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人気のある微分積分 >

integral from 0 to pi/4 of (tan(x))^3

解

\int_{0}^{\frac{π}{4}} (tan (x))^{3} d x

解

\frac{1}{2} - \frac{1}{2} ln (2)

+1

十進法表記

0.15342 \dots

解答ステップ

\int_{0}^{\frac{π}{4}} (tan (x))^{3} d x

三角関数の公式を使用して書き換える

= \int_{0}^{\frac{π}{4}} \frac{( 1 - cos ^{2} ( x ) ) sin ( x )}{cos ^{3} ( x )} d x

u置換積分法を適用する

= \int_{1}^{\frac{2}{2}} - \frac{1 - u ^{2}}{u ^{3}} d u

拡張

- \frac{1 - u ^{2}}{u ^{3}} : - \frac{1}{u ^{3}} + \frac{1}{u}

= \int_{1}^{\frac{2}{2}} - \frac{1}{u ^{3}} + \frac{1}{u} d u

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

= - \int_{\frac{2}{2}}^{1} - \frac{1}{u ^{3}} + \frac{1}{u} d u

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

= - (- \int_{\frac{2}{2}}^{1} \frac{1}{u ^{3}} d u + \int_{\frac{2}{2}}^{1} \frac{1}{u} d u)

\int_{\frac{2}{2}}^{1} \frac{1}{u ^{3}} d u = \frac{1}{2}

\int_{\frac{2}{2}}^{1} \frac{1}{u} d u = \frac{1}{2} ln (2)

= - (- \frac{1}{2} + \frac{1}{2} ln (2))

簡素化

= \frac{1}{2} - \frac{1}{2} ln (2)

グラフ

人気の例

integral from 2 to y/2 of 6x^2y

\int_{2}^{\frac{y}{2}} 6 x^{2} y d x

integral from 1 to 3 of x+1

\int_{1}^{3} x + 1 d x

integral from 2 to 5 of (x^2+4)

\int_{2}^{5} (x^{2} + 4) d x

integral from 0 to 4 of (2x+1)

\int_{0}^{4} (2 x + 1) d x

integral from-pi to pi of pi^2-x^2

\int_{- π}^{π} π^{2} - x^{2} d x