Oct
24
2012

Integral Trigonometri Bagian 2

Melanjutkan dari Integral Trigonometri Bagian 1, ini adalah sepuluh contoh soal integral trigonometri dengan berbagai variasi.

Masih sama dengan bagian 1, soal integral kali ini juga berkisar mengenai trigonometri yang melibatkan penjumlahan sudut dan teknik substitusi. Selain itu ada tambahan soal mengenai fungsi invers trigonometri yang bentuknya seperti integral aljabar biasa tapi penyelesaiannya menggunakan integral trigonometri.

Selamat berlatih ! :)

  1. \int \tan^3 x \cdot \sec^6 x \: \mathrm{d}x = \dots
    Lihat Jawaban

        \begin{align*}         \int \tan^3 x \cdot \sec^6 x \: \mathrm{d}x &= \int \tan^3 x \cdot \sec^4 x \cdot \sec^2 x \: \mathrm{d}x \\         &= \int \tan^3 x \cdot (\tan^2 x + 1)^2 \cdot \sec^2 x \: \mathrm{d}x \\               \end{align*}

    Misalkan :

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

    Lakukan Substitusi :

        \begin{align*}         & \int \tan^3 x \cdot (\tan^2 x + 1)^2 \cdot \sec^2 x \: \mathrm{d}x \\          &= \int u^3(u^2+1)^2 \: \mathrm{d}u \\         &= \int u^3(u^4 + 2u^2 + 1) \: \mathrm{d}u \\         &= \int u^7 + u^5 + u^3 \: \mathrm{d}u \\         &= \frac{1}{8}u^8 + \frac{1}{6}u^6 + \frac{1}{4}u^4 + C \\         &= \frac{1}{8}\tan^8 x+ \frac{1}{6}\tan^6 x + \frac{1}{4}\tan^4 x + C \\       \end{align*}

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

        \begin{align*}         \int \frac{\sin^3 x - \cos^3 x}{\sin^2 x \cos^2 x} \: \mathrm{d}x &= \int \frac{\sin^3 x}{\sin^2 x  \cos^2 x} - \frac{\cos^3 x}{\sin^2 x \cos^2 x} \: \mathrm{d}x \\         &= \int \frac{\sin x}{\cos^2 x} - \frac{\cos x}{\sin^2 x} \: \mathrm{d}x \\         &= \int \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} - \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} \: \mathrm{d}x \\         &= \int \tan x \cdot \sec x - \cot x \csc x \: \mathrm{d}x \\         &= \sec x + \csc x + C       \end{align*}

  3. \int \frac{\sin 2x}{\sqrt{1+2\sin x}}  \: \mathrm{d}x = \dots
    Lihat Jawaban

        \begin{align*}         \int \frac{\sin 2x}{\sqrt{1+2\sin x}}  \: \mathrm{d}x &= \int \frac{2 \sin x \cos x}{\sqrt{1+2\sin x}} \: \mathrm{d}x \\       \end{align*}

    Misalkan :

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

    Misalkan sekali lagi :

        \begin{align*}         u &= 1 + 2\sin x \\         u-1 &= 2\sin x \\                2\sin x &= u-1       \end{align*}

    Lakukan substitusi :

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

  4. \int \cos x \: \cos 2x \: \cos 3x  \: \mathrm{d}x = \dots
    Lihat Jawaban

        \begin{align*}         & \int \cos x \: \cos 2x \: \cos 3x  \: \mathrm{d}x  \\         &= \int {\color{blue}\cos 3x \: \cos 2x} \: \cos x  \: \mathrm{d}x \\         &= \int {\color{blue} \frac{1}{2} (\cos 5x + \cos x)} \: \cos x \: \mathrm{d}x \\         &= \frac{1}{2} \int \cos 5x \cdot \cos x + \cos^2 x \: \mathrm{d}x \\         &= \frac{1}{2} \left[\int \cos 5x \cdot \cos x \: \mathrm{d}x + \int \cos^2 x \: \mathrm{d}x \right]\\         &= \frac{1}{2} \left[\int \frac{1}{2} (\cos 6x + \cos 4x) \: \mathrm{d}x + \int \frac{1+\cos 2x}{2} \: \mathrm{d}x \right] \\         &= \frac{1}{2} \left[ \frac{1}{2} \int \cos 6x + \cos 4x \: \mathrm{d}x + \frac{1}{2} \int 1 + \cos 2x \right] \\         &= \frac{1}{2} \left[ \frac{1}{2} \left(\frac{1}{6} \sin 6x + \frac{1}{4} \sin 4x \right) + \frac{1}{2} \left(x + \frac{1}{2} \sin 2x \right) \right] + C \\        &= \frac{1}{2} \left[ \frac{1}{12}\sin 6x + \frac{1}{8} \sin 4x + \frac{1}{2}x + \frac{1}{4} \sin 2x \right] + C \\        &= \frac{1}{24} \sin 6x + \frac{1}{16} \sin 4x + \frac{1}{4}x + \frac{1}{8} \sin 2x + C          \end{align*}

