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人気のある微分積分 >

integral of (sin(x))/(sin^3(x)+cos^3(x))

解

\int \frac{sin ( x )}{sin ^{3} ( x ) + cos ^{3} ( x )} d x

解

- \frac{1}{3} ln ∣ tan (x) + 1 ∣ + \frac{1}{6} (ln (4 - tan (x) + sec^{2} (x)) + 23 arctan (\frac{1}{3} (2 tan (x) - 1))) + C

解答ステップ

\int \frac{sin ( x )}{sin ^{3} ( x ) + cos ^{3} ( x )} d x

三角関数による置換を適用する

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

以下の部分分数を得る:

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

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

\int \frac{1}{3 ( u + 1 )} d u = \frac{1}{3} ln ∣ u + 1 ∣

\int \frac{u + 1}{3 ( u ^{2} - u + 1 )} d u = \frac{1}{6} (ln 4 u^{2} - 4 u + 4 + 23 arctan (\frac{2 u - 1}{3}))

= - \frac{1}{3} ln ∣ u + 1 ∣ + \frac{1}{6} (ln 4 u^{2} - 4 u + 4 + 23 arctan (\frac{2 u - 1}{3}))

代用を戻す

u = tan (x)

= - \frac{1}{3} ln ∣ tan (x) + 1 ∣ + \frac{1}{6} (ln 4 tan^{2} (x) - 4 tan (x) + 4 + 23 arctan (\frac{2 tan ( x ) - 1}{3}))

簡素化

- \frac{1}{3} ln ∣ tan (x) + 1 ∣ + \frac{1}{6} (ln 4 tan^{2} (x) - 4 tan (x) + 4 + 23 arctan (\frac{2 tan ( x ) - 1}{3})) : - \frac{1}{3} ln ∣ tan (x) + 1 ∣ + \frac{1}{6} (ln (4 - tan (x) + sec^{2} (x)) + 23 arctan (\frac{1}{3} (2 tan (x) - 1)))

= - \frac{1}{3} ln ∣ tan (x) + 1 ∣ + \frac{1}{6} (ln (4 - tan (x) + sec^{2} (x)) + 23 arctan (\frac{1}{3} (2 tan (x) - 1)))

定数を解答に追加する

= - \frac{1}{3} ln ∣ tan (x) + 1 ∣ + \frac{1}{6} (ln (4 - tan (x) + sec^{2} (x)) + 23 arctan (\frac{1}{3} (2 tan (x) - 1))) + C

グラフ

人気の例

integral of 1/((x^2+2x+5)^2)

\int \frac{1}{( x ^{2} + 2 x + 5 ) ^{2}} d x

integral of 1/(3+2cos(x))

\int \frac{1}{3 + 2 cos ( x )} d x

integral of sin(2x)sqrt(cos(2x))

\int sin (2 x) cos (2 x) d x

integral of x\sqrt[4]{4x^2+5}

\int x 4 4 x^{2} + 5 d x

integral of x^3sqrt(x^2+13)

\int x^{3} x^{2} + 13 d x