Pembahasan Soal UN Integral

Soal 1, UN SMA Tapel 2015/2016 Program Studi IPA
Hasil $\int {2x(5-x)^{3}dx}=...$
A. $-\frac{1}{10}(4x+5)(5-x)^{4}+C$
B. $-\frac{1}{10}(6x+5)(5-x)^{4}+C$
C. $-\frac{1}{10}(x+5)(5-x)^{4}+C$
D. $\frac{1}{10}(4x+5)(5-x)^{4}+C$
E. $\frac{1}{2}(5+x)^{4}+C$

Konsep dasar yang digunakan dalam perhitungan:
Integral Parsial: $\int {udv}=uv-\int {vdu}$

Pembahasan
Misal:
$u=2x$
$du=2dx$
$dv=(5-x)^{3}dx$
$v=\int {(5-x)^{3}dx}$

$=\int {(5-x)^{3}d(5-x)}$

$=-\frac{1}{4}(5-x)^{4}$

Maka, dengan menggunakan integral parsial di dapat
$\int {udv}=uv-\int {vdu}$

$=2x\left ( -\frac{1}{4}(5-x)^{4} \right )-\int {-\frac{1}{4}(5-x)^{4}2dx}$

$=-\frac{1}{2}x(5-x)^{4}+\frac{1}{2}\int {(5-x)^{4}dx}$

$=-\frac{1}{2}x(5-x)^{4}+\left ( -\frac{1}{2}.\frac{1}{5}(5-x)^{5} \right)+C$

$=-\frac{1}{2}x(5-x)^{4}-\frac{1}{10}(5-x)^{5}+C$

Catatan:$(A \times B)+(A \times C)=A \times (B+C)$

$=-\frac{1}{10}(5-x)^{4}(5x+(5-x))+C$

$=-\frac{1}{10}(5-x)^{4}(5x+5-x)+C$

$=-\frac{1}{10}(5-x)^{4}(4x+5)+C$

$=-\frac{1}{10}(4x+5)(5-x)^{4}+C$

Jawaban A

Soal 2, UN SMA Tapel 2015/2016 Program Studi IPA
Hasil dari $\int {sin^{5}\left ( 2x \right )cos \left ( 2x \right )dx}=...$
A. $-\frac{1}{5}sin^{6}\left ( 2x \right )+C$
B. $-\frac{1}{10}sin^{6}\left ( 2x \right )+C$
C. $-\frac{1}{12}sin^{6}\left ( 2x \right )+C$
D. $\frac{1}{12}sin^{6}\left ( 2x \right )+C$
E. $\frac{1}{10}sin^{6}\left ( 2x \right )+C$

Konsep dasar yang digunakan dalam perhitungan:

• $\int {ku^{n}du}=k\frac{1}{(n+1)}u^{n+1}+C$; dengan $k$ adalah konstanta
• $\frac{du}{dx}$ merupakan turunan $u$ terhadap $x$
• $u=sin \left ( ax+b \right )\Rightarrow \frac{du}{dx}=a.cos \left ( ax+b \right )$

Pembahasan
$\int {sin^{5}\left ( 2x \right )cos \left ( 2x \right )dx}=$
Misal:

$u=sin \left ( 2x \right )$

$\frac{du}{dx}=2 cos \left ( 2x \right )$

$du = 2cos \left ( 2x \right )dx$

$\frac{1}{2}du = cos \left ( 2x \right )dx$

Maka:
$\int {sin^{5}\left ( 2x \right )cos \left ( 2x \right )dx}$

$=\int {u^{5}.\frac{1}{2}du}$

$=\frac{1}{2}\int {u^{5}du}$

$=\frac{1}{2}.\frac{1}{6}u^{6}+C$

$=\frac{1}{12}sin^{6}\left ( 2x \right )+C$

Jawaban D

Soal 3, UN SMA Tapel 2015/2016 Program Studi IPA
Hasil dari $\int {\frac{x^{2}-2}{\sqrt{6x-x^{3}}}}dx=...$
A. $-\frac{3}{2}\sqrt{6x-x^{3}}+C$
B. $-\frac{2}{3}\sqrt{6x-x^{3}}+C$
C. $-\frac{1}{6}\sqrt{6x-x^{3}}+C$
D. $\frac{1}{6}\sqrt{6x-x^{3}}+C$
E. $\frac{2}{3}\sqrt{6x-x^{3}}+C$

Konsep dasar yang digunakan dalam perhitungan:
$\int {ku^{n}du}=k\frac{1}{(n+1)}u^{n+1}+C$; dengan $k$ adalah konstanta

Pembahasan
$\int {\frac{x^{2}-2}{\sqrt{6x-x^{3}}}}dx=...$
Misal:
$u=6x-x^{3}$
$\frac{du}{dx}=6-3x^{2}$
$du=(6-3x^{2})dx$
$-\frac{1}{3}du=(x^{2}-2)dx$
Jadi
$\int {\frac{1}{\sqrt{6x-x^{3}}}(x^{2}-2)}dx$
$=\int {\frac{1}{\sqrt{u}}\left ( -\frac{1}{3}du \right )}$
$= -\frac{1}{3}\int (u)^{-\frac{1}{2}}du$
$=-\frac{1}{3}\frac{1}{\frac{1}{2}}u^{\frac{1}{2}}+C$
$=-\frac{1}{3}.2u^{\frac{1}{2}}+C$
$=-\frac{2}{3}\sqrt{u}+C$
$=-\frac{2}{3}\sqrt{6x-x^{3}}+C$
Jawaban B