Des 2 2012

Pembahasan Latihan Soal Integral (2) UN SMA


  1. \int x sin 2x dx = …

    A. \frac{1}{4} sin 2x – \frac{1}{2} x cos 2x + C

    B. \frac{1}{4} sin 2x + \frac{1}{2} x cos 2x + C

    C. \frac{1}{4} sin 2x – \frac{1}{2} x cos 2x + C

    D. –\frac{1}{4} cos 2x – \frac{1}{2} x sin 2x + C

    E. \frac{1}{4} cos 2x + \frac{1}{2} x sin 2x + C

    PEMBAHASAN :

    misal u = x \Rightarrow du = dx

    dv = sin 2x dx \Rightarrow v = –\frac{1}{2} cos 2x

    \int u dv = uv – \int v du

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

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

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

    JAWABAN : C

  2. \int_0^{\frac{\pi}{2}} (sin2 x – cos2 x) dx = …

    A. –\frac{1}{2}

    B. –\frac{1}{2} \pi

    C. 0

    D. \frac{1}{2}

    E. \frac{1}{2} \pi

    PEMBAHASAN :

    \int_0^{\frac{\pi}{2}} (sin2 x – cos2 x) dx = \int_0^{\frac{\pi}{2}} cos 2x dx

    [ingat Sifat Dasar Trigonometri]

    misal u = 2x \Rightarrow du = 2 dx

    = \int_0^{\frac{\pi}{2}} cos u \frac{du}{2}

    = \frac{1}{2} sin u \mid_0^{\frac{\pi}{2}}

    Substitusi kembali u = 2x

    = \frac{1}{2} sin 2x \mid_0^{\frac{\pi}{2}}

    = \frac{1}{2} [sin 2\frac{\pi}{2} – sin 2.0]

    = 0

    JAWABAN : C

  3. Hasil \int 2x cos \frac{1}{2} x dx = …

    A. 4x sin \frac{1}{2} x + 8 cos \frac{1}{2} x + C

    B. 4x sin \frac{1}{2} x – 8 cos \frac{1}{2} x + C

    C. 4x sin \frac{1}{2} x + 4 cos \frac{1}{2} x + C

    D. 4x sin \frac{1}{2} x – 8 cos \frac{1}{2} x + C

    E. 4x sin \frac{1}{2} x + 2 cos \frac{1}{2} x + C

    PEMBAHASAN :

    disini akan digunakan Integral Parsial

    misal u = 2x \Rightarrow du = 2 dx

    dv = cos \frac{1}{2} x \Rightarrow v = 2 sin \frac{1}{2} x

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

    = 4x sin \frac{1}{2} x – 4 (-2 cos \frac{1}{2} x) + C

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

    JAWABAN : A

  4. Hasil \int x\sqrt{9-x^2} dx = …

    A. –\frac{1}{3} (9 – x2) x\sqrt{9-x^2} + C

    B. –\frac{2}{3} (9 – x2) x\sqrt{9-x^2} + C

    C. \frac{2}{3} (9 – x2) x\sqrt{9-x^2} + C

    D. \frac{2}{3} (9 – x2) x\sqrt{9-x^2} + \frac{2}{9} (9 – x2) x\sqrt{9-x^2} + C

    E. \frac{1}{3} (9 – x2) x\sqrt{9-x^2} + \frac{1}{9} (9 – x2) x\sqrt{9-x^2} + C

    PEMBAHASAN :

    misal u = 9 – x2 \Rightarrow du = -2x dx

    = \int \sqrt{u} \frac{du}{-2}

    = \int -\frac{1}{2} u1/2 du

    = -\frac{1}{2} .\frac{2}{3} u3/2 + C

    substitusi u = 9 – x2, sehingga diperoleh

    = -\frac{1}{3} (9 – x2)3/2 + C

    = -\frac{1}{3} (9 – x2) \sqrt{9-x^2} + C

    JAWABAN :

  5. Nilai \int_0^1 5x(1 – x)6 dx = …

    A. 75/56

    B. 10/56

    C. 5/56

    D. -7/56

    E. -10/56

    PEMBAHASAN :

    misal u = 5x \Rightarrow du = 5 dx

    dv = (1 – x)6 dx \Rightarrow v = -\frac{1}{7}(1 – x)7

    \int_0^1 5x(1 – x)6 dx = 5x -\frac{1}{7}(1 – x) – \int_0^1 -\frac{1}{7}(1 – x)7 5 dx

    = -\frac{5}{7}x(1 – x) + \int_0^1 \frac{1}{7}.\frac{1}{8} (1 – x)8 5 dx

    = (-\frac{5}{7}x(1 – x)7 + \frac{5}{56}(1 – x)8) \mid_0^1

    = (-\frac{5}{7}.1.(1 – 1)7 + \frac{5}{56}(1 – 1)8) – (-\frac{5}{7}.0.(1 – 0)7 + \frac{5}{56}(1 – 0)8)

    = (0 + 0) – (0 + \frac{5}{56})

    = \frac{5}{56}

    JAWABAN : C

  6. Hasil dari \int cos x cos 4x dx = …

    A. –\frac{1}{5} sin 5x – \frac{1}{3} x sin 3x + C

    B. \frac{1}{10} sin 5x + \frac{1}{6} x sin 3x + C

    C. \frac{2}{5} sin 5x + \frac{2}{3} x sin 3x + C

    D. \frac{1}{2} cos 5x + \frac{1}{2} x cos 3x + C

    E. –\frac{1}{2} sin 5x – \frac{1}{2} x sin 3x + C

    PEMBAHASAN :

