Sep
16
2012

Integral Trigonometri Bagian 1

Melanjutkan posting-posting sebelumnya mengenai integral, berikut ini adalah lanjutan soal-soal integral yang khusus mengenai integral trigonometri. Soal integral trigonometri kali ini akan dibagi menjadi 2 posting, masing-masing sepuluh nomor karena banyaknya variasi soal.

Untuk membantu menyelesaikan soal, mungkin dibutuhkan rumus-rumus dasar integral untuk trigonometri. Gunakan rumus tersebut beserta dengan rumus penjumlahan sudut maupun teknik integral substitusi untuk mengerjakan soal-soal berikut ini. Selamat berlatih :)

  1. \int (\sin x + 2 \cos x) \: \mathrm{d}x = \dots
    Lihat Jawaban

        \begin{align*}          & \int (\sin x + 2 \cos x) \: \mathrm{d}x \\          &= -\cos x + 2 \sin x + C       \end{align*}

  2. \int \sec^2 (\frac{\pi}{2} - 4) \: \mathrm{d}x = \dots
    Lihat Jawaban

        \begin{align*}          & \int \sec^2 (\frac{\pi}{2} - 4x) \: \mathrm{d}x \\          &= - \frac{1}{4} \tan (\frac{\pi}{2} - 4x) + C       \end{align*}

  3. \int \frac{\cos^2 x - \sin^2 x}{\cos x - \sin x} \: \mathrm{d}x = \dots
    Lihat Jawaban

        \begin{align*}          & \int \frac{\cos^2 x - \sin^2 x}{\cos x - \sin x} \: \mathrm{d}x \\           &= \int \frac{\cancel{(\cos x - \sin x)}(\cos x + \sin x)}{\cancel{\cos x - \sin x}} \: \mathrm{d}x \\          &= \int \cos x + \sin x \: \mathrm{d}x \\          &= \sin x - \cos x + C       \end{align*}

  4. \int (\tan x - \sec x)^2 \: \mathrm{d}x = \dots
    Lihat Jawaban

        \begin{align*}         \int (\tan x - \sec x)^2 \: \mathrm{d}x &= \int \tan^2 x - 2 \tan x \sec x + \sec^2 x \: \mathrm{d}x \\         &= \int (\sec^2 x - 1 - 2 \tan x \sec x + \sec^2 x) \: \mathrm{d}x \\         &= \int (2 \sec^2 x - 1 - 2 \tan x \sec x) \: \mathrm{d}x \\         &= 2 \tan x - x - 2 \sec x + C \\         &= 2 \tan x - 2 \sec x - x + C       \end{align*}

  5. \int -\cos^2x \sin x \: \mathrm{d}x = \dots
    Lihat Jawaban

    Gunakan cara substitusi, jadi lakukan permisalan dahulu :

        \begin{align*}          u &= \cos x \\         \mathrm{d}u &= - \sin x \: \mathrm{d}x \\         - \sin x \: \mathrm{d}x &= \mathrm{d}u       \end{align*}

    Lakukan substitusi

        \begin{align*}          \int -\cos^2x \sin x \: \mathrm{d}x &= \int \cos^2x (- \sin x ) \: \mathrm{d}x \\          &= \int u^2 \: \mathrm{d}u \\          &= \frac{1}{3} u^3 + C \\          &= \frac{1}{3} \cos^3 x + C       \end{align*}

  6. \int x \sin (x^2 + 1) \: \mathrm{d}x = \dots
    Lihat Jawaban

    Misalkan:

        \begin{align*}          u &= x^2 + 1 \\          \mathrm{d}u &= 2x \: \mathrm{d}x \\           x \: \mathrm{d}x &= \frac{\mathrm{d}u}{2}        \end{align*}

    Lakukan substitusi

        \begin{align*}          \int x \sin (x^2 + 1) \: \mathrm{d}x &= \int \sin (x^2 + 1) x \: \mathrm{d}x \\           &= \int \sin u \: \frac{\mathrm{d}u}{2} \\           &= \frac{1}{2} \int \sin u \: \mathrm{d}u \\           &= - \frac{1}{2} \cos u + C \\           &= - \frac{1}{2} \cos (x^2+1) + C       \end{align*}

  7. \int \tan^2 x \sec^2 x \: \mathrm{d}x = \dots
    Lihat Jawaban

    Misalkan

        \begin{align*}        u &= \tan x \\       \mathrm{d}u &= \sec^2 x \: \mathrm{d}x      \end{align*}