  5. \int \frac{\mathrm{d}x}{9+x^2} = \dots
    Lihat Jawaban

    Bentuk ini adalah integral fungsi invers trigonometri. Terlihat dari bentuk \int \frac{\mathrm{d}x}{a^2+x^2}

        \begin{align*}          \int \frac{\mathrm{d}x}{9+x^2} &= \int \frac{\mathrm{d}x}{3^2+x^2} \\       \end{align*}

    Misalkan :

        \begin{align*}          x &= 3 \tan \theta \\         \mathrm{d}x &= 3 \sec^2 \theta \mathrm{d}\theta \\                \end{align*}

    Misalkan juga :

        \begin{align*}          x &= 3 \tan \theta \\          \frac{x}{3} &= \tan \theta \\          \theta &= \arctan \frac{x}{3}                 \end{align*}

    Lakukan substitusi :

        \begin{align*}          \int \frac{\mathrm{d}x}{3^2+x^2} &= \int \frac{3 \sec^2 \theta \: \mathrm{d}\theta}{9+(3 \tan \theta)^2} \\          &= \int \frac{3 \sec^2 \theta \: \mathrm{d}\theta}{9+9 \tan^2 \theta} \\          &= \int \frac{3 \sec^2 \theta \: \mathrm{d}\theta}{9(1 +\tan^2 \theta)} \\          &= \frac{1}{3} \int \frac{\sec^2 \theta \: \mathrm{d}\theta}{1 +\tan^2 \theta} \\          &= \frac{1}{3} \int \frac{\sec^2 \theta \: \mathrm{d}\theta}{\sec^2 \theta} \\              &= \frac{1}{3} \int \mathrm{d}\theta \\            &= \frac{1}{3} \: \theta + C \\          &= \frac{1}{3} \arctan \frac{x}{3} + C       \end{align*}

  6. \int \frac{\mathrm{d}x}{x^2+2x+5} = \dots
    Lihat Jawaban

        \begin{align*}         \int \frac{\mathrm{d}x}{x^2+2x+5} &= \int \frac{\mathrm{d}x}{x^2+2x+1+4} \\         &= \int \frac{\mathrm{d}x}{(x+1)^2+2^2}       \end{align*}

    Lakukan permisalan

        \begin{align*}         x+1 &= 2 \tan \theta \\         \mathrm{d}x &= 2 \sec^2 \theta \: \mathrm{d}\theta       \end{align*}

    Lakukan permisalan untuk \theta

        \begin{align*}         x+1 &= 2 \tan \theta \\         \frac{x+1}{2} &= \tan \theta \\         \theta &= \arctan \frac{x+1}{2}       \end{align*}

    Lakukan substitusi

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

  7. \int \cot x \sqrt{\sin^2 x - \sin^4 x} \: \mathrm{d}x = \dots
    Lihat Jawaban

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

  8. \int \frac{\mathrm{d}x}{\csc 2x - \cot 2x} = \dots
    Lihat Jawaban

        \begin{align*}         \int \frac{\mathrm{d}x}{\csc 2x - \cot 2x} &= \int \frac{\mathrm{d}x}{\frac{1}{\sin 2x} - \frac{\cos 2x}{\sin 2x}} \\          &= \int \frac{\mathrm{d}x}{\frac{1 - \cos 2x}{\sin 2x}} \\          &= \int \frac{\sin 2x \: \mathrm{d}x}{1 - \cos 2x}       \end{align*}

    Lakukan permisalan

        \begin{align*}          u &= 1 - \cos 2x \\          \mathrm{d}u &= 2 \sin 2x \: \mathrm{d}x \\          \frac{1}{2} \mathrm{d}u &= \sin 2x \: \mathrm{d}x \\       \end{align*}

    Lakukan substitusi

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

  9. \int \frac{1}{2x^2+2x+5} \: \mathrm{d}x = \dots
    Lihat Jawaban

        \begin{align*}         \int \frac{1}{2x^2+2x+5} \: \mathrm{d}x &= \int \frac{1}{2(x^2+x+\frac{5}{2})} \: \mathrm{d}x \\         &= \frac{1}{2} \int \frac{1}{x^2+x+\frac{5}{2}} \: \mathrm{d}x\\         &= \frac{1}{2} \int \frac{1}{x^2+x+\frac{1}{4} + \frac{9}{4}} \: \mathrm{d}x\\         &= \frac{1}{2} \int \frac{1}{(x+\frac{1}{2})^2 + (\frac{3}{2})^2} \: \mathrm{d}x \\       \end{align*}

    Misalkan

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

    Misalkan sekali lagi untuk mencari \theta

        \begin{align*}         x+\frac{1}{2} &= \frac{3}{2} \tan \theta \\         \frac{2}{3} (x+\frac{1}{2}) &=  \tan \theta \\         \tan \theta &= \frac{2x + 1}{3} \\         \theta &= \arctan \frac{2x + 1}{3} \\         \end{align*}

