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
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)
(A) $0$
(B) $1$
(C) $2$
(D) $3$
(E) $4$
PEMBAHASAN:
$\int_{-4}^{4}f(x)sin \ x \ dx + \int_{-4}^{4}f(x)dx=8$
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)
(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:
$\int_{0}^{\frac{\pi}{4}}sin \ 5x \ . \ sin \ x \ dx$
$=\int_{0}^{\frac{\pi}{4}}\left (-\frac{1}{2}cos \ 6x \ + \frac{1}{2} \ cos \ 4x \ \right )dx$
$=\left [ -\frac{1}{2} \ . \ \frac{1}{6} \ sin \ 6x+ \frac{1}{2} \ . \frac{1}{4} \ sin \ 4x \right ]_{0}^{\frac{\pi}{4}}$
$=\left [-\frac{1}{12} \ sin \ 270^0+\frac{1}{8} \ sin \ 180^0 \right ]-\left [-\frac{1}{12} \ sin \ 0^0+\frac{1}{8} \ sin \ 0^0 \right ]$
$72=(k-9)(k-11)(k+12)$
$k^3-8k^2-141k+1116=0$
$(k-12)(k^2+4k+93)=0$
$k=12$
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 $(0,0)$
$\int_{-4}^{4}f(x)sin \ x \ dx + \int_{-4}^{4}f(x)dx=8$
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:
$\int_{0}^{\frac{\pi}{4}}sin \ 5x \ . \ sin \ x \ dx$
$=\int_{0}^{\frac{\pi}{4}}\left (-\frac{1}{2}cos \ 6x \ + \frac{1}{2} \ cos \ 4x \ \right )dx$
$=\left [ -\frac{1}{2} \ . \ \frac{1}{6} \ sin \ 6x+ \frac{1}{2} \ . \frac{1}{4} \ sin \ 4x \right ]_{0}^{\frac{\pi}{4}}$
$=\left [-\frac{1}{12} \ sin \ 270^0+\frac{1}{8} \ sin \ 180^0 \right ]-\left [-\frac{1}{12} \ sin \ 0^0+\frac{1}{8} \ sin \ 0^0 \right ]$
$=\left [-\frac{1}{12} \ (-1)+\frac{1}{8} \ (0) \right ]-\left [-\frac{1}{12} \ (0)+\frac{1}{8} \ (0) \right ]$
$=\frac{1}{12}$
$=\frac{1}{12}$
Pilihan jawabannya adalah (C)
(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
$\int xsin(x^2+1)dx$
$=\int sin(x^2+1)xdx$
$=\int xsin(u) \frac{1}{2}du$
$=\frac{1}{2}\int xsin(u) du$
$=\frac{1}{2}(-cos \ u)+C$
$=-\frac{1}{2}cos (x^2+1)+C$
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)
(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) $-\frac{1}{2}sin \ 5x \ - \frac{1}{2}sin \ 3x \ +C$
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)
(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)
(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)
(A) $\frac{2}{3}$
(B) $\frac{1}{3}$
(C) $0$
(D) $-\frac{1}{2}$
(E) $-\frac{2}{3}$
PEMBAHASAN:
$\int_{0}^{\frac{\pi}{6}}(sin \ 3x \ + cos \ 3x \ )dx$
$=\left [ -\frac{1}{3} \ . \ cos \ 3x \ +\frac{1}{3} . sin \ 3x \ \right ]_{0}^{\frac{\pi}{6}}$
$=\left [ -\frac{1}{3} \ . \ cos \ 3\left ( \frac{\pi}{6} \right ) \ +\frac{1}{3} . sin \ 3\left ( \frac{\pi}{6} \right ) \ \right ] -\left [ -\frac{1}{3} \ . \ cos \ 3(0) \ +\frac{1}{3} . sin \ 3 (0) \ \right ]$
$=\left [ -\frac{1}{3} \ . \ cos \ \left ( \frac{\pi}{2} \right ) \ +\frac{1}{3} . sin \ \left ( \frac{\pi}{2} \right ) \ \right ] -\left [ -\frac{1}{3} \ . \ cos \ (0) \ +\frac{1}{3} . sin \ (0) \ \right ]$
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)
(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)
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)
Post a Comment for "Soal dan Pembahasan Integral Fungsi Trigonometri"
Sobat Ayo Sekolah Matematika! Berikan Komentar di kolom komentar dengan bahasa yang sopan dan sesuai isi konten...Terimasih untuk kunjunganmu di blog ini, semoga bermanfaat!