BLANTERVIO103

Pembahasan Soal Un Integral Fungsi Trigonometri

Pembahasan Soal Un Integral Fungsi Trigonometri
10/15/2018

Pembahasan soal Ujian Nasional (UN) Matematika IPA jenjang pendidikan Sekolah Menengan Atas untuk pokok bahasan Integral Fungsi Trigonometri.

Berikut beberapa konsep yang dipakai dalam pembahasan.

Integral Fungsi Trigonometri
1. ∫ sin x dx = −cos x + C
2. ∫ cos x dx = sin x + C
3. ∫ sin (ax+b) dx = \(\mathrm{-\frac{1}{a}}\)cos (ax+b) + C
4. ∫ cos (ax+b) dx = \(\mathrm{\frac{1}{a}}\)sin (ax+b) + C

Integral Substitusi
∫ sinnax cos ax = \(\mathrm{\frac{1}{{\color{Green} a}({\color{Red} n}+1)}}\)sinn+1ax + C
∫ cosnax sin ax = \(\mathrm{-\frac{1}{{\color{Green} a}({\color{Red} n}+1)}}\)cosn+1ax + C

Sudut istimewa dalam radian

\(\frac{\pi}{6}\)
\(\frac{\pi}{4}\)
\(\frac{\pi}{3}\)
sin(θ)
\(\frac{1}{2}\) 
\(\frac{1}{2}\)√2 
\(\frac{1}{2}\)√3 
cos(θ)
 \(\frac{1}{2}\)√3
\(\frac{1}{2}\)√2 
\(\frac{1}{2}\) 
tan(θ)
 \(\frac{1}{3}\)√3
 1
√3     

Sudut kuadrantal dalam radian

0
\(\frac{\pi}{2}\)
Ï€
\(\frac{3\pi}{2}\)
2Ï€
sin(θ)
0
1
0
−1
0
cos(θ)
1
0
−1
0
1
tan(θ)
0
tdf
0
tdf
0
*tdf = tidak terdefinisi


Identitias dan sifat-sifat trigonometri
sin2x + cos2x = 1
sin2x = 1 − cos2x
cos2x = 1 − sin2x

sin 2A = 2 sin A cos A
cos 2A = cos2A − sin2A

sin2x = \(\frac{1}{2}\) − \(\frac{1}{2}\)cos 2x
cos2x = \(\frac{1}{2}\) + \(\frac{1}{2}\)cos 2x

sin A cos B = \(\frac{1}{2}\)(sin (A + B) + sin (A − B))
cos A sin B = \(\frac{1}{2}\)(sin (A + B) − sin (A − B))

cos (−x) = cos x
sin (−x) = −sin x



1.  UN 2005
Hasil dari ∫ cos5x dx = ...
A.  \(\mathrm{-\frac{1}{6}}\)cos6x sin x + C
B.  \(\mathrm{\frac{1}{6}}\)cos6x sin x + C
C.  −sin x + \(\mathrm{\frac{2}{3}}\)sin3\(\mathrm{\frac{1}{5}}\)sin5x + C
D.  sin x − \(\mathrm{\frac{2}{3}}\)sin3\(\mathrm{\frac{1}{5}}\)sin5x + C
E.  sin x + \(\mathrm{\frac{2}{3}}\)sin3\(\mathrm{\frac{1}{5}}\)sin5x + C

Pembahasan :
∫ cos5x dx
⇒ ∫ (cos2x)2 cos x dx
⇒ ∫ (1 − sin2x)2 cos x dx

Misalkan :
u = sin x
du = cosx dx

⇒ ∫ (1 − u2)2 du
⇒ ∫ (1 − 2u2 + u4) du
= u − \(\mathrm{\frac{2}{3}}\)u3 \(\mathrm{\frac{1}{5}}\)u5 + C
= sin x − \(\mathrm{\frac{2}{3}}\)sin3\(\mathrm{\frac{1}{5}}\)sin5x + C

Jawaban : D


2.  UN 2006
Nilai \(\mathrm{\int_{0}^{\pi }}\)sin 2x cos x dx = ...
A.  \(-\frac{4}{3}\)
B.  \(-\frac{1}{3}\)
C.  \(\frac{1}{3}\)
D.  \(\frac{2}{3}\)
E.  \(\frac{4}{3}\)

