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