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人気のある微分積分 >

integral from 0 to 2pi of sqrt(θ^2+1)

解

\int_{0}^{2 π} θ^{2} + 1 d θ

解

π 4 π^{2} + 1 + \frac{1}{2} ln (2 π + 4 π^{2} + 1)

+1

十進法表記

21.25629 \dots

解答ステップ

\int_{0}^{2 π} θ^{2} + 1 d θ

三角関数による置換を適用する

= \int_{0}^{a r c t a n (2 π)} sec^{3} (u) d u

積分換算を適用する

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

\int_{0}^{a r c t a n (2 π)} sec (u) d u = ln (2 π + 4 π^{2} + 1)

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

簡素化

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

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

限度を計算する:

π 4 π^{2} + 1

= π 4 π^{2} + 1 + \frac{1}{2} ln (2 π + 4 π^{2} + 1)

グラフ

人気の例

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