Problem (3) : Integral Trigonometri


soal dikirim via sms

Buktikan \int sin4 x dx = \frac{3}{8}x – \frac{1}{4} sin 2x + \frac{1}{32} sin 4x dx + c

PENYELESAIAN :

\int sin4 x dx = \int (\frac{1-cos \quad 2x}{2})^2

= \frac{1}{4} \int (1 – 2 cos 2x + cos2 2x) dx

= \frac{1}{4} (\int 1 dx – 2 \int cos 2x dx + \int \frac{1+cos \quad 4x}{2} dx)

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

misal 2x = u maka 2 dx = du (untuk cos 2x)

4x = v maka 4 dx = dv (untuk cos 4x)

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

= \frac{1}{4} (\int 1 dx – \int cos u du + \int \frac{1}{2} dx + \int \frac{1}{8} cos v dv)

= \frac{1}{4} (x – sin u + \frac{1}{2} x + \frac{1}{8} sin v) + C

substitusi u = 2x dan v = 4x, diperoleh

= \frac{1}{4} (\frac{3}{2} x – sin 2x +  \frac{1}{8} sin 4x) + C

= \frac{3}{8} x – \frac{1}{4} sin 2x + \frac{1}{32} sin 4x + C


NOTE :

cos 2x = cos2 x – sin2 x

cos 2x = 1 – 2sin2 x

sin2 x = \frac{1-cos \quad 2x}{2}

cos 4x = cos2 2x – sin2 2x

cos 4x = 2cos2 2x – 1

cos2 2x = \frac{1+cos \quad 4x}{2}

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