soal integral tolong caranya difoto

semoga membantu dan bermanfaat
[tex] \int\limits^ {}4sin3xcosx \, dx [/tex]
keluarkan konstanta
[tex] 4\int\limits^ {}sin3xcosx \, dx [/tex]
gunakan identitas
[tex]cos(a)sin(b)= \frac{sin(a+b)+sin(a-b)}{2} [/tex]
[tex]=4 \int\limits^ {} \frac{sin(3x+x)+sin(3x-x)}{2} \, dx [/tex]
[tex]=4. \frac{1}{2} \int\limits^ {}sin(3x+x)+sin(3x-x) \, dx [/tex]
[tex]=2 \int\limits^ {}sin(3x+x)+sin(3x-x) \, dx [/tex]
[tex]=2 \int\limits^ {}sin(3x+x) \, dx + \int\limits^ {}sin(3x-x) \, dx [/tex]
[tex]=2.( -\frac{1}{4}cos4x- \frac{1}{2}cos2x) [/tex]
[tex]=- \frac{1}{2}cos4x-cos2x+C [/tex]

Integral Fungsi Trigonometri.

∫ 4 sin 3x cos x dx
= 2 ∫ [sin (3x + x) + sin (3x - x)] dx
= 2 ∫ (sin 4x + sin 2x) dx
= 2 (-1/4 cos 4x - 1/2 cos 2x) + C
= -1/2 cos 4x - cos 2x + C