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受欢迎的微积分 >

integral from 0 to pi/4 of xtan(x^2)

解答

\int_{0}^{\frac{π}{4}} x tan (x^{2}) d x

解答

- \frac{1}{2} ln (cos (\frac{π ^{2}}{16}))

+1

十进制

0.10185 \dots

求解步骤

\int_{0}^{\frac{π}{4}} x tan (x^{2}) d x

使用换元积分法

= \int_{0}^{\frac{π ^{2}}{16}} \frac{tan ( u )}{2} d u

提出常数:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{π ^{2}}{16}} tan (u) d u

使用三角恒等式改写

= \frac{1}{2} \cdot \int_{0}^{\frac{π ^{2}}{16}} \frac{sin ( u )}{cos ( u )} d u

使用换元积分法

= \frac{1}{2} \cdot \int_{1}^{c o s (\frac{π ^{2}}{16})} - \frac{1}{v} d v

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= \frac{1}{2} (- \int_{c o s (\frac{π ^{2}}{16})}^{1} - \frac{1}{v} d v)

提出常数:

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

= \frac{1}{2} (- (- \int_{c o s (\frac{π ^{2}}{16})}^{1} \frac{1}{v} d v))

使用积分常见定则:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} (- (- [ln ∣ v ∣]_{c o s (\frac{π ^{2}}{16})}^{1}))

化简

= \frac{1}{2} [ln ∣ v ∣]_{c o s (\frac{π ^{2}}{16})}^{1}

带入上下限计算后得:

- ln (cos (\frac{π ^{2}}{16}))

= \frac{1}{2} (- ln (cos (\frac{π ^{2}}{16})))

化简

= - \frac{1}{2} ln (cos (\frac{π ^{2}}{16}))

作图

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