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Exercise Zone : Integral Trigonometri Tak Tentu

Written By Anonymous Saturday, March 28, 2020 Add Comment Edit
Berikut ini adalah kumpulan soal mengenai Integral Trigonometri Tak Tentu tingkat dasar. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

No. 1

Hasil dari ∫sin⁡3x dx\displaystyle\int\sin^3x\ dx adalah ....
  1. −13cos⁡3x+C{-\dfrac13\cos^3x+C}
  2. −cos⁡x−13sin⁡3x+C{-\cos x-\dfrac13\sin^3x+C}
  3. cos⁡x−13sin⁡3x+C{\cos x-\dfrac13\sin^3x+C}
  1. −cos⁡x+13cos⁡3x+C{-\cos x+\dfrac13\cos^3x+C}
  2. −13sin⁡3x+C{-\dfrac13\sin^3x+C}
∫ sin ⁡3x dx=∫sin⁡x sin ⁡2x dx=∫sin⁡x(1− cos ⁡2x) dx=∫(sin⁡x−sin⁡x cos ⁡2x) dx=∫sin⁡x dx−∫sin⁡x cos ⁡2x dx\begin{aligned}\displaystyle\int\sin^3x\ dx&=\displaystyle\int\sin x\sin^2x\ dx\\&=\displaystyle\int\sin x\left(1-\cos^2x\right)\ dx\\&=\displaystyle\int\left(\sin x-\sin x\cos^2x\right)\ dx\\&=\displaystyle\int \sin x\ dx-\displaystyle\int\sin x\cos^2x\ dx\\\end{aligned}
Misal
u=cos⁡xdu=−sin⁡x dxdx=du−sin⁡x\begin{aligned}u&=\cos x\\du&=-\sin x\ dx\\dx&=\dfrac{du}{-\sin x}\end{aligned}

∫sin⁡x dx−∫sin⁡x cos ⁡2x dx=−cos⁡x−∫sin⁡x⋅u2 du−sin⁡x=−cos⁡x+∫u2 du=−cos⁡x+13u3+C=−cos⁡x+13 cos ⁡3x+C\begin{aligned}\displaystyle\int \sin x\ dx-\displaystyle\int\sin x\cos^2x\ dx&=-\cos x-\displaystyle\int\sin x\cdot u^2\ \dfrac{du}{-\sin x}\\[8pt]&=-\cos x+\displaystyle\int u^2\ du\\&=-\cos x+\dfrac13u^3+C\\&=\boxed{\boxed{-\cos x+\dfrac13\cos^3x+C}}\end{aligned}
Jadi, ∫sin⁡3x dx=−cos⁡x+13cos⁡3x+C\displaystyle\int\sin^3x\ dx=-\cos x+\dfrac13\cos^3x+C.
JAWAB: D

No. 2

∫sin⁡x(1−sin⁡2x) dx=\displaystyle\int\sin x\left(1-\sin^2x\right)\ dx= ....
  1. cos⁡x−13cos⁡3x+C{\cos x-\dfrac13\cos^3x+C}
  2. cos⁡x−13sin⁡3x+C{\cos x-\dfrac13\sin^3x+C}
  3. −13cos⁡3x+C{-\dfrac13\cos^3x+C}
  1. −13sin⁡3x+C{-\dfrac13\sin^3x+C}
  2. −cos⁡x−13sin⁡3x+C{-\cos x-\dfrac13\sin^3x+C}
∫sin⁡x(1−sin⁡2x) dx=∫sin⁡xcos⁡2x dx\displaystyle\int\sin x\left(1-\sin^2x\right)\ dx=\displaystyle\int\sin x\cos^2x\ dx
Misal
u=cos⁡xdu=−sin⁡x dxdx=du−sin⁡x\begin{aligned}u&=\cos x\\du&=-\sin x\ dx\\dx&=\dfrac{du}{-\sin x}\end{aligned}

Related:


∫sin⁡x cos ⁡2x dx=∫sin⁡x⋅u2 du−sin⁡x=−∫⋅u2 du=−13u3+C=−13 cos ⁡3x+C\begin{aligned}\displaystyle\int\sin x\cos^2x\ dx&=\displaystyle\int\sin x\cdot u^2\ \dfrac{du}{-\sin x}\\[8pt]&=-\displaystyle\int\cdot u^2\ du\\&=-\dfrac13u^3+C\\&=\boxed{\boxed{-\dfrac13\cos^3x+C}}\end{aligned}
Jadi, ∫sin⁡x(1−sin⁡2x) dx=−13cos⁡3x+C\displaystyle\int\sin x\left(1-\sin^2x\right)\ dx=-\dfrac13\cos^3x+C.
JAWAB: C

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