Setelah mempelajari "Integral Tak Tentu Fungsi Aljabar", kita akan lanjutkan lagi materi integral yang berkaitan dengan Integral Tak Tentu Fungsi Trigonometri. Sifat-sifat integral tak tentu juga berlaku pada integral fungsi trigonometri. Untuk memudahkan, silahkan baca materi "Turunan Fungsi Trigonometri" terlebih dahulu karena integral adalah kebalikan dari turunan.
Contoh soal integral fungsi trigonometri :
1). Tentukan hasil integral berikut ini :
a). $ \int 2\sin x dx $
b). $ \int \sin x - \cos x dx $
c). $ \int \frac{\sin x + \csc x}{\tan x } dx $
d). $ \int \frac{\tan x + \cot x}{\sin 2 x } dx $
Penyelesaian :
*). Rumus dasar trigonometri :
$ \sec x = \frac{1}{\cos x }, \, \csc x = \frac{1}{\sin x}, \, \tan x = \frac{\sin x}{\cos x}, \, \cot x = \frac{\cos x}{\sin x} $.
*). Soal yang ada kita arahkan menjadi bentuk rumus dasar integral trigonometri di atas.
a). $ \int 2\sin x dx = 2 \int \sin x dx = 2(-\cos x ) + c = -2\cos x + c $
b). $ \int \sin x - \cos x dx = -\cos x - \sin x + c $
c). Kita sederhanakan soalnya :
$ \begin{align} \int \frac{\sin x + \csc x}{\tan x } dx & = \int \frac{\sin x + \csc x}{ \frac{\sin x}{\cos x} } dx \\ & = \int (\sin x + \csc x) \times \frac{\cos x}{\sin x} dx \\ & = \int ( \sin x . \times \frac{\cos x}{\sin x} + \csc x \times \frac{\cos x}{\sin x} ) dx \\ & = \int ( \cos x + \csc x \cot x ) dx \\ & = \sin x - \csc x + c \end{align} $
Sehingga, hasil dari $ \int \frac{\sin x + \csc x}{\tan x } dx = \sin x - \csc x + c $.
d). Kita sederhanakan soalnya : $ \sin 2x = 2\sin x \cos x $
$ \begin{align} \int \frac{\tan x + \cot x}{\sin 2 x } dx & = \int \frac{\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} }{2 \sin x \cos x } dx \\ & = \int \frac{\frac{\sin x}{\cos x} }{2 \sin x \cos x } + \frac{ \frac{\cos x}{\sin x} }{2 \sin x \cos x } dx \\ & = \int \frac{\sin x}{\cos x . 2 \sin x \cos x} + \frac{\cos x}{\sin x . 2 \sin x \cos x} dx \\ & = \int \frac{1}{\cos x . 2 \cos x} + \frac{1}{\sin x . 2 \sin x } dx \\ & = \int \frac{1}{2} ( \frac{1}{\cos ^2 x } + \frac{1}{\sin ^2 x } ) dx \\ & = \frac{1}{2} \int \sec ^2 x + \csc ^2 x dx \\ & = \frac{1}{2} (\tan x - \cot x ) + c \end{align} $
Sehingga, hasil dari $ \int \frac{\tan x + \cot x}{\sin 2 x } dx = \frac{1}{2} (\tan x - \cot x ) + c $.
