Skorr

∫ 2cos x sin(1-2x) dx=…

Kalkulus Matematika · PTN hard

∫2cos⁡x sin⁡(1−2x) dx=…\displaystyle\int 2\cos x\,\sin(1-2x)\,dx=\ldots

  1. A
    cos⁡(x−1)+13cos⁡(3x−1)+C\cos(x-1)+\frac13\cos(3x-1)+C
    ✓
  2. B
    cos⁡(x−1)−13cos⁡(3x−1)+C\cos(x-1)-\frac13\cos(3x-1)+C
  3. C
    −sin⁡(x−1)+13sin⁡(3x−1)+C-\sin(x-1)+\frac13\sin(3x-1)+C
  4. D
    −sin⁡(x−1)−13sin⁡(3x−1)+C-\sin(x-1)-\frac13\sin(3x-1)+C
  5. E
    sin⁡(x−1)+13sin⁡(3x−1)+C\sin(x-1)+\frac13\sin(3x-1)+C

Kunci jawaban

A

Penjelasan singkat

2cos⁡xsin⁡(1−2x)=sin⁡(1−x)−sin⁡(3x−1)2\cos x\sin(1-2x)=\sin(1-x)-\sin(3x-1). Integralnya cos⁡(x−1)+13cos⁡(3x−1)+C\cos(x-1)+\frac13\cos(3x-1)+C.

Gunakan 2cos⁡Asin⁡B=sin⁡(A+B)−sin⁡(A−B)2\cos A\sin B=\sin(A+B)-\sin(A-B) dengan A=x, B=1−2xA=x,\ B=1-2x:

2cos⁡xsin⁡(1−2x)=sin⁡(1−x)−sin⁡(3x−1)2\cos x\sin(1-2x)=\sin(1-x)-\sin(3x-1).

∫sin⁡(1−x) dx=cos⁡(x−1)\int\sin(1-x)\,dx=\cos(x-1) dan −∫sin⁡(3x−1) dx=13cos⁡(3x−1)-\int\sin(3x-1)\,dx=\frac13\cos(3x-1). Hasil cos⁡(x−1)+13cos⁡(3x−1)+C\cos(x-1)+\frac13\cos(3x-1)+C. Jawabannya A.

⚡ Latihan Matematika PTN