    \int cos x cos 4x dx = \int \frac{1}{2}(cos 5x + cos 3x) dx

    = \int \frac{1}{2} cos 5x dx + \int \frac{1}{2} cos 3x dx

    misal u = 5x \Rightarrow du = 5 dx

    v = 3x \Rightarrow dv = 3 dx

    substitusi, sehingga

    = \int \frac{1}{2} cos u \frac{du}{5} + \int \frac{1}{2} cos v \frac{dv}{3}

    = \frac{1}{10} sin u + \frac{1}{6} sin v + C

    substitusi kembali u = 5x dan v = 3x

    = \frac{1}{10} sin 5x + \frac{1}{6} sin 3x + C

    JAWABAN : B

  7. Hasil dari \int cos4 2x sin 2x dx = …

    A. -\frac{1}{10} sin5 2x + C

    B. -\frac{1}{10} cos5 2x + C

    C. -\frac{1}{5} cos5 2x + C

    D. \frac{1}{5} cos5 2x + C

    E. \frac{1}{10} sin5 2x + C

    PEMBAHASAN :

    misal u = cos 2x \Rightarrow du = -2 sin 2x dx

    \int cos4 2x sin 2x dx = \int u4 -\frac{du}{2}

    = -\frac{1}{2} \frac{1}{5} u5 + C

    substitusi kembali u = cos 2x

    = -\frac{1}{10} cos5 2x + C

    JAWABAN : B

  8. Hasil dari \int 4 sin 5x cos 3x dx = …

    A. -2 cos 8x – 2 cos 2x + C

    B. -\frac{1}{4} cos 8x – 2 cos 2x + C

    C. \frac{1}{4} cos 8x + 2 cos 2x + C

    D. -\frac{1}{2} cos 8x – 2 cos 2x + C

    E. \frac{1}{2} cos 8x – 2 cos 2x + C

    PEMBAHASAN :

    \int 4 sin 5x cos 3x dx = \int 2(sin 8x + sin 2x) dx

    = \int 2(sin 8x + sin 2x) dx

    = \int 2 sin 8x dx + \int 2 sin 2x dx

    misal u = 8x \Rightarrow du = 8

    v = 2x \Rightarrow du = 2

    substitusi, sehingga

    = \int 2 sin u \frac{du}{8} + \int 2 sin v \frac{dv}{2}

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

    = -\frac{1}{4} cos u – cos v + C

    substitusi kembali u = 8x dan v = 2x

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

    JAWABAN : B

  9. Hasil dari \int x2 sin 2x dx = …

    A. -\frac{1}{2} x2 cos 2x – \frac{1}{2} x sin 2x + \frac{1}{4} cos 2x + C

    B. -\frac{1}{2} x2 cos 2x + \frac{1}{2} x sin 2x – \frac{1}{4} cos 2x + C

    C. -\frac{1}{2} x2 cos 2x + \frac{1}{2} x sin 2x + \frac{1}{4} cos 2x + C

    D. \frac{1}{2} x2 cos 2x – \frac{1}{2} x sin 2x – \frac{1}{4} cos 2x + C

    E. \frac{1}{2} x2 cos 2x – \frac{1}{2} x sin 2x + \frac{1}{4} cos 2x + C

    PEMBAHASAN :

    disini kita akan menggunakan Integral Parsial

    misal u = x2 \Rightarrow du = 2x dx

    dv = sin 2x dx \Rightarrow v = -\frac{1}{2} cos 2x

    \int x2 sin 2x dx = (x2) -\frac{1}{2} cos 2x – \int -\frac{1}{2} cos 2x 2x dx

    = -\frac{1}{2} x2 cos 2x + \int x cos 2x dx

    Integral Parsial lagi

    misal u = x \Rightarrow du = dx

    dv = cos 2x dx \Rightarrow v = \frac{1}{2} sin 2x

    = -\frac{1}{2} x2 cos 2x + [x \frac{1}{2} sin 2x – \int \frac{1}{2} sin 2x dx]

    = -\frac{1}{2} x2 cos 2x + \frac{1}{2} x sin 2x + \frac{1}{4} cos 2x

    JAWABAN : C

  10. Hasil dari \int sin2 x cos x dx = …

    A. \frac{1}{3} cos3 x + C

    B. -\frac{1}{3} cos3 x + C

    C. -\frac{1}{3} sin3 x + C

    D. \frac{1}{3} sin3 x + C

    E. 3 sin3 x + C

    PEMBAHASAN :

    misal u = sin x \Rightarrow du = cos x du,

    kemudian substitusi, sehingga

    \int sin2 x cos x dx = \int u2 du

    = \frac{1}{3} u3 + C

    substitusi kembali u = sin x,

    = \frac{1}{3} sin3 x + C

    JAWABAN : D

NOTE : silahkan dikoreksi dan berikan komentar jika ada kesalahan atau masih ada keambiguan dalam penyelesaian soal-soal ini.

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2 comments on “Pembahasan Latihan Soal Integral (2) UN SMA”

  1. Bisa jelaskan ini? Pada baris ketiga dan terakhir, mengapa bisa begitu?
    Int sin^5 2x cos 2x dx
    = 1/2 int (sin^5 2x)(2 cos 2x) dx
    = 1/2 int sin^5 2x d(sin 2x)
    = 1/12 sin^6 2x + C

    Balas

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