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人気のある微分積分 >

integral from 0 to pi/4 of (3cos^2(θ))/2

解

\int_{0}^{\frac{π}{4}} \frac{3 cos ^{2} ( θ )}{2} d θ

解

\frac{3 ( π + 2 )}{16}

+1

十進法表記

0.96404 \dots

解答ステップ

\int_{0}^{\frac{π}{4}} \frac{3 cos ^{2} ( θ )}{2} d θ

定数を除く:

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

= \frac{3}{2} \cdot \int_{0}^{\frac{π}{4}} cos^{2} (θ) d θ

三角関数の公式を使用して書き換える

= \frac{3}{2} \cdot \int_{0}^{\frac{π}{4}} \frac{1 + cos ( 2 θ )}{2} d θ

定数を除く:

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

= \frac{3}{2} \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} 1 + cos (2 θ) d θ

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{3}{2} \cdot \frac{1}{2} (\int_{0}^{\frac{π}{4}} 1 d θ + \int_{0}^{\frac{π}{4}} cos (2 θ) d θ)

\int_{0}^{\frac{π}{4}} 1 d θ = \frac{π}{4}

\int_{0}^{\frac{π}{4}} cos (2 θ) d θ = \frac{1}{2}

= \frac{3}{2} \cdot \frac{1}{2} (\frac{π}{4} + \frac{1}{2})

簡素化

\frac{3}{2} \cdot \frac{1}{2} (\frac{π}{4} + \frac{1}{2}) : \frac{3 ( π + 2 )}{16}

= \frac{3 ( π + 2 )}{16}

グラフ

人気の例

integral from 2 to 6 of 2/x-4/(x^2)

\int_{2}^{6} \frac{2}{x} - \frac{4}{x ^{2}} d x

integral from 0 to 5 of (10)/(sqrt(x))

\int_{0}^{5} \frac{10}{x} d x

integral from 0 to 6 of (x^2+2)

\int_{0}^{6} (x^{2} + 2) d x

integral from 0 to pi/2 of (1-cos(2θ))/2

\int_{0}^{\frac{π}{2}} \frac{1 - cos ( 2 θ )}{2} d θ

integral from-1 to x of e^t

\int_{- 1}^{x} e^{t} d t