Exercise Zone : Integral Trigonometri Tak Tentu

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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}
   

$\begin{array}{rl}{\displaystyle \int {\sin }^{3}xdx}& ={\displaystyle \int \sin x{\sin }^{2}xdx}\\ & ={\displaystyle \int \sin x(1-{\cos }^{2}x)dx}\\ & ={\displaystyle \int (\sin x-\sin x{\cos }^{2}x)dx}\\ & ={\displaystyle \int \sin xdx-{\displaystyle \int \sin x{\cos }^{2}xdx}}\end{array}$
Misal
$\begin{array}{rl}u& =\cos x\\ du& =-\sin xdx\\ dx& ={\displaystyle \frac{du}{-\sin x}}\end{array}$

$\begin{array}{rl}{\displaystyle \int \sin xdx-{\displaystyle \int \sin x{\cos }^{2}xdx}}& =-\cos x-{\displaystyle \int \sin x\cdot u^2{\displaystyle \frac{du}{-\sin x}}}\\ & =-\cos x+{\displaystyle \int u^{2}du}\\ & =-\cos x+{\displaystyle \frac{1}{3}}{u}^3+C\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -\cos x+{\displaystyle \frac{1}{3}}{\cos }^{3}x+C}}}}\end{array}$

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
$\begin{array}{rl}u& =\cos x\\ du& =-\sin xdx\\ dx& ={\displaystyle \frac{du}{-\sin x}}\end{array}$

$\begin{array}{rl}{\displaystyle \int \sin x{\cos }^{2}xdx}& ={\displaystyle \int \sin x\cdot u^2{\displaystyle \frac{du}{-\sin x}}}\\ & =-{\displaystyle \int \cdot u^{2}du}\\ & =-{\displaystyle \frac{1}{3}}{u}^3+C\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{1}{3}}{\cos }^{3}x+C}}}}\end{array}$

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

$\begin{array}{rl}{\displaystyle \int 6{\sec }^2(2\pi -4x)dx}& =6\cdot {\displaystyle \frac{1}{-4}}\tan (2\pi -4x)+C\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{3}{2}}\tan (2\pi -4x)+C}}}}\end{array}$

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