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Pembahasan Soal UN Integral Fungsi Trigonometri


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

Berikut beberapa konsep yang digunakan 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
\(\frac{\pi}{2}\)
π
\(\frac{3\pi}{2}\)
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\(]_{0}^{\pi }\)
= \(\mathrm{-\frac{2}{3}}\){cos3π − 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 merupakan ...
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 merupakan ...
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 merupakan ...
     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 merupakan ...
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 − π) cos (\(\mathrm{\frac{1}{2}}\)x − π) dx = ...
A.  −2 cos (x − 2π) + C
B.  \(-\frac{1}{2}\)cos (x − 2π) + C
C.  \(\frac{1}{2}\)cos (x − 2π) + C
D.  cos (x − 2π) + C
E.  2 cos (x − 2π) + C

Pembahasan :
∫ sin (\(\mathrm{\frac{1}{2}}\)x − π) cos (\(\mathrm{\frac{1}{2}}\)x − π) dx
⇒ ∫ \(\frac{1}{2}\). 2 sin (\(\mathrm{\frac{1}{2}}\)x − π) cos (\(\mathrm{\frac{1}{2}}\)x − π) dx
⇒ ∫ \(\mathrm{\frac{1}{2}}\) sin 2(\(\mathrm{\frac{1}{2}}\)x − π) dx
⇒ \(\mathrm{\frac{1}{2}}\) ∫ sin (x − 2π) dx
= \(\frac{1}{2}\)(−cos (x − 2π)) + C
= \(-\frac{1}{2}\)cos (x − 2π) + 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 3π + sin π) − (\(-\frac{1}{3}\)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 merupakan...
     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 − π) dx = ...
A.  −2
B.  −1
C.  0
D.  2
E.  4

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