Soal dan Pembahasan Integral Fungsi Trigonometri

Soal 1: SBMPTN 2018 Kode 527

Hasil dari $\int_{0}^{\frac{\pi}{6}}cos \ 2x \ . \ cos \ x \ dx=...$ 
(A) $\frac{5}{6}$
(B) $\frac{4}{6}$
(C) $\frac{5}{12}$
(D) $-\frac{5}{12}$
(E) $-\frac{5}{6}$

PEMBAHASAN:
Kita gunakan indentitas trigonometri:
$cos \ A \ . \ cos \ B \ = \frac{1}{2} cos(A+B)+\frac{1}{2}cos (A-B)$

Sifat trigonometri di atas kita gunakan untuk menyederahanakan bentuk soal menjadi seperti berikut ini:

$\int_{0}^{\frac{\pi}{6}}cos \ 2x \ . \ cos \ x \ dx$
$=\int_{0}^{\frac{\pi}{6}} \frac{1}{2}(cos \ (3x) \ + \ cos \ (x) \ )dx$
$=\frac{1}{2}\left [\frac{1}{3}sin \ (3x) \ + sin \ x \  \right ]_{0}^{\frac{\pi}{6}}$
$=\frac{1}{2}\left ( \left [\frac{1}{3}sin \ (3.\frac{\pi}{6}) \ + sin \ \frac{\pi}{6} \  \right ] -\left [\frac{1}{3}sin \ (3\ . \ 0 \ ) \ + sin \ 0 \  \right ]  \right )$
$=\frac{1}{2}\left ( \frac{1}{3} +\frac{1}{2}-0\right )$
$=\frac{1}{2} \ . \ \frac{5}{6}=\frac{5}{12}$

Pilihan jawabannya adalah (C)

Soal 2: SIMAK UI 2018 Kode 421

Jika $\int_{-2}^{0}\left ( cos\left ( \pi+ \frac{\pi kx}{2} \right )+ \frac{9x^2-10x+14}{k+12}\right )dx=(k-9)(k-11)$ untuk nilai $k$ bilangan bulat, maka $k^2-14=...$
(A) $140$
(B) $135$
(C) $130$
(D) $125$
(E) $120$

PEMBAHASAN:
$\int_{-2}^{0}\left ( cos\left ( \pi+ \frac{\pi kx}{2} \right )+ \frac{9x^2-10x+14}{k+12}\right )dx=(k-9)(k-11)$

$\int_{-2}^{0}\left ( cos\left ( \pi+ \frac{\pi kx}{2} \right )+ \frac{9x^2-10x+14}{k+12}\right )dx$
$\left [\frac{2}{\pi k}sin\left ( \pi+\frac{\pi kx}{2} \right )+\frac{3x^3-5x^2+14x}{k+12}  \right ]_{-2}^{0}$
$=\left [\frac{2}{\pi k}sin\left ( \pi+\frac{\pi k(0)}{2} \right )+\frac{3(0)^3-5(0)^2+140(0)}{k+12}  \right ]_{-2}^{0}$
$-\left [\frac{2}{\pi k}sin\left ( \pi+\frac{\pi k(-2)}{2} \right )+\frac{3(-2)^3-5(-2)^2+140(-2)}{k+12}  \right ]_{-2}^{0}$
$=\left [ \frac{2}{\pi k}sin \ \pi \ \right ]-\left [ \frac{2}{\pi k}sin \ (\pi-\pi k)+ \frac{-24-20-28}{k+12} \ \right ]$

$=[0]-\left [ \frac{2}{\pi k}sin \ \pi (1-k)+ \frac{-72}{k+12} \ \right ]$

$=-\left [0- \frac{72}{k+12} \ \right ]$

$=\frac{72}{k+12}$

sehingga

$\frac{72}{k+12}=(k-9)(k-11)$
$72=(k-9)(k-11)(k+12)$

$k^3-8k^2-141k+1116=0$
$(k-12)(k^2+4k+93)=0$
$k=12$

maka $k^2-14=144-14=130$

Pilihan jawabannya adalah (C)

