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人気のある微分積分 >

integral of 5sec^{-2}(xta)n^3x

解

\int 5 sec^{- 2} (x t a) n^{3} x d x

解

\frac{5 n ^{3}}{8 a ^{2} t ^{2}} (2 a^{2} t^{2} x^{2} + 2 a t x sin (2 a t x) + cos (2 a t x)) + C

解答ステップ

\int 5 sec^{- 2} (x t a) n^{3} x d x

定数を除く:

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

= 5 n^{3} \cdot \int sec^{- 2} (a t x) x d x

簡素化

sec^{- 2} (a t x) x : x cos^{2} (a t x)

= 5 n^{3} \cdot \int x cos^{2} (a t x) d x

u置換積分法を適用する

= 5 n^{3} \cdot \int \frac{u cos ^{2} ( u )}{a ^{2} t ^{2}} d u

定数を除く:

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

= 5 n^{3} \frac{1}{a ^{2} t ^{2}} \cdot \int u cos^{2} (u) d u

部分積分を適用する

= 5 n^{3} \frac{1}{a ^{2} t ^{2}} (\frac{1}{2} u (u + \frac{1}{2} sin (2 u)) - \int \frac{1}{2} (u + \frac{1}{2} sin (2 u)) d u)

\int \frac{1}{2} (u + \frac{1}{2} sin (2 u)) d u = \frac{1}{2} (\frac{u ^{2}}{2} - \frac{1}{4} cos (2 u))

= 5 n^{3} \frac{1}{a ^{2} t ^{2}} (\frac{1}{2} u (u + \frac{1}{2} sin (2 u)) - \frac{1}{2} (\frac{u ^{2}}{2} - \frac{1}{4} cos (2 u)))

代用を戻す

u = a t x

= 5 n^{3} \frac{1}{a ^{2} t ^{2}} (\frac{1}{2} a t x (a t x + \frac{1}{2} sin (2 a t x)) - \frac{1}{2} (\frac{( a t x ) ^{2}}{2} - \frac{1}{4} cos (2 a t x)))

簡素化

5 n^{3} \frac{1}{a ^{2} t ^{2}} (\frac{1}{2} a t x (a t x + \frac{1}{2} sin (2 a t x)) - \frac{1}{2} (\frac{( a t x ) ^{2}}{2} - \frac{1}{4} cos (2 a t x))) : \frac{5 n ^{3}}{8 a ^{2} t ^{2}} (2 a^{2} t^{2} x^{2} + 2 a t x sin (2 a t x) + cos (2 a t x))

= \frac{5 n ^{3}}{8 a ^{2} t ^{2}} (2 a^{2} t^{2} x^{2} + 2 a t x sin (2 a t x) + cos (2 a t x))

定数を解答に追加する

= \frac{5 n ^{3}}{8 a ^{2} t ^{2}} (2 a^{2} t^{2} x^{2} + 2 a t x sin (2 a t x) + cos (2 a t x)) + C

グラフ

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