Contoh soal integral fungsi trigonometri :
2). Tentukan hasil integral dari :
a). $ \int \sin (2x + 3) dx $
b). $ \int 6\sin (1-3x) dx $
c). $ \int \sec ^2 (4x) dx $
d). $ \int \csc ^2 (-2x + 1) + \sin (2x) dx $
e). $ \int \sec (x - 1) \tan (x-1) dx $
f). $ \int \csc (5x - 3) \cot (5x - 3) dx $
Penyelesaian :
a). $ \int \sin (2x + 3) dx = -\frac{1}{2} \cos (2x + 3) + c $
b). $ \int 6\cos (1-3x) dx = \frac{6}{-3} \sin (1-3x) + c = -2\sin (1-3x) + c $
c). $ \int \sec ^2 (4x) dx = \frac{1}{4} \tan (4x) + c $
d). $ \int \csc ^2 (-2x + 1) + \sin (2x) dx = -\frac{1}{-2} \cot (-2x + 1) - \frac{1}{2} \cos (2x) + c $
$ = \frac{1}{2} \cot (-2x + 1) - \frac{1}{2} \cos (2x) + c $
e). $ \int \sec (x - 1) \tan (x-1) dx = \sec (x-1) + c $
f). $ \int \csc (5x - 3) \cot (5x - 3) dx = -\frac{1}{5} \csc (5x-3) + c $
Contoh Soal :
3). Tentukan hasil integral fungsi trigonometri berikut ini :
a). $ \int \sin 5x \cos 2x dx $
b). $ \int 4\cos 7x \sin 4x dx $
c). $ \int 3cos (3x - 1) \cos (2x + 2) dx $
d). $ \int \cos ^2 3x dx $
e). $ \int \sin ^4 5x dx $
Penyelesaian :
a). Gunakan rumus : $ \, \sin A \cos B = \frac{1}{2} [\sin (A+B) + \sin (A-B)] $
$ \begin{align} \int \sin 5x \cos 2x dx & = \int \frac{1}{2} [ \sin (5x+2x) + \sin (5x-2x)] dx \\ & = \int \frac{1}{2} [ \sin (7x) + \sin (3x)] dx \\ & = \frac{1}{2} \int [ \sin (7x) + \sin (3x)] dx \\ & = \frac{1}{2} [ -\frac{1}{7}\cos (7x) - \frac{1}{3} \cos (3x)] + c \\ & = -\frac{1}{14}\cos (7x) - \frac{1}{6} \cos (3x) + c \end{align} $
b). Gunakan rumus : $ \, \cos A \sin B = \frac{1}{2} [\sin (A+B) - \sin (A-B)] $
$ \begin{align} \int 4\cos 7x \sin 4x dx & = \int 4 . \frac{1}{2} [ \sin (7x + 4x ) - \sin (7x - 4x)] dx \\ & = \int 2 [ \sin (11x ) - \sin (3x)] dx \\ & = 2 [ - \frac{1}{11} \cos (11x ) - (-\frac{1}{3}\cos (3x)) + c \\ & = 2 [ - \frac{1}{11} \cos (11x ) + \frac{1}{3}\cos (3x) + c \\ & = - \frac{2}{11} \cos (11x ) + \frac{2}{3}\cos (3x) + c \end{align} $
c). Gunakan rumus : $ \, \cos A \cos B = \frac{1}{2} [\cos (A+B) + \cos (A-B)] $
$ \begin{align} & \int 3cos (3x - 1) \cos (2x + 2) dx \\ & = \int 3 . \frac{1}{2} [ \cos ((3x - 1) + (2x + 2)) + \cos ((3x - 1) - (2x + 2))] dx \\ & = \int \frac{3}{2} [ \cos (5x + 1) + \cos (x - 3)] dx \\ & = \frac{3}{2} [ \frac{1}{5} \sin (5x + 1) + \sin (x - 3)] + c \\ & = \frac{3}{10} \sin (5x + 1) + \frac{3}{2} \sin (x - 3) + c \end{align} $
d). Gunakan rumus : $ \, \cos ^2 p f(x) = \frac{1}{2} [ 1 + \cos 2p f(x) ] $
$ \int \cos ^2 3x dx $
$ \begin{align} \int \cos ^2 3x dx & = \int \frac{1}{2} [ 1 + \cos 2 . 3x ] dx \\ & = \int \frac{1}{2} [ 1 + \cos 6x ] dx \\ & = \frac{1}{2} [ x + \frac{1}{6} \sin 6x ] + c \\ & = \frac{1}{2}x + \frac{1}{12} \sin 6x + c \end{align} $
e). Gunakan rumus : $ \, \sin ^2 p f(x) = \frac{1}{2} [ 1 - \cos 2p f(x) ] $