Soal 3: SBMPTN 2017 Kode 106

Jika $\int_{-4}^{4}f(x)(sin \ x \ +1)dx=8$, dengan $f(x)$ fungsi genap dan $\int_{-2}^{4}f(x)dx=4$, maka $\int_{-4}^{0}f(x)dx=...$
(A) $0$
(B) $1$
(C) $2$
(D) $3$
(E) $4$

PEMBAHASAN:

Konsep: Pada fungsi genap berlaku:

• $f(-x)=f(x)$
• Ciri fungsi genap pada integral adalah: $\int_{-a}^{a}f(x)dx=2\int_{0}^{a}f(x)dx$
• Bentuk grafik fungsi simetris dengan pusat sumbu $y$

Konsep: Pada fungsi ganjil berlaku:

• $f(-x)=-f(x)$
• Ciri fungsi ganjil pada integral adalah: $\int_{-a}^{a}f(x)dx=0$
• Bentuk grafik fungsi simetris dengan pusat sumbu $(0,0)$

$\int_{-4}^{4}f(x)(sin \ x \ +1)dx=8$
$\int_{-4}^{4}f(x)sin \ x \ dx + \int_{-4}^{4}f(x)dx=8$

karena fungsi $f(x)$ fungsi genap dan $sin \ x \ $ fungsi ganjil maka $f(x)sin \ x \ $ merupajan fungsi ganjil sehingga berlaku $\int_{-4}^{4}f(x)sin \ x \ dx =0$ dan $\int_{-4}^{4}f(x)dx=2\int_{0}^{4}f(x)dx$

$\int_{-4}^{4}f(x)sin \ x \ dx + \int_{-4}^{4}f(x)dx=8$
$0 + \int_{-4}^{4}f(x)dx=8$
$2\int_{0}^{4}f(x)dx=8$
$\int_{0}^{4}f(x)dx=4$

$\int_{-2}^{4}f(x)dx=4$
$\int_{-2}^{0}f(x)dx + \int_{0}^{4}f(x)dx=4$
$\int_{-2}^{0}f(x)dx + 4=4$
$\int_{-2}^{0}f(x)dx=0$

Pilihan yang sesuai adalah (A)

Soal 4: EBTANAS Matematika SMA IPA 2003

Nilai dari $\int_{0}^{\frac{\pi}{4}}sin \ 5x \ . \ sin \ x \ dx=...$ 
(A) $-\frac{1}{2}$
(B) $-\frac{1}{6}$
(C) $\frac{1}{12}$
(D) $\frac{1}{8}$
(E) $\frac{2}{12}$

PENYELESAIAN:
Kita gunakan indentitas trigonometri:
$sin \ A \ . \ cos \ B \ = -\frac{1}{2} cos(A+B)+\frac{1}{2}cos (A-B)$

Sifat trigonometri di atas kita gunakan untuk menyederahanakan bentuk soal menjadi seperti berikut ini:

$\int_{0}^{\frac{\pi}{4}}sin \ 5x \ . \ sin \ x \ dx$
$=\int_{0}^{\frac{\pi}{4}}\left (-\frac{1}{2}cos \ 6x \ + \frac{1}{2} \ cos \ 4x \ \right )dx$
$=\left [ -\frac{1}{2} \ . \ \frac{1}{6} \ sin \ 6x+ \frac{1}{2} \ . \frac{1}{4} \ sin \ 4x \right ]_{0}^{\frac{\pi}{4}}$
$=\left [-\frac{1}{12} \ sin \ 270^0+\frac{1}{8} \ sin \ 180^0 \right ]-\left [-\frac{1}{12} \ sin \ 0^0+\frac{1}{8} \ sin \ 0^0 \right ]$

$=\left [-\frac{1}{12} \ (-1)+\frac{1}{8} \ (0) \right ]-\left [-\frac{1}{12} \ (0)+\frac{1}{8} \ (0) \right ]$
$=\frac{1}{12}$

Pilihan jawabannya adalah (C)

Soal 5: EBTANAS Matematika SMA IPA 1997

Nilai $\int xsin(x^2+1)dx=...$ 
(A) $-cos(x^2+1)+C$
(B) $-cos(x^2+1)+C$
(C) $-\frac{1}{2}cos (x^2+1)+C$
(D) $\frac{1}{2}cos(x^2+1)+C$
(E) $-2cos(x^2+1+C)$

