 x ) ^{2}} d x

integral of (ln(cos(x)))/((cos(x))^2)

\int \frac{ln ( cos ( x ) )}{( cos ( x ) ) ^{2}} d x

integral from-infinity to 0 of 2^r

\int_{- \infty}^{0} 2^{r} d r

integral of (-9x+3)^4

\int (- 9 x + 3)^{4} d x