integral from 0 to 2pi of cos^2(xsi)n^2x

解答

\int_{0}^{2 π} cos^{2} (x s i) n^{2} x d x

\frac{n ^{2} ( 8 π ^{2} s ^{2} i ^{2} + 4 π s i sin ( 4 π s i ) + cos ( 4 π s i ) - 1 )}{8 s ^{2} i ^{2}}

求解步骤

\int_{0}^{2 π} cos^{2} (x s i) n^{2} x d x

提出常数:

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

= n^{2} \cdot \int_{0}^{2 π} cos^{2} (s i x) x d x

使用换元积分法

= n^{2} \cdot \int_{0}^{s i \cdot 2 π} \frac{u cos ^{2} ( u )}{s ^{2} i ^{2}} d u

提出常数:

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

= n^{2} \frac{1}{s ^{2} i ^{2}} \cdot \int_{0}^{s i \cdot 2 π} u cos^{2} (u) d u

使用分布积分法

= n^{2} \frac{1}{s ^{2} i ^{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]_{0}^{2 π s i}

\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))

= n^{2} \frac{1}{s ^{2} i ^{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))]_{0}^{2 π s i}

化简

n^{2} \frac{1}{s ^{2} i ^{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))]_{0}^{2 π s i} : \frac{n ^{2}}{s ^{2} i ^{2}} [\frac{2 u ^{2} + 2 u sin ( 2 u ) + cos ( 2 u )}{8}]_{0}^{2 π s i}

= \frac{n ^{2}}{s ^{2} i ^{2}} [\frac{2 u ^{2} + 2 u sin ( 2 u ) + cos ( 2 u )}{8}]_{0}^{2 π s i}

带入上下限计算后得:

\frac{8 π ^{2} s ^{2} i ^{2} + 4 π s i sin ( 4 π s i ) + cos ( 4 π s i )}{8} - \frac{1}{8}

= \frac{n ^{2}}{s ^{2} i ^{2}} (\frac{8 π ^{2} s ^{2} i ^{2} + 4 π s i sin ( 4 π s i ) + cos ( 4 π s i )}{8} - \frac{1}{8})

化简

= \frac{n ^{2} ( 8 π ^{2} s ^{2} i ^{2} + 4 π s i sin ( 4 π s i ) + cos ( 4 π s i ) - 1 )}{8 s ^{2} i ^{2}}

\int ln (\frac{1}{x - 2}) d x

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\frac{d}{d x} (2 x^{3} - 15 x^{2} + 36 x - 10)

\int (t - \frac{1}{t}) d t

\frac{d}{d x} (sin (7 x + ln (5 x)))