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人気のある微分積分 >

integral of (cos(x))/(sin^4(x)-16)

解

\int \frac{cos ( x )}{sin ^{4} ( x ) - 16} d x

解

- \frac{1}{16} arctan (\frac{sin ( x )}{2}) - \frac{1}{32} ln ∣ sin (x) + 2 ∣ + \frac{1}{32} ln ∣ sin (x) - 2 ∣ + C

解答ステップ

\int \frac{cos ( x )}{sin ^{4} ( x ) - 16} d x

u置換積分法を適用する

= \int \frac{1}{u ^{4} - 16} d u

以下の部分分数を得る:

\frac{1}{u ^{4} - 16} : - \frac{1}{8 ( u ^{2} + 4 )} - \frac{1}{32 ( u + 2 )} + \frac{1}{32 ( u - 2 )}

= \int - \frac{1}{8 ( u ^{2} + 4 )} - \frac{1}{32 ( u + 2 )} + \frac{1}{32 ( u - 2 )} 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}{8 ( u ^{2} + 4 )} d u - \int \frac{1}{32 ( u + 2 )} d u + \int \frac{1}{32 ( u - 2 )} d u

\int \frac{1}{8 ( u ^{2} + 4 )} d u = \frac{1}{16} arctan (\frac{u}{2})

\int \frac{1}{32 ( u + 2 )} d u = \frac{1}{32} ln ∣ u + 2 ∣

\int \frac{1}{32 ( u - 2 )} d u = \frac{1}{32} ln ∣ u - 2 ∣

= - \frac{1}{16} arctan (\frac{u}{2}) - \frac{1}{32} ln ∣ u + 2 ∣ + \frac{1}{32} ln ∣ u - 2 ∣

代用を戻す

u = sin (x)

= - \frac{1}{16} arctan (\frac{sin ( x )}{2}) - \frac{1}{32} ln ∣ sin (x) + 2 ∣ + \frac{1}{32} ln ∣ sin (x) - 2 ∣

定数を解答に追加する

= - \frac{1}{16} arctan (\frac{sin ( x )}{2}) - \frac{1}{32} ln ∣ sin (x) + 2 ∣ + \frac{1}{32} ln ∣ sin (x) - 2 ∣ + C

グラフ

人気の例

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

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

derivative y=sqrt(x)(x-14)

d er i v a t i v e y = x (x - 14)

integral of sqrt(x)(2+xsqrt(x))^5

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

derivative of-e^{-x}x^2+e^{-x}x

\frac{d}{d x} (- e^{- x} x^{2} + e^{- x} x)

(dy)/(dx)=sqrt(xy)

\frac{d y}{d x} = x y