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受欢迎的微积分 >

integral of 1/(64sec(x)tan^2(x))

解答

\int \frac{1}{64 sec ( x ) tan ^{2} ( x )} d x

解答

\frac{1}{64} (- csc (x) - sin (x)) + C

求解步骤

\int \frac{1}{64 sec ( x ) tan ^{2} ( x )} d x

提出常数:

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

= \frac{1}{64} \cdot \int \frac{1}{sec ( x ) tan ^{2} ( x )} d x

使用三角恒等式改写

= \frac{1}{64} \cdot \int \frac{( 1 - sin ^{2} ( x ) ) cos ( x )}{sin ^{2} ( x )} d x

使用换元积分法

= \frac{1}{64} \cdot \int \frac{1 - u ^{2}}{u ^{2}} d u

乘开

\frac{1 - u ^{2}}{u ^{2}} : \frac{1}{u ^{2}} - 1

= \frac{1}{64} \cdot \int \frac{1}{u ^{2}} - 1 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

= \frac{1}{64} (\int \frac{1}{u ^{2}} d u - \int 1 d u)

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

\int 1 d u = u

= \frac{1}{64} (- \frac{1}{u} - u)

u = sin (x)

代回

= \frac{1}{64} (- \frac{1}{sin ( x )} - sin (x))

使用基本三角恒等式:

\frac{1}{sin ( x )} = csc (x)

= \frac{1}{64} (- csc (x) - sin (x))

解答补常数

= \frac{1}{64} (- csc (x) - sin (x)) + C

作图

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