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

Baca lebih lanjut

Problem (23) : Integral


\int [tan2(2x-1) + sec(2x-1)]2 dx

= \int [tan2(2x-1) tan2(2x-1) + 2 tan2(2x-1) sec(2x-1) + sec2(2x-1)] dx

= \int tan2(2x-1) tan2(2x-1) dx + 2 \int tan2(2x-1) sec(2x-1) dx + \int sec2(2x-1) dx

= \int (sec2(2x-1) – 1) tan2(2x-1)) dx + 2 \int (sec2(2x-1) – 1)sec(2x-1) dx + \int sec2(2x-1) dx

= \int sec2(2x-1) tan2(2x-1) dx – \int tan2(2x-1) dx + 2 \int sec3(2x-1) dx – 2 \int sec(2x-1) dx + \int sec2(2x-1) dx

perhatikan :

 .

\int sec2(2x-1) tan2(2x-1) dx

misal : u = tan(2x-1) \Rightarrow du = 2 sec2(2x-1) dx

= \frac{1}{2} \int u2 du

= \frac{1}{6} u3

\int sec2(2x-1) tan2(2x-1) dx = \frac{1}{6} tan3(2x-1)

 .

\int tan2(2x-1) dx = \frac{1}{2} tan (2x-1) – \frac{1}{2} (2x-1) [bukti]

 .

\int sec2(2x-1) dx = \frac{1}{2} tan(2x-1) [bukti]

 .

\int sec3(2x-1) dx = \int sec2(2x-1) sec x dx

misal : u = sec(2x-1) \Rightarrow du = 2 sec(2x-1) tan(2x-1) dx [bukti]

dv = sec2(2x-1) dx \Rightarrow v = \frac{1}{2} tan(2x-1)

= \frac{1}{2} sec(2x-1) tan(2x-1) – \int tan2(2x-1) sec(2x-1) dx

= \frac{1}{2} sec(2x-1) tan(2x-1) – \int (sec2(2x-1) – 1)sec(2x-1) dx

= \frac{1}{2} sec(2x-1) tan(2x-1) – \int sec3(2x-1) dx – \int sec(2x-1) dx

2 \int sec3(2x-1) dx = \frac{1}{2} sec(2x-1) tan(2x-1) – \frac{1}{2} ln|sec(2x-1) –tan(2x-1)|

 .

\int sec(2x-1) = \frac{1}{2} ln|sec(2x-1) – tan(2x-1)| [bukti]

 .

\int [tan2(2x-1) + sec(2x-1)]2 dx = \int sec2(2x-1) tan2(2x-1) dx – \int tan2(2x-1) dx + 2 \int sec3(2x-1) dx – 2 \int sec(2x-1) dx + \int sec2(2x-1) dx

= [\frac{1}{6} tan3(2x-1)] – [\frac{1}{2} tan (2x-1) – \frac{1}{2} (2x-1)] + [\frac{1}{2} sec(2x-1) tan(2x-1) – \frac{1}{2} ln|sec(2x-1) – tan(2x-1)|] – 2[\frac{1}{2} ln|sec(2x-1) –tan(2x-1)|] + [\frac{1}{2} tan(2x-1)]

= \frac{1}{6} tan3(2x-1) – \frac{1}{2} tan (2x-1) – \frac{1}{2} (2x-1) + \frac{1}{2} sec(2x-1) tan(2x-1) – \frac{1}{2} ln|sec(2x-1) – tan(2x-1)| – ln|sec(2x-1) –tan(2x-1)| + \frac{1}{2} tan(2x-1)

= \frac{1}{6} tan3(2x-1) + \frac{1}{2} sec(2x-1) tan(2x-1) – \frac{3}{2} ln|sec(2x-1) – tan(2x-1)| – \frac{1}{2} (2x-1)

Problem (22)


soal dikirim via email by khaaliifaah@yahoo.com

  1. Himpunan penyelesaian dari -8 < 2(x – 1) < 6 adalah…

    A. {x │-3 < x < 4 , x € R}

    B. { x │3 < x < -4 , x € R}

    C. { x │-3 < x < -4 , x € R}

    D. { x │4 < x < 3 , x € R}

    E. { x │4 < x < -3 , x € R}

    Penyelesaian :

    -8 < 2(x – 1) < 6

    Pandang sebagai dua pertidaksamaan yaitu -8 < 2(x – 1) dan 2(x – 1) < 6

    -8 < 2(x – 1)

    -8 < 2x – 2

    -6 < 2x

    -3 < x … (i) Baca lebih lanjut

Problem(20) : Trigonometri


soal dikirim via email

Buktikan :

\frac{sin(A)+sin(2A)+sin(3A)}{cos(A)+cos(2A)+cos(3A)} = tan 2A

Penyelesaian :

pandang pembilang :

sin A + sin 2A + sin 3A = (sin A + sin 3A) + sin 2A

= 2 sin ½(4A) cos ½(2A) + sin 2A

= 2 sin 2A cos A + sin 2A

= sin 2A (2 cos A + 1)

pandang penyebut :

cos A + cos 2A + cos 3A = (cos A + cos 3A) + cos 2A

= 2 cos ½(4A) cos ½(2A) + cos 2A

= 2 cos 2A cos A + cos 2A

= cos 2A (2 cos A + 1)

sehingga diperoleh :

\frac{sin(2A)[cos(A)+1]}{cos(2A)[cos(A)+1]}

= \frac{sin(2A)}{cos(2A)}

= tan 2A