Pembahasan Soal Un Integral Fungsi Trigonometri

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	\(\frac{\pi}{2}\)	π	\(\frac{3\pi}{2}\)	2π
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 + \(\mathrm{\frac{2}{3}}\)sin³x + \(\mathrm{\frac{1}{5}}\)sin⁵x + C
D.  sin x − \(\mathrm{\frac{2}{3}}\)sin³x + \(\mathrm{\frac{1}{5}}\)sin⁵x + C
E.  sin x + \(\mathrm{\frac{2}{3}}\)sin³x + \(\mathrm{\frac{1}{5}}\)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 − \(\mathrm{\frac{2}{3}}\)sin³x + \(\mathrm{\frac{1}{5}}\)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 adalah ...
A.  sin x \(\mathrm{-\frac{1}{3}}\)sin³x + C
B.  \(\mathrm{\frac{1}{4}}\)cos⁴x + C
C.  3 cos²x sin x + C
D.  \(\mathrm{\frac{1}{3}}\)sin³x − sin x + C
E.  sin x − 3 sin³x + 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 − \(\mathrm{\frac{1}{3}}\)sin³x + 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 − 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 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 − 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 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 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 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 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

Sumber http://smatika.blogspot.com

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