$ \begin{align} \int \sin ^4 5x dx & = \int \sin ^2 5x \sin ^2 5x dx \\ & = \int (\sin ^2 5x)^2 dx \\ & = \int (\frac{1}{2} [ 1 - \cos 2 . 5x ])^2 dx \\ & = \int \frac{1}{4} [ 1 - \cos 10x ]^2 dx \\ & = \frac{1}{4} \int [ 1 - 2 \cos 10 x + \cos ^2 10 x ] dx \\ & = \frac{1}{4} \int [ 1 - 2 \cos 10 x + \frac{1}{2} [ 1 + \cos 2 . 10x ] ] dx \\ & = \frac{1}{4} \int [ 1 - 2 \cos 10 x + \frac{1}{2} [ 1 + \cos 20x ] ] dx \\ & = \frac{1}{4} \int [ 1 - 2 \cos 10 x + \frac{1}{2} + \frac{1}{2} \cos 20x ] dx \\ & = \frac{1}{4} [ x - \frac{2}{10} \sin 10 x + \frac{1}{2}x + \frac{1}{2} . \frac{1}{20} \sin 20x ] + c \\ & = \frac{1}{4} [ \frac{3}{2}x - \frac{2}{10} \sin 10 x + \frac{1}{40} \sin 20x ] + c \\ & = \frac{3}{8}x - \frac{2}{40} \sin 10 x + \frac{1}{160} \sin 20x ] + c \\ & = \frac{3}{8}x - \frac{1}{20} \sin 10 x + \frac{1}{160} \sin 20x ] + c \end{align} $
Integral Tak Tentu Fungsi Trigonometri (Rumus Dasar)
Berdasarkan pengertian integral, $ \int f^\prime (x) dx = f(x) + c $, dimana $ f^\prime (x) \, $ adalah turuan dari fungsi $ f(x) $ :
Rumus integral Trigonometri :
1). $ f(x) = \sin x \rightarrow f^\prime (x) = \cos x $
artinya $ \int \cos x dx = \sin x + c $
2). $ f(x) = \cos x \rightarrow f^\prime (x) = -\sin x $
artinya $ \int - \sin x dx = \cos x + c \, $ atau $ \, \int \sin x dx = -\cos x + c $
3). $ f(x) = \tan x \rightarrow f^\prime (x) = \sec ^2 x $
artinya $ \int \sec ^2 x dx = \tan x + c $
4). $ f(x) = \cot x \rightarrow f^\prime (x) = -\csc ^2 x $
artinya $ \int - \csc ^2 x dx = \cot x + c \, $ atau $ \, \int \csc ^2 x dx = -\cot x + c $
5). $ f(x) = \sec x \rightarrow f^\prime (x) = \sec x \tan x $
artinya $ \int \sec x \tan x dx = \sec x + c $
6). $ f(x) = \csc x \rightarrow f^\prime (x) = -\csc x \cot x $
artinya $ \int -\csc x \cot x dx = \csc x + c \, $
atau $ \, \int \csc x \cot x dx = -\csc x + c $
Rumus integral Trigonometri :
1). $ f(x) = \sin x \rightarrow f^\prime (x) = \cos x $
artinya $ \int \cos x dx = \sin x + c $
2). $ f(x) = \cos x \rightarrow f^\prime (x) = -\sin x $
artinya $ \int - \sin x dx = \cos x + c \, $ atau $ \, \int \sin x dx = -\cos x + c $
3). $ f(x) = \tan x \rightarrow f^\prime (x) = \sec ^2 x $
artinya $ \int \sec ^2 x dx = \tan x + c $
4). $ f(x) = \cot x \rightarrow f^\prime (x) = -\csc ^2 x $
artinya $ \int - \csc ^2 x dx = \cot x + c \, $ atau $ \, \int \csc ^2 x dx = -\cot x + c $
5). $ f(x) = \sec x \rightarrow f^\prime (x) = \sec x \tan x $
artinya $ \int \sec x \tan x dx = \sec x + c $
6). $ f(x) = \csc x \rightarrow f^\prime (x) = -\csc x \cot x $
artinya $ \int -\csc x \cot x dx = \csc x + c \, $
atau $ \, \int \csc x \cot x dx = -\csc x + c $
1). Tentukan hasil integral berikut ini :
a). $ \int 2\sin x dx $
b). $ \int \sin x - \cos x dx $
c). $ \int \frac{\sin x + \csc x}{\tan x } dx $
d). $ \int \frac{\tan x + \cot x}{\sin 2 x } dx $
Penyelesaian :
*). Rumus dasar trigonometri :
$ \sec x = \frac{1}{\cos x }, \, \csc x = \frac{1}{\sin x}, \, \tan x = \frac{\sin x}{\cos x}, \, \cot x = \frac{\cos x}{\sin x} $.
