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人気のある微分積分 >

integral of tan^3(x)sin^2(x)

解

\int tan^{3} (x) sin^{2} (x) d x

解

\frac{1}{2} sec^{2} (x) + 2 ln ∣ cos (x) ∣ - \frac{1}{2} cos^{2} (x) + C

解答ステップ

\int tan^{3} (x) sin^{2} (x) d x

三角関数の公式を使用して書き換える

= \int \frac{( 1 - cos ^{2} ( x ) ) ^{2} sin ( x )}{cos ^{3} ( x )} d x

u置換積分法を適用する

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

拡張

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

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

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

\int \frac{2}{u} d u = 2 ln ∣ u ∣

\int u d u = \frac{u ^{2}}{2}

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

代用を戻す

u = cos (x)

= - (- \frac{1}{2 cos ^{2} ( x )}) + 2 ln ∣ cos (x) ∣ - \frac{cos ^{2} ( x )}{2}

簡素化

- (- \frac{1}{2 cos ^{2} ( x )}) + 2 ln ∣ cos (x) ∣ - \frac{cos ^{2} ( x )}{2} : \frac{1}{2} sec^{2} (x) + 2 ln ∣ cos (x) ∣ - \frac{1}{2} cos^{2} (x)

= \frac{1}{2} sec^{2} (x) + 2 ln ∣ cos (x) ∣ - \frac{1}{2} cos^{2} (x)

定数を解答に追加する

= \frac{1}{2} sec^{2} (x) + 2 ln ∣ cos (x) ∣ - \frac{1}{2} cos^{2} (x) + C

グラフ

人気の例

integral of (t^3-(e^{-t}+4)/(e^{-t)})

\int (t^{3} - \frac{e ^{- t} + 4}{e ^{- t}}) d t

integral of (x^3+2x^2-1)

\int (x^{3} + 2 x^{2} - 1) d x

integral of (100-5/(x^2))

\int (100 - \frac{5}{x ^{2}}) d x

integral of ((tan(x))/(sec(x)))^5

\int (\frac{tan ( x )}{sec ( x )})^{5} d x

integral of-8xcos(7x)

\int - 8 x cos (7 x) d x