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受欢迎的微积分 >

integral of (sin^7(6x))/(cos^4(6x))

解答

\int \frac{sin ^{7} ( 6 x )}{cos ^{4} ( 6 x )} d x

解答

\frac{1}{18} (sec^{3} (6 x) - 9 sec (6 x) - 9 cos (6 x) + cos^{3} (6 x)) + C

求解步骤

\int \frac{sin ^{7} ( 6 x )}{cos ^{4} ( 6 x )} d x

使用换元积分法

= \int \frac{sin ^{7} ( u )}{6 cos ^{4} ( u )} d u

提出常数:

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

= \frac{1}{6} \cdot \int \frac{sin ^{7} ( u )}{cos ^{4} ( u )} d u

使用三角恒等式改写

= \frac{1}{6} \cdot \int \frac{( 1 - cos ^{2} ( u ) ) ^{3} sin ( u )}{cos ^{4} ( u )} d u

使用换元积分法

= \frac{1}{6} \cdot \int - \frac{( 1 - v ^{2} ) ^{3}}{v ^{4}} d v

乘开

- \frac{( 1 - v ^{2} ) ^{3}}{v ^{4}} : - \frac{1}{v ^{4}} + \frac{3}{v ^{2}} - 3 + v^{2}

= \frac{1}{6} \cdot \int - \frac{1}{v ^{4}} + \frac{3}{v ^{2}} - 3 + v^{2} d v

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{6} (- \int \frac{1}{v ^{4}} d v + \int \frac{3}{v ^{2}} d v - \int 3 d v + \int v^{2} d v)

\int \frac{1}{v ^{4}} d v = - \frac{1}{3 v ^{3}}

\int \frac{3}{v ^{2}} d v = - \frac{3}{v}

\int 3 d v = 3 v

\int v^{2} d v = \frac{v ^{3}}{3}

= \frac{1}{6} (- (- \frac{1}{3 v ^{3}}) - \frac{3}{v} - 3 v + \frac{v ^{3}}{3})

代回

= \frac{1}{6} (- (- \frac{1}{3 cos ^{3} ( 6 x )}) - \frac{3}{cos ( 6 x )} - 3 cos (6 x) + \frac{cos ^{3} ( 6 x )}{3})

化简

\frac{1}{6} (- (- \frac{1}{3 cos ^{3} ( 6 x )}) - \frac{3}{cos ( 6 x )} - 3 cos (6 x) + \frac{cos ^{3} ( 6 x )}{3}) : \frac{1}{18} (sec^{3} (6 x) - 9 sec (6 x) - 9 cos (6 x) + cos^{3} (6 x))

= \frac{1}{18} (sec^{3} (6 x) - 9 sec (6 x) - 9 cos (6 x) + cos^{3} (6 x))

解答补常数

= \frac{1}{18} (sec^{3} (6 x) - 9 sec (6 x) - 9 cos (6 x) + cos^{3} (6 x)) + C

作图

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