Pembahasan :
\(\mathrm{\int_{0}^{\pi }}\)sin 2x cos x dx
⇒ \(\mathrm{\int_{0}^{\pi }}\)2sin x cos x cos x dx
⇒ 2\(\mathrm{\int_{0}^{\pi }}\)cos2x sin x dx
= 2\(\mathrm{\left [ -\frac{1}{({\color{Red} 2}+1)}cos^{{\color{Red} 2}+1}x\right ]_{0}^{\pi }}\)
= \(\mathrm{-\frac{2}{3}}\)[cos3x\(]_{0}^{\pi }\)
= \(\mathrm{-\frac{2}{3}}\){cos3Ï€ − cos30}
= \(\mathrm{-\frac{2}{3}}\){(−1)3 − (1)3}
= \(\mathrm{\frac{4}{3}}\)

Jawaban : E


3.  UN 2008
Hasil dari ∫ cos2x sin x dx = ...
A.  \(\mathrm{\frac{1}{3}}\)cos3x + C
B.  \(\mathrm{-\frac{1}{3}}\)cos3x + C
C.  \(\mathrm{-\frac{1}{3}}\)sin3x + C
D.  \(\mathrm{\frac{1}{3}}\)sin3x + C
E.  3 sin3x + C

Pembahasan :
∫ cos2x sin x dx
= \(\mathrm{-\frac{1}{({\color{Red} 2}+1)}}\)cos2+1x + C
= \(\mathrm{-\frac{1}{3}}\)cos3x + C

Jawaban : B


4. UN 2009
Hasil ∫ cos3x dx adalah ...
A.  sin x \(\mathrm{-\frac{1}{3}}\)sin3+ C
B.  \(\mathrm{\frac{1}{4}}\)cos4+ C
C.  3 cos2x sin x + C
D.  \(\mathrm{\frac{1}{3}}\)sin3x − sin x + C
E.  sin x − 3 sin3+ C

Pembahasan :
∫ cos3x dx
⇒ ∫ cos2x cos x dx
⇒ ∫ (1 − sin2x) cos x dx

Misalkan :
u = sin x
du = cos x dx

⇒ ∫ (1 − u2) du
= u − \(\mathrm{\frac{1}{3}}\)u3 + C
= sin x − \(\mathrm{\frac{1}{3}}\)sin3+ C

Jawaban : A


5.  UN 2009
Hasil dari ∫ sin 3x cos x dx adalah ...
A.  \(\mathrm{-\frac{1}{8}}\)cos 4x \(\mathrm{-\frac{1}{4}}\)cos 2x + C
B.  \(\mathrm{\frac{1}{8}}\)cos 4x + \(\mathrm{\frac{1}{4}}\)cos 2x + C
C.  \(\mathrm{-\frac{1}{4}}\)cos 4x \(\mathrm{-\frac{1}{2}}\)cos 2x + C
D.  \(\mathrm{\frac{1}{4}}\)cos 4x + \(\mathrm{\frac{1}{2}}\)cos 2x + C
E.  −4 cos 4x − 2 sin 2x + C

Pembahasan :
∫ sin 3x cos x dx
⇒ ∫\(\frac{1}{2}\)(sin (3x + x) + sin (3x − x)) dx
⇒ \(\frac{1}{2}\) ∫ (sin 4x + sin 2x) dx
= \(\mathrm{\frac{1}{2}\left ( -\frac{1}{4}cos\:4x+\left ( -\frac{1}{2}cos\:2x \right ) \right )+C}\)
= \(\mathrm{-\frac{1}{8}}\)cos 4x \(\mathrm{-\frac{1}{4}}\)cos 2x + C

Jawaban : A


6.  UN 2010
Hasil dari \(\mathrm{\int_{0}^{\frac{\pi }{6}}}\)(sin 3x + cos 3x) dx adalah ...
     A.  \(\frac{2}{3}\)
     B.  \(\frac{1}{3}\)
     C.  \(0\)
     D.  \(-\frac{1}{3}\)
     E.  \(-\frac{2}{3}\)

Pembahasan :
\(\mathrm{\int_{0}^{\frac{\pi }{6}}}\)(sin 3x + cos 3x) dx
= \(\mathrm{\left [ -\frac{1}{3}cos\:3x+\frac{1}{3}sin\:3x \right ]_{0}^{\frac{\pi }{6}}}\)
= \(\mathrm{\frac{1}{3}\left [ sin\,3x-cos\,3x \right ]_{0}^{\frac{\pi }{6}}}\)
= \(\frac{1}{3}\){(sin \(\frac{\pi}{2}\) − cos \(\frac{\pi}{2}\)) − (sin 0 − cos 0)}
= \(\mathrm{\frac{1}{3}}\)((1 − 0) − (0 − 1))
= \(\mathrm{\frac{2}{3}}\)