*). Soal yang ada kita arahkan menjadi bentuk rumus dasar integral trigonometri di atas.
a). $ \int 2\sin x dx = 2 \int \sin x dx = 2(-\cos x ) + c = -2\cos x + c $
b). $ \int \sin x - \cos x dx = -\cos x - \sin x + c $
c). Kita sederhanakan soalnya :
$ \begin{align} \int \frac{\sin x + \csc x}{\tan x } dx & = \int \frac{\sin x + \csc x}{ \frac{\sin x}{\cos x} } dx \\ & = \int (\sin x + \csc x) \times \frac{\cos x}{\sin x} dx \\ & = \int ( \sin x . \times \frac{\cos x}{\sin x} + \csc x \times \frac{\cos x}{\sin x} ) dx \\ & = \int ( \cos x + \csc x \cot x ) dx \\ & = \sin x - \csc x + c \end{align} $
Sehingga, hasil dari $ \int \frac{\sin x + \csc x}{\tan x } dx = \sin x - \csc x + c $.
d). Kita sederhanakan soalnya : $ \sin 2x = 2\sin x \cos x $
$ \begin{align} \int \frac{\tan x + \cot x}{\sin 2 x } dx & = \int \frac{\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} }{2 \sin x \cos x } dx \\ & = \int \frac{\frac{\sin x}{\cos x} }{2 \sin x \cos x } + \frac{ \frac{\cos x}{\sin x} }{2 \sin x \cos x } dx \\ & = \int \frac{\sin x}{\cos x . 2 \sin x \cos x} + \frac{\cos x}{\sin x . 2 \sin x \cos x} dx \\ & = \int \frac{1}{\cos x . 2 \cos x} + \frac{1}{\sin x . 2 \sin x } dx \\ & = \int \frac{1}{2} ( \frac{1}{\cos ^2 x } + \frac{1}{\sin ^2 x } ) dx \\ & = \frac{1}{2} \int \sec ^2 x + \csc ^2 x dx \\ & = \frac{1}{2} (\tan x - \cot x ) + c \end{align} $
Sehingga, hasil dari $ \int \frac{\tan x + \cot x}{\sin 2 x } dx = \frac{1}{2} (\tan x - \cot x ) + c $.
Integral Tak Tentu Fungsi Trigonometri (Rumus Dasar II)
Untuk berikut ini, kita akan pelajari rumus integral trigonometri dengan sudut yang lebih kompleks.
