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人気のある微分積分 >

integral of (cos(x))/(sin(x)+2cos(x))

解

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

解

\frac{1}{5} ln ∣ tan (x) + 2 ∣ + \frac{1}{5} (- \frac{1}{2} ln (sec^{2} (x)) + 2 arctan (tan (x))) + C

解答ステップ

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

三角関数による置換を適用する

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

以下の部分分数を得る:

\frac{1}{( u + 2 ) ( 1 + u ^{2} )} : \frac{1}{5 ( u + 2 )} + \frac{- u + 2}{5 ( u ^{2} + 1 )}

= \int \frac{1}{5 ( u + 2 )} + \frac{- u + 2}{5 ( u ^{2} + 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}{5 ( u + 2 )} d u + \int \frac{- u + 2}{5 ( u ^{2} + 1 )} d u

\int \frac{1}{5 ( u + 2 )} d u = \frac{1}{5} ln ∣ u + 2 ∣

\int \frac{- u + 2}{5 ( u ^{2} + 1 )} d u = \frac{1}{5} (- \frac{1}{2} ln u^{2} + 1 + 2 arctan (u))

= \frac{1}{5} ln ∣ u + 2 ∣ + \frac{1}{5} (- \frac{1}{2} ln u^{2} + 1 + 2 arctan (u))

代用を戻す

u = tan (x)

= \frac{1}{5} ln ∣ tan (x) + 2 ∣ + \frac{1}{5} (- \frac{1}{2} ln tan^{2} (x) + 1 + 2 arctan (tan (x)))

\frac{1}{2} ln tan^{2} (x) + 1 = \frac{1}{2} ln (sec^{2} (x))

= \frac{1}{5} ln ∣ tan (x) + 2 ∣ + \frac{1}{5} (- \frac{1}{2} ln (sec^{2} (x)) + 2 arctan (tan (x)))

定数を解答に追加する

= \frac{1}{5} ln ∣ tan (x) + 2 ∣ + \frac{1}{5} (- \frac{1}{2} ln (sec^{2} (x)) + 2 arctan (tan (x))) + C

グラフ

人気の例

integral of t^{5/2}(t^2-4t+23)

\int t^{\frac{5}{2}} (t^{2} - 4 t + 23) d t

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

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

integral of (4x^2-9)^2

\int (4 x^{2} - 9)^{2} d x

integral of ((7x+3))/(1+x^2)

\int \frac{( 7 x + 3 )}{1 + x ^{2}} d x

integral of xe^x-xe^{x^2}

\int x e^{x} - x e^{x^{2}} d x