    Substitusi hasil permisalan diatas

        \begin{align*}        \int \tan^2 x \sec^2 x \: \mathrm{d}x &= \int u^2 \: \mathrm{d}u \\        &= \frac{1}{3} u^3 + C \\        &= \frac{1}{3} \tan^3 x + C     \end{align*}

  8. \int \sin 7x \sin x \: \mathrm{d}x = \dots
    Lihat Jawaban

    Ingat rumus penjumlahan sudut

        \begin{align*}          \boxed{\sin ax \cdot \sin bx = -\frac{1}{2} \left[ \cos(a+b)x - \cos(a-b)x \right]}       \end{align*}

        \begin{align*}          \int \sin 7x \sin x \: \mathrm{d}x &= \int -\frac{1}{2} \left[ \cos(7+1)x - \cos(7-1)x \right] \: \mathrm{d}x \\          &= -\frac{1}{2} \int \cos 8x - \cos 6x \: \mathrm{d}x \\          &= -\frac{1}{2} \left( \frac{1}{8} \sin 8x - \frac{1}{6} \sin 6x \right) + C \\          &= - \frac{1}{16} \sin 8x + \frac{1}{12} \sin 6x + C       \end{align*}

  9. \int \sin^4 x \: \mathrm{d}x = \dots
    Lihat Jawaban

    Ingat rumus trigonometri: \boxed{\sin^2 x = \frac{1-\cos 2x}{2}} dan \boxed{\cos^2 x = \frac{1+\cos 2x}{2}}

        \begin{align*}           \int \sin^4 x \: \mathrm{d}x &= \int (\sin^2 x)^2 \: \mathrm{d}x \\           &= \int \left(\frac{1-\cos 2x}{2}\right)^2 \: \mathrm{d}x \\           &= \int \frac{1-2\cos 2x + \cos^2 2x}{4} \: \mathrm{d}x \\           &= \int \frac{1}{4} \: \mathrm{d}x - \int \frac{2\cos 2x}{4} \: \mathrm{d}x + \int \frac{\cos^2 2x}{4} \: \mathrm{d}x \\           &= \frac{1}{4}x - \frac{2}{4} \cdot \frac{1}{2} \sin 2x + \frac{1}{4} \int \cos^2 2x \: \mathrm{d}x \\           &= \frac{1}{4}x - \frac{1}{4} \sin 2x + \frac{1}{4} \int \frac{1-\cos 4x}{2} \: \mathrm{d}x \\           &= \frac{1}{4}x - \frac{1}{4} \sin 2x + \frac{1}{4} \left(\int \frac{1}{2} \: \mathrm{d}x - \int \frac{1}{2} \cos 4x \: \mathrm{d}x \right) \\           &= \frac{1}{4}x - \frac{1}{4} \sin 2x + \frac{1}{4} \left(\frac{1}{2}x - \frac{1}{2} \cdot \frac{1}{4} (-\sin 4x) \right) + C \\             &= \frac{1}{4}x - \frac{1}{4} \sin 2x + \frac{1}{8}x + \frac{1}{32} \sin 4x + C \\             &= \frac{3}{8}x - \frac{1}{4} \sin 2x + \frac{1}{32} \sin 4x + C \\          \end{align*}

  10. \int \cos^5 x \: \mathrm{d}x = \dots
    Lihat Jawaban

    Ubah dulu ke bentuk yang memungkinkan untuk permisalan dan substitusi

        \begin{align*}       \int \cos^5 x \: \mathrm{d}x &= \int \cos x(\cos^4x) \: \mathrm{d}x \\       &= \int \cos x(\cos^2 x)^2 \: \mathrm{d}x \\       &= \int \cos x(1 - \sin^2 x)^2 \: \mathrm{d}x \\           \end{align*}

    Misalkan

        \begin{align*}       u &= \sin x \\      \mathrm{d}u &= \cos x \: \mathrm{d}x     \end{align*}

    Substitusi dan selesaikan integral nya

        \begin{align*}       & \int \cos x(1 - \sin^2 x)^2 \: \mathrm{d}x \\        &= \int (1-\sin^2x)^2 \cos x \: \mathrm{d}x \\       &= \int (1-u^2)^2 \: \mathrm{d}u \\       &= \int 1 - 2u^2 + u^4 \: \mathrm{d}u \\       &= u - \frac{2}{3}u^3 + \frac{1}{5}u^5 + C \\       &= \sin x - \frac{2}{3}\sin^3 x + \frac{1}{5}\sin^5 x + C \\       &= \frac{1}{5}\sin^5 x - \frac{2}{3}\sin^3 x + \sin x +  + C \\     \end{align*}

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