Jawaban : A


7. UN 2010
Hasil dari ∫ (sin2x − cos2x) dx adalah ...
A.  \(\mathrm{\frac{1}{2}}\)cos 2x + C
B.  −2 cos 2x + C
C.  −2 sin 2x + C
D.  \(\mathrm{\frac{1}{2}}\)sin 2x + C
E.  \(\mathrm{-\frac{1}{2}}\)sin 2x + C

Pembahasan :
∫ (sin2x − cos2x) dx
⇒ ∫ −(cos2x − sin2x) dx
⇒ − ∫ cos 2x dx
= \(\mathrm{-\frac{1}{2}}\)sin 2x + C

Jawaban : E


8.  UN 2010
Hasil dari ∫ sin (\(\mathrm{\frac{1}{2}}\)x − Ï€) cos (\(\mathrm{\frac{1}{2}}\)x − Ï€) dx = ...
A.  −2 cos (x − 2Ï€) + C
B.  \(-\frac{1}{2}\)cos (x − 2Ï€) + C
C.  \(\frac{1}{2}\)cos (x − 2Ï€) + C
D.  cos (x − 2Ï€) + C
E.  2 cos (x − 2Ï€) + C

Pembahasan :
∫ sin (\(\mathrm{\frac{1}{2}}\)x − Ï€) cos (\(\mathrm{\frac{1}{2}}\)x − Ï€) dx
⇒ ∫ \(\frac{1}{2}\). 2 sin (\(\mathrm{\frac{1}{2}}\)x − Ï€) cos (\(\mathrm{\frac{1}{2}}\)x − Ï€) dx
⇒ ∫ \(\mathrm{\frac{1}{2}}\) sin 2(\(\mathrm{\frac{1}{2}}\)x − Ï€) dx
⇒ \(\mathrm{\frac{1}{2}}\) ∫ sin (x − 2Ï€) dx
= \(\frac{1}{2}\)(−cos (x − 2Ï€)) + C
= \(-\frac{1}{2}\)cos (x − 2Ï€) + C

Jawaban : B


9.  UN 2011
Hasil \(\mathrm{\int_{0}^{\pi }}\)(sin 3x + cos x) dx = ...
A.  \(\frac{10}{3}\)
B.  \(\frac{8}{3}\)
C.  \(\frac{4}{3}\)
D.  \(\frac{2}{3}\)
E.  \(-\frac{4}{3}\)

Pembahasan :
\(\mathrm{\int_{0}^{\pi }}\)(sin 3x + cos x) dx
= \(\mathrm{\left [ -\frac{1}{3}cos\,3x+sin\,x \right ]_{0}^{\pi }}\)
= (\(-\frac{1}{3}\)cos 3Ï€ + sin Ï€) − (\(-\frac{1}{3}\)cos 0 + sin 0)
= (\(-\frac{1}{3}\)(−1) + 0) − (\(-\frac{1}{3}\)(1) + 0)
= \(\frac{1}{3}\) + \(\frac{1}{3}\)
= \(\frac{2}{3}\)

Jawaban : D


10. UN 2011

Nilai dari ∫ cos42x sin 2x dx adalah...
     A.  \(\mathrm{-\frac{1}{10}}\)sin52x + C
     B.  \(\mathrm{-\frac{1}{10}}\)cos52x + C
     C.  \(\mathrm{-\frac{1}{5}}\)cos52x + C
     D.  \(\mathrm{\frac{1}{5}}\)cos52x + C
     E.  \(\mathrm{\frac{1}{10}}\)sin52x + C

Pembahasan :
∫ cos42x sin 2x dx
= \(\mathrm{-\frac{1}{{\color{Green} 2}({\color{Red} 4}+1)}}\)cos4+12x + C
= \(\mathrm{-\frac{1}{10}}\)cos52x + C

Jawaban : B


11.  UN 2012
Nilai dari \(\mathrm{\int_{0}^{\frac{\pi }{2}}}\)sin (2x − Ï€) dx = ...
A.  −2
B.  −1
C.  0
D.  2
E.  4

Pembahasan :
\(\mathrm{\int_{0}^{\frac{\pi }{2}}}\)sin (2x − Ï€) dx
= \(\mathrm{\left [ -\frac{1}{2}cos(2x-\pi ) \right ]_{0}^{\frac{\pi }{2}}}\)
= \(-\frac{1}{2}\){cos (2.\(\frac{\pi}{2}\) − Ï€) − cos (2.0 − Ï€)}
= \(\mathrm{-\frac{1}{2}}\){cos 0 − cos (−Ï€)}
= \(\mathrm{-\frac{1}{2}}\){1 − (−1)}
=  −1

