Problem (23) : Integral Trigonometri


  1. \int \dfrac{\sin x}{\sqrt{2-\cos^2 x}} dx

    Penyelesaian :

    \int \dfrac{\sin x}{\sqrt{2-\cos^2 x}} dx = \int \dfrac{\sin x}{\sqrt{2-(1-\sin^2 x)}} dx

    = \int \dfrac{\sin x}{\sqrt{1+\sin^2 x}} dx

    = \int \dfrac{\sin x}{\sqrt{\sin^2x+\cos^2x+\sin^2 x}} dx

    misal : u = sin2 x \Rightarrow du = 2 sin x cos x dx

    sin x dx = \dfrac{2 \cdot du}{\cos x}

    = \int \dfrac{2 \cdot du}{\cos x \sqrt{1+u^2~x}}

    jika kita manfaatkan segitiga siku-siku dan teorema Pythagoras, maka diperoleh sisi miringnya, yaitu sebagai berikut

    = \int \dfrac{du}{\cos x \sqrt{1+u^2~x}}

  2. \int \dfrac{dx}{\sin^3 x}

    Penyelesaian :

    \int \dfrac{dx}{\sin^3 x} = \int csc3 x dx

    = \int csc x csc2 x dx

    kemudian gunakan Integral Parsial, misal

    u = csc x

    u = \dfrac{1}{\sin x}

    \dfrac{du}{dx} = \dfrac{0 \cdot (\sin x)-1 \cdot (\cos x)}{(\sin x)^2}

    = \dfrac{-\cos x}{\sin^2 x}

    du = -cot x csc x dx

    dv = csc2 x dx

    v = –cot x [bukti]

    = uv – \int v du

    = csc x (–cot x) – \int –cot x (-cot x csc x dx)

    = -csc x cot x – \int cot2 x csc x dx

    = -csc x cot x – \int (csc2 x – 1) csc x dx

    = -csc x cot x + \int csc x dx – \int csc3 x dx

    2 \int csc3 x dx = -csc x cot x + \int csc x dx

    INGAT : \int csc x dx = ln|csc x – cot x| + C [bukti]

    2 \int csc3 x dx = -csc x cot x + ln|csc x – cot x| + C

    \int csc3 x dx = -1/2 csc x cot x + 1/2 ln|csc x – cot x| + K

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