integral from 0 to a of sqrt(a^2+x^2)

解答

\int_{0}^{a} a^{2} + x^{2} d x

\frac{( 2 + ln ( 1 + 2 ) ) a ^{2}}{2}

求解步骤

\int_{0}^{a} a^{2} + x^{2} d x

使用三角换元法

= \int_{0}^{\frac{π}{4}} a^{2} sec^{3} (u) d u

提出常数:

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

= a^{2} \cdot \int_{0}^{\frac{π}{4}} sec^{3} (u) d u

使用降次积分法

= a^{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} sec (u) d u)

\int_{0}^{\frac{π}{4}} sec (u) d u = ln (1 + 2)

= a^{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

化简

a^{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)) : a^{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

= a^{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

带入上下限计算后得:

\frac{1}{2}

= a^{2} (\frac{1}{2} + \frac{1}{2} ln (1 + 2))

化简

= \frac{( 2 + ln ( 1 + 2 ) ) a ^{2}}{2}

n = 0 \sum \infty \frac{n}{3}

\int_{0}^{\frac{π}{2}} cos^{2} (t) d t

x \to 0 + lim (x^{2} + 4)

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