Problem (6) : Integral Trigonometri


soal dikirim via sms

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

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

    = \int \sqrt{2cos^2 \quad x} dx

    = \sqrt{2} \int cos x dx

    = \sqrt{2} sin x + C

    atau

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

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

    = \int \sqrt{2sin^2 \quad x} dx

    = \sqrt{2} \int sin x dx

    = –\sqrt{2} cos x + C

  2. \int sin 5x cos 3x dx = \int \frac{1}{2} (sin 8x + sin 2x) dx (Sifat-Sifat Dasar Trigonometri)

    = \frac{1}{2} \int sin 8x dx + \frac{1}{2} \int sin 2x dx

    misal 8x = u maka 8 dx = du (untuk sin 8x)

    2x = v maka 2 dx = dv (untuk sin 2x)

    = \frac{1}{2} \int sin u (du/8) + \frac{1}{2} \int sin v (dv/2)

    = \frac{1}{16} \int sin u du + \frac{1}{4} \int sin v dv

    = \frac{1}{16} \int (-cos u) + \frac{1}{4} \int (-cos) v + C

    substitusi u = 8x dan v = 2x, diperoleh

    = -\frac{1}{16} \int cos 8x – \frac{1}{4} \int cos 2x + C

  3. \int sin x cos x dx = \int sin x (cos x dx)

    misal sin x = u maka cos x dx = du, kemudian substitusi

    = \int u (du)

    = \frac{1}{2} u2 + C

    substitusi u = sin x, diperoleh

    = \frac{1}{2} sin2 x + C

    atau

    \int sin x cos x dx = \int cos x (sin x dx)

    misal cos x = u maka -sin x dx = du, kemudian substitusi

    = \int u (-du)

    = -\frac{1}{2} u2 + C

    substitusi u = cos x, diperoleh

    = -\frac{1}{2} cos2 x + C

  4. \int sin3 x cos x dx = \int sin3 x (cos x dx)

    misal sin x = u maka cos x dx = du

    = \int u3 du

    = \frac{1}{4} u4 + C

    substitusi u = sin x, diperoleh

    = \frac{1}{4} sin4 x + C

    atau

    \int sin3 x cos x dx = \int sin2 x sin x cos x dx

    = \int \frac{1}{2} (1 – cos2 x) cos x (sin x dx)

    misal cos x = u maka -sin x dx = du

    = \int \frac{1}{2} (1 – u2)u (-du)

    = \int -\frac{1}{2} (u – u3) du

    = \int -\frac{1}{2} u du + \int \frac{1}{2} u3 du

    = -\frac{1}{2}\frac{1}{2} u2 + \frac{1}{2}\frac{1}{4} u4 + C

    = -\frac{1}{4} u2 + \frac{1}{8} u4 + C

    substitusi u = cos x, diperoleh

    = -\frac{1}{4} cos2 x + \frac{1}{8} cos4 x + C

  5. \int 2 sin2 \frac{1}{2}x dx = …

    misal ½ x = u maka ½ dx = du

    = \int 2 sin2 u (2 du)

    = \int 4 sin2 u du

    = 4 -cos2 u + C

    substitusi u = ½ x, diperoleh

    = -4 cos2 \frac{1}{2}x + C

  6. \int 4 cos2 3x dx = …

    misal 3x = u maka 3 dx = du, kemudian substitusi

    = \int 4 cos2 u (du/3)

    = \frac{4}{3} \int cos2 u du

    = \frac{4}{3} sin2 u + C

    atau

    \int 4 cos2 3x dx = \int 4 \frac{1}{2} (1 + cos 6x) dx (Sifat-Sifat Dasar Trigonometri)

    = \int 2 dx + \int 2 cos 6x dx

    misal 6x = u maka 6 dx = du, kemudian substitusi

    = \int 2 dx + \int 2 cos u (du/6)

    = \int 2 dx + \frac{1}{3} \int cos u du

    = 2x + \frac{1}{3} sin u + C

    substitusi u = 6x, diperoleh

    = 2x + \frac{1}{3} sin 6x + C

  7. \int sin2 2x dx = …

    misal 2x = u maka 2 dx = du

    = \int sin2 u (du/2)

    = \frac{1}{2} \int sin2 u du

    = \frac{1}{2} -cos2 u + C

    substitusi u = 2x, diperoleh

    = -\frac{1}{2} cos2 2x + C

    atau

    \int sin2 2x dx = \int \frac{1}{2} (1 – cos 4x) dx (Sifat-Sifat Dasar Trigonometri)

    = \int \frac{1}{2} dx – \int \frac{1}{2} cos 4x dx

    misal 4x = u maka 4 dx = du, kemudian disubstitusi

    = \int \frac{1}{2} dx – \int \frac{1}{2} cos u (du/4)

    = \int \frac{1}{2} dx – \int \frac{1}{8} cos u du

    = \frac{1}{2} x – \int \frac{1}{8} sin u + C

    substitusi u = 4x, diperoleh

    = \frac{1}{2} x – \int \frac{1}{8} sin 4x + C

  8. \int (2 cos2 x – 1) dx = \int (2 cos2 x – [sin2 x + cos2 x]) dx

    = \int (cos2 x – sin2 x) dx

    = \int 2 cos 2x dx

    misal 2x = u maka 2 dx = du, substitusi

    = \int 2 cos u (du/2)

    = \int cos u du

    = sin u + C

    substitusi u = 2x, diperoleh

    = sin 2x + C

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