PEMBAHASAN:
Misalkan:
$u=x^2+1$ maka $\frac{du}{dx}=2x$ atau $du=2xdx$, maka

$\int xsin(x^2+1)dx$
$=\int sin(x^2+1)xdx$
$=\int xsin(u) \frac{1}{2}du$  
$=\frac{1}{2}\int xsin(u) du$
$=\frac{1}{2}(-cos \ u)+C$       
$=-\frac{1}{2}cos (x^2+1)+C$  

Pilihan jawaban yang benar adalah (C)  

Soal 6: EBTANAS Matematika SMA IPA 1990

Hasil dari $\int cos \ x \ cos \ 4x \ dx=...$ 
(A) $-\frac{1}{5}sin \ 5x \ - \frac{1}{3}sin \ 3x \ +C$
(B) $\frac{1}{10}sin \ 5x \ + \frac{1}{6}sin \ 3x \ +C$
(C) $\frac{2}{5}sin \ 5x \ + \frac{2}{5}sin \ 3x \ +C$
(D) $\frac{1}{2}sin \ 5x \ + \frac{1}{2}sin \ 3x \ +C$
(E) $-\frac{1}{2}sin \ 5x \ - \frac{1}{2}sin \ 3x \ +C$

PEMBAHASAN:
Kita gunakan indentitas trigonometri:
$cos \ A \ . \ cos \ B \ = \frac{1}{2} cos(A+B)+\frac{1}{2}cos (A-B)$, maka

$\int cos \ x \ cos \ 4x \ dx$
$=\int (\frac{1}{2}cos \ 5x \ + \frac{1}{2}cos \ 3x \ )dx$
$=\frac{1}{2}.\frac{1}{5}sin \ 5x \ + \frac{1}{2}.\frac{1}{3}sin \ 3x \ +C$
$=\frac{1}{10}sin \ 5x \ + \frac{1}{6}sin \ 3x \ +C$

Pilihan jawabannya adalah (B)

Soal 7: EBTANAS Matematika SMA IPA 1988

Nilai $\int sin^5 \ x \ cos \ x \ dx=...$ 
(A) $\frac{1}{6}sin^6 \ x \ +C$
(B) $\frac{1}{6}cos^6 \ x \ +C$
(C) $-\frac{1}{6}sin^6 \ x \ +C$
(D) $-\frac{1}{6}cos^6 \ x \ +C$
(E) $\frac{1}{4}sin^4 \ x \ +C$

PEMBAHASAN:
Misalkan:
$u=sin \ x \ $ maka $\frac{du}{dx}=cos \ x \ $ atau $du=cos \ x \ dx$

$\int sin^5 \ x \ cos \ x \ dx$
$=\int (sin \ x \ )^5 cos \ x \ dx$  
$=\int (u)^5 du$

$=\frac{1}{5+1}(u)^{5+1} +C$
$=\frac{1}{6}(u)^{6} +C$
$=\frac{1}{6}(sin \ x \ )^{6} +C$
$=\frac{1}{6}sin^6 \ x \  +C$

Pilihan jawabannya adalah (A)

Soal 8: EBTANAS Matematika SMA IPA 1990

Nilai $\int_{0}^{\frac{\pi}{6}}sin \ (x+\frac{\pi}{3}) \ . \ cos \ (x+\frac{\pi}{3}) \ dx=...$ 
(A) $-\frac{1}{4}$
(B) $-\frac{1}{8}$
(C) $\frac{1}{8}$
(D) $\frac{1}{4}$
(E) $\frac{3}{8}$

PEMBAHASAN:
Kita gunakan indentitas trigonometri:
$sin \ A \ . \ sin \ B \ = \frac{1}{2} sin(A+B)+\frac{1}{2}sin (A-B)$