Jawaban : B


12.  UN 2013
Nilai dari \(\mathrm{\int_{0}^{\frac{\pi }{2}}}\)sin3x dx = ...
A.  −1/3
B.  −1/2
C.  0
D.  1/3
E.  2/3

Pembahasan :
⇒ ∫ sin2x sin x dx
⇒ ∫ (1 − cos2x) sin x dx

Misalkan :
u = cos x
du = −sin x dx
−du = sin x dx

Substitusi :
⇒ ∫ (1 − u2). −du
⇒ ∫ (u2 − 1) du
= \(\frac{1}{3}\)u3 − u + C
= \(\frac{1}{3}\)cos3x − cos x + C

Untuk batas x = 0 hingga x = \(\frac{\pi}{2}\)
= (\(\frac{1}{3}\)cos3\(\frac{\pi}{2}\) − cos \(\frac{\pi}{2}\)) − (\(\frac{1}{3}\)cos30 − cos 0)
= (\(\frac{1}{3}\).03 − 0) − (\(\frac{1}{3}\). 13 − 1)
= 0 − (\(−\frac{2}{3}\))
= \(\frac{2}{3}\)

Jawaban : E


13. UN 2013
Nilai \(\mathrm{\int_{0}^{\frac{\pi }{4}}}\)cos2x dx = ...
     A.  \(\mathrm{\frac{\pi }{8}+\frac{1}{4}}\)
     B.  \(\mathrm{\frac{\pi }{8}+\frac{1}{2}}\)
     C.  \(\mathrm{\frac{\pi }{8}-\frac{1}{4}}\)
     D.  \(\mathrm{\frac{\pi }{4}+\frac{1}{\sqrt{2}}}\)
     E.  \(\mathrm{\frac{\pi }{4}-\frac{1}{\sqrt{2}}}\)

Pembahasan :
\(\mathrm{\int_{0}^{\frac{\pi }{4}}}\)cos2x dx
⇒ \(\mathrm{\int_{0}^{\frac{\pi }{4}}}\)(\(\frac{1}{2}\) + \(\frac{1}{2}\)cos 2x) dx
= \(\mathrm{\left [\frac{1}{2}x+\frac{1}{4}sin\,2x  \right ]_{0}^{\frac{\pi }{4}}}\)
= (\(\mathrm{ \frac{\pi }{8}+\frac{1}{4} }\)sin \(\frac{\pi}{2}\)) − (0 + \(\frac{1}{4}\)sin 0)
= (\(\mathrm{ \frac{\pi }{8}+\frac{1}{4} }\).1) − (0 + \(\frac{1}{4}\).0)
= \(\mathrm{ \frac{\pi }{8}+\frac{1}{4} }\)

Jawaban : A


14.  UN 2014
Hasil dari ∫ sin23x cos 3x dx = ...
A.  −sin33x + C
B.  \(\mathrm{-\frac{1}{3}}\)sin33x + C
C.  \(\mathrm{-\frac{1}{9}}\)sin33x + C
D.  \(\mathrm{\frac{1}{9}}\)sin33x + C
E.  \(\mathrm{\frac{1}{3}}\)sin33x + C

Pembahasan :
∫ sin23x cos 3x dx
= \(\mathrm{\frac{1}{{\color{Green} 3}({\color{Red} 2}+1)}}\)sin2+13x + C
= \(\mathrm{\frac{1}{9}}\)sin33x + C

Jawaban : D


15.  UN 2014
Hasil dari \(\mathrm{\int_{0}^{\frac{\pi }{6}}}\)sin 4x cos 2x dx = ...
A.  \(\frac{4}{3}\)
B.  \(\frac{2}{3}\)
C.  \(\frac{1}{3}\)
D.  \(\frac{7}{24}\)
E.  \(-\frac{1}{3}\)

