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
|
\(\frac{3\pi}{2}\)
|
2π
| |
sin(θ)
|
0
|
1
|
0
|
−1
|
0
|
cos(θ)
|
1
|
0
|
−1
|
0
|
1
|
tan(θ)
|
0
|
tdf
|
0
|
tdf
|
0
|
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
D. sin x − \(\mathrm{\frac{2}{3}}\)sin3x
E. sin x
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
= sin x − \(\mathrm{\frac{2}{3}}\)sin3x
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
B. \(\mathrm{\frac{1}{4}}\)cos4x
C. 3 cos2x sin x
D. \(\mathrm{\frac{1}{3}}\)sin3x − sin x
E. sin x − 3 sin3x
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
= sin x − \(\mathrm{\frac{1}{3}}\)sin3x
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
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 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 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
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)}
= \(-\frac{1}{4}\){\(-\frac{7}{6}\)}
= \(\frac{7}{24}\)
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
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
EmoticonEmoticon