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受欢迎的微积分 >

integral of (sin^2(x))/(sin(x)+2cos(x))

解答

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

解答

4 (\frac{- 1 - 2 arctan ( tan ( \frac{x}{2} ) ) - 2 tan ( \frac{x}{2} ) - 2 tan ^{2} ( \frac{x}{2} ) arctan ( tan ( \frac{x}{2} ) )}{10 sec ^{2} ( \frac{x}{2} )} + \frac{1}{5} arctan (tan (\frac{x}{2})) + \frac{1}{5 5} (- ln - 1 + \frac{1}{5} (- 1 + 2 tan (\frac{x}{2})) + ln 1 + \frac{1}{5} (- 1 + 2 tan (\frac{x}{2})))) + C

求解步骤

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

使用换元积分法

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

提出常数:

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

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

将

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

用部份分式展开:

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

= 4 \cdot \int - \frac{1}{5 ( u ^{2} - u - 1 )} + \frac{1}{5 ( u ^{2} + 1 )} + \frac{u - 2}{5 ( u ^{2} + 1 ) ^{2}} d u

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

= 4 (- \int \frac{1}{5 ( u ^{2} - u - 1 )} d u + \int \frac{1}{5 ( u ^{2} + 1 )} d u + \int \frac{u - 2}{5 ( u ^{2} + 1 ) ^{2}} d u)

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

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

\int \frac{u - 2}{5 ( u ^{2} + 1 ) ^{2}} d u = \frac{- 2 u ^{2} arctan ( u ) - 2 u - 2 arctan ( u ) - 1}{10 ( u ^{2} + 1 )}

= 4 (- (- \frac{1}{5 5} (ln \frac{2 u - 1}{5} + 1 - ln \frac{2 u - 1}{5} - 1)) + \frac{1}{5} arctan (u) + \frac{- 2 u ^{2} arctan ( u ) - 2 u - 2 arctan ( u ) - 1}{10 ( u ^{2} + 1 )})

u = tan (\frac{x}{2})

代回

= 4 (- (- \frac{1}{5 5} (ln \frac{2 tan ( \frac{x}{2} ) - 1}{5} + 1 - ln \frac{2 tan ( \frac{x}{2} ) - 1}{5} - 1)) + \frac{1}{5} arctan (tan (\frac{x}{2})) + \frac{- 2 tan ^{2} ( \frac{x}{2} ) arctan ( tan ( \frac{x}{2} ) ) - 2 tan ( \frac{x}{2} ) - 2 arctan ( tan ( \frac{x}{2} ) ) - 1}{10 ( tan ^{2} ( \frac{x}{2} ) + 1 )})

化简

4 (- (- \frac{1}{5 5} (ln \frac{2 tan ( \frac{x}{2} ) - 1}{5} + 1 - ln \frac{2 tan ( \frac{x}{2} ) - 1}{5} - 1)) + \frac{1}{5} arctan (tan (\frac{x}{2})) + \frac{- 2 tan ^{2} ( \frac{x}{2} ) arctan ( tan ( \frac{x}{2} ) ) - 2 tan ( \frac{x}{2} ) - 2 arctan ( tan ( \frac{x}{2} ) ) - 1}{10 ( tan ^{2} ( \frac{x}{2} ) + 1 )}) : 4 (\frac{- 1 - 2 arctan ( tan ( \frac{x}{2} ) ) - 2 tan ( \frac{x}{2} ) - 2 tan ^{2} ( \frac{x}{2} ) arctan ( tan ( \frac{x}{2} ) )}{10 sec ^{2} ( \frac{x}{2} )} + \frac{1}{5} arctan (tan (\frac{x}{2})) + \frac{1}{5 5} (- ln - 1 + \frac{1}{5} (- 1 + 2 tan (\frac{x}{2})) + ln 1 + \frac{1}{5} (- 1 + 2 tan (\frac{x}{2}))))

= 4 (\frac{- 1 - 2 arctan ( tan ( \frac{x}{2} ) ) - 2 tan ( \frac{x}{2} ) - 2 tan ^{2} ( \frac{x}{2} ) arctan ( tan ( \frac{x}{2} ) )}{10 sec ^{2} ( \frac{x}{2} )} + \frac{1}{5} arctan (tan (\frac{x}{2})) + \frac{1}{5 5} (- ln - 1 + \frac{1}{5} (- 1 + 2 tan (\frac{x}{2})) + ln 1 + \frac{1}{5} (- 1 + 2 tan (\frac{x}{2}))))

解答补常数

= 4 (\frac{- 1 - 2 arctan ( tan ( \frac{x}{2} ) ) - 2 tan ( \frac{x}{2} ) - 2 tan ^{2} ( \frac{x}{2} ) arctan ( tan ( \frac{x}{2} ) )}{10 sec ^{2} ( \frac{x}{2} )} + \frac{1}{5} arctan (tan (\frac{x}{2})) + \frac{1}{5 5} (- ln - 1 + \frac{1}{5} (- 1 + 2 tan (\frac{x}{2})) + ln 1 + \frac{1}{5} (- 1 + 2 tan (\frac{x}{2})))) + C

作图

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