$\int_{0}^{\frac{\pi}{6}}sin \ (x+\frac{\pi}{3}) \ . \ cos \ (x+\frac{\pi}{3}) \ dx$
$=\int_{0}^{\frac{\pi}{6}}[\frac{1}{2}sin \ (2x+\frac{\pi}{3}) \ + \frac{1}{2} \ sin \ (0) \ ] dx$
$=\int_{0}^{\frac{\pi}{6}}[\frac{1}{2}sin \ (2x+\frac{2\pi}{3}) \ + \frac{1}{2} \ sin \ (0) \ ] dx$
$=\int_{0}^{\frac{\pi}{6}}\frac{1}{2}sin \ (2x+\frac{2\pi}{3}) \ dx$
$=\frac{1}{2}\left [ -\frac{1}{2}cos(2x+\frac{2\pi}{3}) \right ]_{0}^{\frac{\pi}{6}}$
$=-\frac{1}{4}\left [ cos(2.\frac{\pi}{6}+\frac{2\pi}{3})-cos(0+\frac{2\pi}{3}) \right ]$
$=-\frac{1}{4}\left [ cos \ \pi \ - cos \ \frac{2 \pi}{3} \right ]$
$=-\frac{1}{4}\left [ -1 - (-\frac{1}{2})\right]$
$=-\frac{1}{4}\left [-\frac{1}{2} \right]=\frac{1}{8}$

Pilihan jawabannya adalah (C)

Soal 9: EBTANAS Matematika SMA IPA 1990

Nilai $\int_{0}^{\frac{\pi}{6}}(sin \ 3x \ + cos \ 3x \ )dx=...$ 
(A) $\frac{2}{3}$
(B) $\frac{1}{3}$
(C) $0$
(D) $-\frac{1}{2}$
(E) $-\frac{2}{3}$

PEMBAHASAN:
$\int_{0}^{\frac{\pi}{6}}(sin \ 3x \ + cos \ 3x \ )dx$ 
$=\left [ -\frac{1}{3} \ . \ cos \ 3x \ +\frac{1}{3} . sin \ 3x \ \right ]_{0}^{\frac{\pi}{6}}$ 
$=\left [ -\frac{1}{3} \ . \ cos \ 3\left ( \frac{\pi}{6} \right ) \ +\frac{1}{3} . sin \ 3\left ( \frac{\pi}{6} \right ) \ \right ] -\left [ -\frac{1}{3} \ . \ cos \ 3(0) \ +\frac{1}{3} . sin \ 3 (0) \ \right ]$
$=\left [ -\frac{1}{3} \ . \ cos \ \left ( \frac{\pi}{2} \right ) \ +\frac{1}{3} . sin \ \left ( \frac{\pi}{2} \right ) \ \right ] -\left [ -\frac{1}{3} \ . \ cos \ (0) \ +\frac{1}{3} . sin \ (0) \ \right ]$

$=\frac{1}{3}+\frac{1}{3}$

$=\frac{2}{3}$

Pilihan jawabannya adalah (A)

Soal 10: EBTANAS Matematika SMA IPA 2004

Hasil dari $\int cos \ x \ cos \ 4x \ dx=...$ 
(A) $-8(2x+6) sin \ 2x \ - 4sin \ 2x \ +C$
(B) $-8(2x+6) sin \ 2x \ + 4sin \ 2x \ +C$
(C) $-8(x+3) sin \ 2x \ - 4sin \ 2x \ +C$
(D) $-8(x+3) sin \ 2x \ + 4sin \ 2x \ +C$
(E) $8(x+3) sin \ 2x \ - 4sin \ 2x \ +C$

PEMBAHASAN:
Kita gunakan indentitas trigonometri:
$cos \ A \ . \ cos \ B \ = \frac{1}{2} cos(A+B)+\frac{1}{2}cos (A-B)$

Sifat trigonometri di atas kita gunakan untuk menyederahanakan bentuk soal menjadi seperti berikut ini:

$\int cos \ x \ . \ cos \ 4x \ dx$
$=\int \frac{1}{2}(cos \ (5x) \ + \ cos \ (3x) \ )dx$
$=\frac{1}{2}\left [\frac{1}{5}sin \ (5x) \ + \frac{1}{3}sin \ 3x \  \right ] + C$
$=\frac{1}{10} \sin \ 5x \ + \frac{1}{6} \sin \ 3x \ + C$

Pilihan jawabannya adalah (C)

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