Pembahasan :
\(\int_{0}^{\frac{\pi }{6}}\) sin 4x cos 2x dx
⇒ \(\int_{0}^{\frac{\pi }{6}}\)\(\frac{1}{2}\)(sin (4x + 2x) + sin (4x − 2x)) dx
⇒ \(\frac{1}{2}\) \(\int_{0}^{\frac{\pi }{6}}\) (sin 6x + sin 2x) dx
= \(\mathrm{\frac{1}{2}\left [ -\frac{1}{6}cos\,6x+\left ( -\frac{1}{2}cos\,2x \right ) \right ]_{0}^{\frac{\pi }{6}}}\)
= \(\mathrm{-\frac{1}{4}\left [ \frac{1}{3}cos\,6x+cos\,2x \right ]_{0}^{\frac{\pi }{6}}}\)
= \(-\frac{1}{4}\){(\(\frac{1}{3}\)cos Ï€ + cos \(\frac{\pi}{3}\)) − (\(\frac{1}{3}\)cos 0 + cos 0)}
= \(-\frac{1}{4}\){(\(\frac{1}{3}\)(−1) + \(\frac{1}{2}\)) − (\(\frac{1}{3}\)(1) + 1)}
= \(-\frac{1}{4}\){\(-\frac{7}{6}\)}
= \(\frac{7}{24}\)

Jawaban : D


16.  UN 2014
Hasil dari ∫ sin3x cos x dx = ...
A.  \(\mathrm{\frac{1}{2}}\)sin4x + C
B.  \(\mathrm{\frac{1}{4}}\)sin4x + C
C.  \(\mathrm{\frac{1}{8}}\)sin4x + C
D.  \(\mathrm{-\frac{1}{8}}\)sin4x + C
E.  \(\mathrm{-\frac{1}{2}}\)sin4x + C

Pembahasan :
∫ sin3x cos x dx
= \(\mathrm{\frac{1}{({\color{Red} 3}+1)}}\)sin3+1x + C
= \(\mathrm{\frac{1}{4}}\)sin4x + C

Jawaban : B


17.  UN 2015
Hasil ∫ 4 sin 4x cos 2x dx adalah...
A.  \(\mathrm{-\frac{1}{6}}\)cos 6x − \(\mathrm{\frac{1}{2}}\)cos 2x + C
B.  \(\mathrm{-\frac{1}{3}}\)cos 6x − cos 2x + C
C.  \(\mathrm{\frac{1}{6}}\)cos 6x − \(\mathrm{\frac{1}{2}}\)cos 2x + C
D.  \(\mathrm{\frac{1}{6}}\)cos 6x + \(\mathrm{\frac{1}{2}}\)cos 2x + C
E.  \(\mathrm{\frac{1}{3}}\)cos 6x + cos 2x + C

Pembahasan :
4 ∫ sin 4x cos 2x dx
⇒ 4 ∫ \(\frac{1}{2}\)(sin (4x + 2x) + sin (4x − 2x)) dx
⇒ 4.\(\frac{1}{2}\) ∫ (sin 6x + sin 2x) dx
= 2 (\(-\frac{1}{6}\)cos 6x + (\(-\frac{1}{2}\)cos 2x)) + C
= \(-\frac{1}{3}\)cos 6x − cos 2x + C

Jawaban : B


18.  UN 2015
Nilai dari \(\mathrm{\int_{\frac{\pi }{4}}^{\pi }}\)(16 sin 2x − 2 cos 2x) dx = ...
A.  −9
B.  −8
C.  −7
D.  −4
E.  −2

Pembahasan :
\(\int_{\frac{\pi }{4}}^{\pi }\) (16 sin 2x − 2 cos 2x) dx
= \(\mathrm{\left [ -\frac{16}{2}cos\,2x-\frac{2}{2}sin\,2x \right ]_{\frac{\pi }{4}}^{\pi }}\)
= \(\mathrm{\left [ -8\,cos\,2x-sin\,2x \right ]_{\frac{\pi }{4}}^{\pi }}\)
= (−8cos 2Ï€ − sin 2Ï€)  − (−8cos \(\frac{\pi }{2}\) − sin \(\frac{\pi }{2}\))
= (−8(1) − 0) − (−8(0) − 1)
= −8 + 1
= −7

Jawaban : C


19.  UN 2016
Hasil dari ∫ sin52x cos 2x dx = ...
A.  \(\mathrm{-\frac{1}{5}}\)sin62x + C
B.  \(\mathrm{-\frac{1}{10}}\)sin62x + C
C.  \(\mathrm{-\frac{1}{12}}\)sin62x + C
D.  \(\mathrm{\frac{1}{12}}\)sin62x + C
E.  \(\mathrm{\frac{1}{10}}\)sin62x + C

Pembahasan :
∫ sin52x cos 2x dx
= \(\mathrm{\frac{1}{{\color{Green} 2}({\color{Red} 5}+1)}}\)sin5+12x + C
= \(\mathrm{\frac{1}{12}}\)sin62x + C

Jawaban : D



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