    Substitusikan

        \begin{align*}          & \frac{1}{2} \int \frac{1}{(x+\frac{1}{2})^2 + (\frac{3}{2})^2} \: \mathrm{d}x \\         &= \frac{1}{2} \int \frac{1}{(\frac{3}{2} \tan \theta)^2 + (\frac{3}{2})^2} \: \frac{3}{2} \sec^2 \theta \: \mathrm{d}\theta \\         &= \frac{1}{2} \int \frac{1}{\frac{9}{4} \tan^2 \theta + \frac{9}{4}} \: \frac{3}{2} \sec^2 \theta \: \mathrm{d}\theta \\         &= \frac{1}{2} \int \frac{1}{\frac{9}{4} (\tan^2 \theta + 1)} \: \frac{3}{2} \sec^2 \theta \: \mathrm{d}\theta \\         &= \frac{1}{2} \int \frac{4}{9 \cancel{sec^2 \theta}} \: \frac{3}{2} \cancel{\sec^2 \theta} \: \mathrm{d}\theta \\         &= \frac{1}{2} \int \frac{2}{3} \: \mathrm{d}\theta \\         &= \frac{1}{3} \int \mathrm{d}\theta \\         &= \frac{1}{3} \theta + C \\         &= \frac{1}{3} \arctan \frac{2x + 1}{3} + C       \end{align*}

  10. \int \sqrt{\frac{a+x}{a-x}} \: \mathrm{d}x = \dots
    Lihat Jawaban

    Kalikan dengan \frac{\sqrt{a-x}}{\sqrt{a-x}} untuk mendapat bentuk fungsi invers trigonometri

        \begin{align*}         \int \sqrt{\frac{a+x}{a-x}} \: \mathrm{d}x &= \int \frac{\sqrt{a+x}}{\sqrt{a-x}} \cdot \frac{\sqrt{a-x}}{\sqrt{a-x}} \: \mathrm{d}x \\         &= \int \frac{\sqrt{(a+x) \cdot (a-x)}}{\sqrt{(a-x) \cdot (a-x)}} \: \mathrm{d}x \\         &= \int \frac{\sqrt{a^2-x^2}}{a-x} \: \mathrm{d}x        \end{align*}

    Lakukan permisalan

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

    Lakukan permisalan sekali lagi untuk mencari \theta

        \begin{align*}          x &= a \sin \theta \\         \frac{x}{a} &= \sin \theta \\         \theta &= \arcsin \frac{x}{a}       \end{align*}

    Lakukan substitusi

        \begin{align*}         & \int \frac{\sqrt{a^2-x^2}}{a-x} \: \mathrm{d}x \\         &= \int \frac{\sqrt{a^2-(a \sin \theta)^2}}{a-a \sin \theta} a \cos \theta \: \mathrm{d}\theta \\         &= \int \frac{\sqrt{a^2-a^2 \sin^2 \theta}}{a-a \sin \theta} a \cos \theta \: \mathrm{d}\theta \\         &= \int \frac{\sqrt{a^2(1- \sin^2 \theta)}}{a(1 - \sin \theta)} a \cos \theta \: \mathrm{d}\theta \\         &= \int \frac{\sqrt{a^2(1- \sin^2 \theta)}}{\cancel{a}(1 - \sin \theta)} \cancel{a} \cos \theta \: \mathrm{d}\theta \\         &= \int \frac{\sqrt{a^2 cos^2 \theta}}{1 - \sin \theta} \cos \theta \: \mathrm{d}\theta \\         &= \int \frac{a \cos \theta}{1 - \sin \theta} \cos \theta \: \mathrm{d}\theta \\         &= \int \frac{a \cos^2 \theta}{1 - \sin \theta} \: \mathrm{d}\theta \\         &= \int \frac{a (1 - \sin^2 \theta)}{1 - \sin \theta} \: \mathrm{d}\theta \\         &= \int \frac{a (1 + \sin \theta)\cancel{(1 - \sin \theta)}}{\cancel{1 - \sin \theta}} \: \mathrm{d}\theta \\         &= \int a (1 + \sin \theta) \: \mathrm{d}\theta \\         &= \int a + a \sin \theta \: \mathrm{d}\theta \\         &= a \theta - a \cos \theta  + C \\         &= a \arcsin \frac{x}{a} - a \cos \left(\arcsin \frac{x}{a} \right) + C \\         	&= a \arcsin \frac{x}{a} - a \sqrt{1 - \left(\frac{x}{a} \right)^2} + C  \\         &= a \arcsin \frac{x}{a} - a \sqrt{\frac{a^2-x^2}{a^2}} + C \\         &= a \arcsin \frac{x}{a} - a \cdot \frac{1}{a}\sqrt{a^2 -x^2} + C \\         &= a \arcsin \frac{x}{a} - \sqrt{a^2 -x^2} + C        \end{align*}

    Catatan:

    Dengan rumus segitiga siku-siku, ingat bahwa \cos(\arcsin x) = \sqrt{1-x^2}

2 Comments + Add Comment

  • Hello can I reference some of the material here in this entry if I reference you with a link back to your site?

    • Hello John, sure you can :)