Exercise Zone : Integral Trigonometri Tak Tentu

Berikut ini adalah kumpulan soal mengenai Integral Trigonometri Tak Tentu tipe standar. Jika ingin bertanya soal, silahkan gabung ke grup Facebook atau Telegram.

No. 1

Hasil dari \displaystyle\int\sin^3x\ dx adalah ....

1. {-\dfrac13\cos^3x+C}
2. {-\cos x-\dfrac13\sin^3x+C}
   
3. {\cos x-\dfrac13\sin^3x+C}
   

4. {-\cos x+\dfrac13\cos^3x+C}
5. {-\dfrac13\sin^3x+C}
   

\(\begin{aligned} \displaystyle\int\sin^3x\ dx&=\displaystyle\int\sin x\sin^2x\ dx\\ &=\displaystyle\int\sin x\left(1-\cos^2x\right)\ dx\\ &=\displaystyle\int\left(\sin x-\sin x\cos^2x\right)\ dx\\ &=\displaystyle\int \sin x\ dx-\displaystyle\int\sin x\cos^2x\ dx\\ \end{aligned}\)
Misal
\(\begin{aligned} u&=\cos x\\ du&=-\sin x\ dx\\ dx&=\dfrac{du}{-\sin x} \end{aligned}\)

\(\begin{aligned} \displaystyle\int \sin x\ dx-\displaystyle\int\sin x\cos^2x\ dx&=-\cos x-\displaystyle\int\sin x\cdot u^2\ \dfrac{du}{-\sin x}\\[8pt] &=-\cos x+\displaystyle\int u^2\ du\\ &=-\cos x+\dfrac13u^3+C\\ &=\boxed{\boxed{-\cos x+\dfrac13\cos^3x+C}} \end{aligned}\)

Jadi, \displaystyle\int\sin^3x\ dx=-\cos x+\dfrac13\cos^3x+C.
JAWAB: D

No. 2

\displaystyle\int\sin x\left(1-\sin^2x\right)\ dx= ....

1. {\cos x-\dfrac13\cos^3x+C}
2. {\cos x-\dfrac13\sin^3x+C}
   
3. {-\dfrac13\cos^3x+C}
   

4. {-\dfrac13\sin^3x+C}
5. {-\cos x-\dfrac13\sin^3x+C}
   

\displaystyle\int\sin x\left(1-\sin^2x\right)\ dx=\displaystyle\int\sin x\cos^2x\ dx
Misal
\(\begin{aligned} u&=\cos x\\ du&=-\sin x\ dx\\ dx&=\dfrac{du}{-\sin x} \end{aligned}\)

\(\begin{aligned} \displaystyle\int\sin x\cos^2x\ dx&=\displaystyle\int\sin x\cdot u^2\ \dfrac{du}{-\sin x}\\[8pt] &=-\displaystyle\int\cdot u^2\ du\\ &=-\dfrac13u^3+C\\ &=\boxed{\boxed{-\dfrac13\cos^3x+C}} \end{aligned}\)

Jadi, \displaystyle\int\sin x\left(1-\sin^2x\right)\ dx=-\dfrac13\cos^3x+C.
JAWAB: C

No. 3

\displaystyle\int6\sec^2(2\pi-4x)\ dx= ...

SKB CPNS MATEMATIKA 2020

\(\eqalign{ \displaystyle\int6\sec^2(2\pi-4x)\ dx&=6\cdot\dfrac1{-4}\tan(2\pi-4x)+C\\ &=\boxed{\boxed{-\dfrac32\tan(2\pi-4x)+C}} }\)

Jadi, \displaystyle\int6\sec^2(2\pi-4x)\ dx=-\dfrac32\tan(2\pi-4x)+C.