Rumus integral Trigonometri :
1). $ f(x) = \sin (ax+b) \rightarrow f^\prime (x) = a\cos (ax + b) $
artinya $ \int a\cos (ax + b) dx = \sin (ax+b) + c $
atau $ \int \cos (ax + b) dx = \frac{1}{a} \sin (ax+b) + c $
2). $ f(x) = \cos (ax + b) \rightarrow f^\prime (x) = -a\sin (ax + b) $
artinya $ \int - a\sin (ax + b) dx = \cos (ax + b) + c \, $
atau $ \, \int \sin (ax + b) dx = -\frac{1}{a}\cos (ax + b) + c $
3). $ f(x) = \tan (ax + b) \rightarrow f^\prime (x) = a \sec ^2 (ax + b) $
artinya $ \int a \sec ^2 (ax + b) dx = \tan (ax + b) + c $
atau $ \int \sec ^2 (ax + b) dx = \frac{1}{a} \tan (ax + b) + c $
4). $ f(x) = \cot (ax + b) \rightarrow f^\prime (x) = -a\csc ^2 (ax + b) $
artinya $ \int - a\csc ^2 (ax + b) dx = \cot (ax + b) + c \, $
atau $ \, \int \csc ^2 (ax + b) dx = -\frac{1}{a} \cot (ax + b) + c $
5). $ f(x) = \sec (ax + b) \rightarrow f^\prime (x) = a\sec (ax + b) \tan (ax + b) $
artinya $ \int a\sec (ax + b) \tan (ax + b) dx = \sec (ax + b) + c $
atau $ \int \sec (ax + b) \tan (ax + b) dx = \frac{1}{a} \sec (ax + b) + c $
6). $ f(x) = \csc (ax + b) \rightarrow f^\prime (x) = -a\csc (ax + b) \cot (ax + b) $
artinya $ \int -a\csc (ax + b) \cot (ax + b) dx = \csc (ax + b) + c $
atau $ \int \csc (ax + b) \cot (ax + b) dx = - \frac{1}{a} \csc (ax + b) + c $
Rumus integral Trigonometri :
1). $ f(x) = \sin (ax+b) \rightarrow f^\prime (x) = a\cos (ax + b) $
artinya $ \int a\cos (ax + b) dx = \sin (ax+b) + c $
atau $ \int \cos (ax + b) dx = \frac{1}{a} \sin (ax+b) + c $
2). $ f(x) = \cos (ax + b) \rightarrow f^\prime (x) = -a\sin (ax + b) $
artinya $ \int - a\sin (ax + b) dx = \cos (ax + b) + c \, $
atau $ \, \int \sin (ax + b) dx = -\frac{1}{a}\cos (ax + b) + c $
3). $ f(x) = \tan (ax + b) \rightarrow f^\prime (x) = a \sec ^2 (ax + b) $
artinya $ \int a \sec ^2 (ax + b) dx = \tan (ax + b) + c $
atau $ \int \sec ^2 (ax + b) dx = \frac{1}{a} \tan (ax + b) + c $
4). $ f(x) = \cot (ax + b) \rightarrow f^\prime (x) = -a\csc ^2 (ax + b) $
artinya $ \int - a\csc ^2 (ax + b) dx = \cot (ax + b) + c \, $
atau $ \, \int \csc ^2 (ax + b) dx = -\frac{1}{a} \cot (ax + b) + c $
5). $ f(x) = \sec (ax + b) \rightarrow f^\prime (x) = a\sec (ax + b) \tan (ax + b) $
artinya $ \int a\sec (ax + b) \tan (ax + b) dx = \sec (ax + b) + c $
atau $ \int \sec (ax + b) \tan (ax + b) dx = \frac{1}{a} \sec (ax + b) + c $
6). $ f(x) = \csc (ax + b) \rightarrow f^\prime (x) = -a\csc (ax + b) \cot (ax + b) $
artinya $ \int -a\csc (ax + b) \cot (ax + b) dx = \csc (ax + b) + c $
