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人気のある微分積分 >

integral of 13sin^2(xco)s^2x

解

\int 13 sin^{2} (x co) s^{2} x d x

解

\frac{13 s ^{2}}{8 c ^{2} o ^{2}} (2 c^{2} o^{2} x^{2} - 2 co x sin (2 co x) - cos (2 co x)) + C

解答ステップ

\int 13 sin^{2} (x co) s^{2} x d x

定数を除く:

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

= 13 s^{2} \cdot \int sin^{2} (co x) x d x

u置換積分法を適用する

= 13 s^{2} \cdot \int \frac{u sin ^{2} ( u )}{c ^{2} o ^{2}} d u

定数を除く:

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

= 13 s^{2} \frac{1}{c ^{2} o ^{2}} \cdot \int u sin^{2} (u) d u

部分積分を適用する

= 13 s^{2} \frac{1}{c ^{2} o ^{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))

= 13 s^{2} \frac{1}{c ^{2} o ^{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 = co x

= 13 s^{2} \frac{1}{c ^{2} o ^{2}} (\frac{1}{2} co x (co x - \frac{1}{2} sin (2 co x)) - \frac{1}{2} (\frac{( co x ) ^{2}}{2} + \frac{1}{4} cos (2 co x)))

簡素化

13 s^{2} \frac{1}{c ^{2} o ^{2}} (\frac{1}{2} co x (co x - \frac{1}{2} sin (2 co x)) - \frac{1}{2} (\frac{( co x ) ^{2}}{2} + \frac{1}{4} cos (2 co x))) : \frac{13 s ^{2}}{8 c ^{2} o ^{2}} (2 c^{2} o^{2} x^{2} - 2 co x sin (2 co x) - cos (2 co x))

= \frac{13 s ^{2}}{8 c ^{2} o ^{2}} (2 c^{2} o^{2} x^{2} - 2 co x sin (2 co x) - cos (2 co x))

定数を解答に追加する

= \frac{13 s ^{2}}{8 c ^{2} o ^{2}} (2 c^{2} o^{2} x^{2} - 2 co x sin (2 co x) - cos (2 co x)) + C

グラフ

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