atau $ \int \csc (ax + b) \cot (ax + b) dx = - \frac{1}{a} \csc (ax + b) + c $
2). Tentukan hasil integral dari :
a). $ \int \sin (2x + 3) dx $
b). $ \int 6\sin (1-3x) dx $
c). $ \int \sec ^2 (4x) dx $
d). $ \int \csc ^2 (-2x + 1) + \sin (2x) dx $
e). $ \int \sec (x - 1) \tan (x-1) dx $
f). $ \int \csc (5x - 3) \cot (5x - 3) dx $
Penyelesaian :
a). $ \int \sin (2x + 3) dx = -\frac{1}{2} \cos (2x + 3) + c $
b). $ \int 6\cos (1-3x) dx = \frac{6}{-3} \sin (1-3x) + c = -2\sin (1-3x) + c $
c). $ \int \sec ^2 (4x) dx = \frac{1}{4} \tan (4x) + c $
d). $ \int \csc ^2 (-2x + 1) + \sin (2x) dx = -\frac{1}{-2} \cot (-2x + 1) - \frac{1}{2} \cos (2x) + c $
$ = \frac{1}{2} \cot (-2x + 1) - \frac{1}{2} \cos (2x) + c $
e). $ \int \sec (x - 1) \tan (x-1) dx = \sec (x-1) + c $
f). $ \int \csc (5x - 3) \cot (5x - 3) dx = -\frac{1}{5} \csc (5x-3) + c $
Rumus Perkalian Fungsi yang sering digunakan
Berikut merupakan rumus perkalian fungsi trigonometri yang sering digunakan dalam integral trigonometri :
*). $ 2 \sin A \cos B = \sin (A+B) + \sin (A-B) \, $
atau $ \, \sin A \cos B = \frac{1}{2} [\sin (A+B) + \sin (A-B)] $
*). $ 2 \cos A \sin B = \sin (A+B) - \sin (A-B) \, $
atau $ \, \cos A \sin B = \frac{1}{2} [\sin (A+B) - \sin (A-B)] $
*). $ 2 \cos A \cos B = \cos (A+B) + \cos (A-B) \, $
atau $ \, \cos A \cos B = \frac{1}{2} [\cos (A+B) + \cos (A-B)] $
*). $ -2 \sin A \sin B = \cos (A+B) - \cos (A-B) \, $
atau $ \, \sin A \sin B = -\frac{1}{2} [\cos (A+B) - \cos (A-B)] $
Dari dua rumus terakshir di atas jika $ A + B = f(x) \, $ maka kita peroleh :
*). $ \cos ^2 p f(x) = \frac{1}{2} [ 1 + \cos 2p f(x) ] $
*). $ \sin ^2 p f(x) = \frac{1}{2} [ 1 - \cos 2p f(x) ] $
Untuk materi lebih lengkapnya, silahkan baca : "Rumus Perkalian, Penjumlahan, dan Pengurangan Trigonometri".
*). $ 2 \sin A \cos B = \sin (A+B) + \sin (A-B) \, $
atau $ \, \sin A \cos B = \frac{1}{2} [\sin (A+B) + \sin (A-B)] $
*). $ 2 \cos A \sin B = \sin (A+B) - \sin (A-B) \, $
atau $ \, \cos A \sin B = \frac{1}{2} [\sin (A+B) - \sin (A-B)] $
*). $ 2 \cos A \cos B = \cos (A+B) + \cos (A-B) \, $
atau $ \, \cos A \cos B = \frac{1}{2} [\cos (A+B) + \cos (A-B)] $
*). $ -2 \sin A \sin B = \cos (A+B) - \cos (A-B) \, $
atau $ \, \sin A \sin B = -\frac{1}{2} [\cos (A+B) - \cos (A-B)] $
Dari dua rumus terakshir di atas jika $ A + B = f(x) \, $ maka kita peroleh :
*). $ \cos ^2 p f(x) = \frac{1}{2} [ 1 + \cos 2p f(x) ] $
*). $ \sin ^2 p f(x) = \frac{1}{2} [ 1 - \cos 2p f(x) ] $
Untuk materi lebih lengkapnya, silahkan baca : "Rumus Perkalian, Penjumlahan, dan Pengurangan Trigonometri".
3). Tentukan hasil integral fungsi trigonometri berikut ini :
a). $ \int \sin 5x \cos 2x dx $
b). $ \int 4\cos 7x \sin 4x dx $
c). $ \int 3cos (3x - 1) \cos (2x + 2) dx $
d). $ \int \cos ^2 3x dx $
e). $ \int \sin ^4 5x dx $
Penyelesaian :
a). Gunakan rumus : $ \, \sin A \cos B = \frac{1}{2} [\sin (A+B) + \sin (A-B)] $
$ \begin{align} \int \sin 5x \cos 2x dx & = \int \frac{1}{2} [ \sin (5x+2x) + \sin (5x-2x)] dx \\ & = \int \frac{1}{2} [ \sin (7x) + \sin (3x)] dx \\ & = \frac{1}{2} \int [ \sin (7x) + \sin (3x)] dx \\ & = \frac{1}{2} [ -\frac{1}{7}\cos (7x) - \frac{1}{3} \cos (3x)] + c \\ & = -\frac{1}{14}\cos (7x) - \frac{1}{6} \cos (3x) + c \end{align} $
b). Gunakan rumus : $ \, \cos A \sin B = \frac{1}{2} [\sin (A+B) - \sin (A-B)] $
$ \begin{align} \int 4\cos 7x \sin 4x dx & = \int 4 . \frac{1}{2} [ \sin (7x + 4x ) - \sin (7x - 4x)] dx \\ & = \int 2 [ \sin (11x ) - \sin (3x)] dx \\ & = 2 [ - \frac{1}{11} \cos (11x ) - (-\frac{1}{3}\cos (3x)) + c \\ & = 2 [ - \frac{1}{11} \cos (11x ) + \frac{1}{3}\cos (3x) + c \\ & = - \frac{2}{11} \cos (11x ) + \frac{2}{3}\cos (3x) + c \end{align} $
c). Gunakan rumus : $ \, \cos A \cos B = \frac{1}{2} [\cos (A+B) + \cos (A-B)] $
$ \begin{align} & \int 3cos (3x - 1) \cos (2x + 2) dx \\ & = \int 3 . \frac{1}{2} [ \cos ((3x - 1) + (2x + 2)) + \cos ((3x - 1) - (2x + 2))] dx \\ & = \int \frac{3}{2} [ \cos (5x + 1) + \cos (x - 3)] dx \\ & = \frac{3}{2} [ \frac{1}{5} \sin (5x + 1) + \sin (x - 3)] + c \\ & = \frac{3}{10} \sin (5x + 1) + \frac{3}{2} \sin (x - 3) + c \end{align} $
d). Gunakan rumus : $ \, \cos ^2 p f(x) = \frac{1}{2} [ 1 + \cos 2p f(x) ] $
$ \int \cos ^2 3x dx $
$ \begin{align} \int \cos ^2 3x dx & = \int \frac{1}{2} [ 1 + \cos 2 . 3x ] dx \\ & = \int \frac{1}{2} [ 1 + \cos 6x ] dx \\ & = \frac{1}{2} [ x + \frac{1}{6} \sin 6x ] + c \\ & = \frac{1}{2}x + \frac{1}{12} \sin 6x + c \end{align} $
e). Gunakan rumus : $ \, \sin ^2 p f(x) = \frac{1}{2} [ 1 - \cos 2p f(x) ] $
$ \begin{align} \int \sin ^4 5x dx & = \int \sin ^2 5x \sin ^2 5x dx \\ & = \int (\sin ^2 5x)^2 dx \\ & = \int (\frac{1}{2} [ 1 - \cos 2 . 5x ])^2 dx \\ & = \int \frac{1}{4} [ 1 - \cos 10x ]^2 dx \\ & = \frac{1}{4} \int [ 1 - 2 \cos 10 x + \cos ^2 10 x ] dx \\ & = \frac{1}{4} \int [ 1 - 2 \cos 10 x + \frac{1}{2} [ 1 + \cos 2 . 10x ] ] dx \\ & = \frac{1}{4} \int [ 1 - 2 \cos 10 x + \frac{1}{2} [ 1 + \cos 20x ] ] dx \\ & = \frac{1}{4} \int [ 1 - 2 \cos 10 x + \frac{1}{2} + \frac{1}{2} \cos 20x ] dx \\ & = \frac{1}{4} [ x - \frac{2}{10} \sin 10 x + \frac{1}{2}x + \frac{1}{2} . \frac{1}{20} \sin 20x ] + c \\ & = \frac{1}{4} [ \frac{3}{2}x - \frac{2}{10} \sin 10 x + \frac{1}{40} \sin 20x ] + c \\ & = \frac{3}{8}x - \frac{2}{40} \sin 10 x + \frac{1}{160} \sin 20x ] + c \\ & = \frac{3}{8}x - \frac{1}{20} \sin 10 x + \frac{1}{160} \sin 20x ] + c \end{align} $